8.2: Polar Coordinates
The coordinate system we are most familiar with is called the Cartesian coordinate system, a rectangular plane divided into four quadrants by horizontal and vertical axes.
In earlier chapters, we often found the Cartesian coordinates of a point on a circle at a given angle from the positive horizontal axis. Sometimes that angle, along with the point’s distance from the origin, provides a more useful way of describing the point’s location than conventional Cartesian coordinates.
POLAR COORDINATES
Polar coordinates of a point consist of an ordered pair, (\(r\),\(\theta\)), where \(r\) is the distance from the point to the origin, and \(\theta\) is the angle measured in standard position.
Notice that if we were to “grid” the plane for polar coordinates, it would look like the graph to the right, with circles at incremental radii, and rays drawn at incremental angles.
Example \(\PageIndex{1}\)
Plot the polar point (\(3, \dfrac{5\pi}{6}\)).
Solution
This point will be a distance of 3 from the origin, at an angle of \(\dfrac{5\pi}{6}\). Plotting this
Example \(\PageIndex{2}\)
Plot the polar point (\(-2, \dfrac{\pi}{4}\)).
Solution
Typically we use positive \(r\) values, but occasionally we run into cases where \(r\) is negative. On a regular number line, we measure positive values to the right and negative values to the left. We will plot this point similarly. To start, we rotate to an angle of \(\dfrac{\pi}{4}\).
Moving this direction, into the first quadrant, would be positive r values. For negative r values, we move the opposite direction, into the third quadrant. Plotting this:
Note the resulting point is the same as the polar point (\(2, \dfrac{5\pi}{4}\)). In fact, any Cartesian point can be represented by an infinite number of different polar coordinates by adding or subtracting full rotations to these points. For example, same point could also be represented as (\(2, \dfrac{13\pi}{4}\)).
Exercise \(\PageIndex{1}\)
Plot the following points given in polar coordinates and label them.
a. \(A = (3, \dfrac{\pi}{6})\)
b. \(B = (-2, \dfrac{\pi}{3})\)
c. \(C = (4, \dfrac{3\pi}{4})\)
- Answer
-
Converting Points
To convert between polar coordinates and Cartesian coordinates, we recall the relationships we developed back in Chapter 5.
CONVERTING BETWEEN POLAR AND CARTESIAN COORDINATES
To convert between polar (\(r, \theta\)) and Cartesian (\(x, y\)) coordidates, we use the relationships
\[\cos (\theta )=\frac{x}{r}\quad x=r\cos (\theta )\]
\[\sin (\theta )=\frac{y}{r}\quad y=r\sin (\theta )\]
\[\tan (\theta )=\frac{y}{x}\quad x^{2}+y^{2}=r^{2}\]
From these relationship and our knowledge of the unit circle, if \(r = 1\) and \(\theta = \dfrac{\pi}{3}\), the polar coordinates would be \((r, \theta) = (1, \dfrac{\pi}{3})\), and the corresponding Cartesian coordinates \((x, y) = (\dfrac{1}{2}, \dfrac{\sqrt{3}}{2})\).
Remembering your unit circle values will come in very handy as you convert between Cartesian and polar coordinates.
Example \(\PageIndex{3}\)
Find the Cartesian coordinates of a point with polar coordinates \((r, \theta) = (5, \dfrac{2\pi}{3})\).
Solution
To find the \(x\) and \(y\) coordinates of the point,
\[x = r\text{cos} (\theta) = 5 \cos (\dfrac{2\pi}{3}) = 5(-\dfrac{1}{2}) = -\dfrac{5}{2}\nonumber\]
\[y = r\text{sin} (\theta) = 5 \sin (\dfrac{2\pi}{3}) = 5(-\dfrac{\sqrt{3}}{2}) = -\dfrac{5\sqrt{3}}{2}\nonumber\]
The Cartesian coordinates are (\(-\dfrac{5}{2}, \dfrac{5\sqrt{3}}{2}\)).
Example \(\PageIndex{4}\)
Find the polar coordinates of the point with Cartesian coordinates (−3,−4) .
Solution
We begin by finding the distance \(r\) using the Pythagorean relationship \(x^2 + y^2 = r^2\)
\[(-3)^2 + (-4)^2 = r^2\nonumber\]
\[9 + 16 = r^2\nonumber\]
\[r^2 = 25\nonumber\]
\[r = 5\nonumber\]
Now that we know the radius, we can find the angle using any of the three trig relationships. Keep in mind that any of the relationships will produce two solutions on the circle, and we need to consider the quadrant to determine which solution to accept. Using the cosine, for example:
\[\cos(\theta) = \dfrac{x}{r} = \dfrac{-3}{5}\nonumber\]
\[\theta = \cos^{-1}(\dfrac{-3}{5}) \approx 2.214\nonumber\]By symmetry, there is a second possibility at
\[\theta = 2\pi - 2.24 = 4.069\nonumber\]
Since the point (-3, -4) is located in the \(3^{\text{rd}}\) quadrant, we can determine that the second angle is the one we need. The polar coordinates of this point are \((r, \theta) = (5, 4.069)\).
Exercise \(\PageIndex{2}\)
Convert the following.
a. Convert polar coordinates \((r, \theta) = (2, \pi)\) to (\(x, y)\).
b. Convert Cartesian coordinates \((x, y) = (0, -4)\) to \((r, \theta)\).
- Answer
-
a. \((r, \theta) = (2, \pi)\) converts to \((x, y) = (2\cos(\pi), 2\sin(\pi)) = (-2, 0)\)
b. \((x, y) = (0, -4)\) converts to \((r, \theta) = (4, \dfrac{3\pi}{2})\ or\ (-4, \dfrac{\pi}{2})\)
Polar Equations
Just as a Cartesian equation like \(y = x^2\) describes a relationship between \(x\) and \(y\) values on a Cartesian grid, a polar equation can be written describing a relationship between \(r\) and \(\theta\) values on the polar grid.
Example \(\PageIndex{5}\)
Sketch a graph of the polar equation \(r = \theta\).
Solution
The equation \(r = \theta\) describes all the points for which the radius \(r\) is equal to the angle. To visualize this relationship, we can create a table of values.
| \(\theta\) | 0 | \(\pi/4\) | \(\pi/2\) | \(3\pi/4\) | \(\pi\) | \(5\pi/4\) | \(3\pi/2\) | \(7\pi/4\) | \(2\pi\) |
| \(r\) | 0 | \(\pi/4\) | \(\pi/2\) | \(3\pi/4\) | \(\pi\) | \(5\pi/4\) | \(3\pi/2\) | \(7\pi/4\) | \(2\pi\) |
We can plot these points on the plane, and then sketch a curve that fits the points. The resulting graph is a spiral.
Notice that the resulting graph cannot be the result of a function of the form \(y = f(x)\), as it does not pass the vertical line test, even though it resulted from a function giving \(r\) in terms of \(\theta\).
Although it is nice to see polar equations on polar grids, it is more common for polar graphs to be graphed on the Cartesian coordinate system, and so, the remainder of the polar equations will be graphed accordingly.
The spiral graph above on a Cartesian grid is shown here.
Example \(\PageIndex{6}\)
Sketch a graph of the polar equation \(r = 3\).
Solution
Recall that when a variable does not show up in the equation, it is saying that it does not matter what value that variable has; the output for the equation will remain the same. For example, the Cartesian equation \(y = 3\) describes all the points where \(y = 3\), no matter what the x values are, producing a horizontal line.
Likewise, this polar equation is describing all the points at a distance of 3 from the origin, no matter what the angle is, producing the graph of a circle.
The normal settings on graphing calculators and software graph on the Cartesian coordinate system with \(y\) being a function of \(x\), where the graphing utility asks for \(f(x)\), or simply \(y =\).
To graph polar equations, you may need to change the mode of your calculator to Polar. You will know you have been successful in changing the mode if you now have \(r\) as a function of \(\theta\), where the graphing utility asks for \(r(\theta)\), or simply \(r =\).
Example \(\PageIndex{7}\)
Sketch a graph of the polar equation \(r = 4 \cos(\theta)\), and find an interval on which it completes one cycle.
Solution
While we could again create a table, plot the corresponding points, and connect the dots, we can also turn to technology to directly graph it. Using technology, we produce the graph shown here, a circle passing through the origin.
Since this graph appears to close a loop and repeat itself, we might ask what interval of \(\theta\) values yields the entire graph. At \(\theta = 0\), \(r = 4\cos(0) = 4\), yielding the point (4, 0). We want the next \(\theta\) value when the graph returns to the point (4, 0). Solving for when \(x = 4\) is equivalent to solving \(r\cos(\theta) = 4\).
\[r \cos(\theta) = 4\nonumber\]Substituting the equation for \(r\) gives
\[4\cos(\theta)\cos(\theta) = 4\nonumber\]Dividing by 4 and simplifying
\[\cos^{2}(\theta)= 1\nonumber\]This has solutions when
\[\cos(\theta) = 1\text{ or }\cos(\theta) = -1\nonumber\]Solving these gives solutions
\[\theta = 0\text{ or }\theta = \pi\nonumber\]
This shows us at 0 radians we are at the point (0, 4), and again at \(\pi\) radians we are at the point (0, 4) having finished one complete revolution.
This interval \(0 \le \theta < \pi\) yields one complete iteration of the circle.
Exercise \(\PageIndex{3}\)
Sketch a graph of the polar equation \(r = 3 \sin (\theta)\), and find an interval on which it completes one cycle.
- Answer
-
\[3 \sin(\theta) = 0\text{ at }\theta = 0\text{ and }\theta = \pi\nonumber\]
It completes one cycle on the interval \(0 \le \theta < \pi\).
The last few examples have all been circles. Next, we will consider two other “named” polar equations, limaçons and roses .
Example \(\PageIndex{8}\)
Sketch a graph of the polar equation \(r = 4\sin(\theta) + 2\). What interval of \(\theta\) values corresponds to the inner loop?
Solution
This type of graph is called a
limaçon
.
Using technology, we can draw the graph. The inner loop begins and ends at the origin, where \(r = 0\). We can solve forthe \(\theta\) values for which \(r = 0\).
\[0 = 4\sin(\theta) + 2\nonumber\]
\[-2 = 4\sin(\theta)\nonumber\]
\[\sin(\theta) = -\dfrac{1}{2}\nonumber\]
\[\theta = \dfrac{7\pi}{6}\text{ or }\theta = \dfrac{11\pi}{6}\nonumber\]
This tells us that \(r = 0\), so the graph passes through the origin, twice on the interval \([0, 2\pi)\).
The inner loop arises from the interval \(\dfrac{7\pi}{6} \le \theta \le \dfrac{11\pi}{6}\).
This corresponds to where the function \(r = 4 \sin(\theta) + 2\) takes on negative values, as we could see if we graphed the function in the \(r \theta\) plane.
Example \(\PageIndex{9}\)
Sketch a graph of the polar equation \(r = \cos(3\theta)\). What interval of \(\theta\) values describes one small loop of the graph?
Solution
This type of graph is called a 3 leaf rose.
We can use technology to produce a graph. The interval \([0, \pi)\) yields one cycle of this function. As with the last problem, we can note that there is an interval on which one loop of this graph begins and ends at the origin, where \(r = 0\). Solving for \(\theta\),
\[0 = \cos(3\theta)\nonumber\]Substitute \(u = 3\theta\)
\[0 = \cos(u)\nonumber\]
\[u = \dfrac{\pi}{2}\text{ or }u = \dfrac{3\pi}{2}\text{ or }u = \dfrac{5\pi}{2}\nonumber\]
Undo the substitution,
\[3 \theta = \dfrac{\pi}{2}\text{ or }3 \theta = \dfrac{3\pi}{2}\text{ or }3 \theta = \dfrac{5\pi}{2}\nonumber\]
\[\theta = \dfrac{\pi}{6}\text{ or }\theta = \dfrac{\pi}{2}\text{ or }\theta = \dfrac{5\pi}{6}\nonumber\]
There are 3 solutions on \(0 \le \theta < \pi\) which correspond to the 3 times the graph returns to the origin, but the first two solutions we solved for above are enough to conclude that
one loop corresponds to the interval \(\dfrac{\pi}{6} \le \theta < \dfrac{\pi}{2}\).
If we wanted to get an idea of how the computer drew this graph, consider when \(\theta = 0\).
\(r = \cos(3\theta) = \cos(0) = 1\), so the graph starts at (1, 0). As we found above, at \(\theta = \dfrac{\pi}{6}\) and \(\theta = \dfrac{\pi}{2}\), the graph is at the origin. Looking at the equation, notice that any angle in between \(\dfrac{\pi}{6}\) and \(\dfrac{\pi}{2}\), for example at \(\theta = \dfrac{\pi}{3}\), produces a negative \(r\): \[r = \cos(3 \cdot \dfrac{\pi}{3}) = \cos(\pi) = -1\nonumber\]
Notice that with a negative \(r\) value and an angle with terminal side in the first quadrant, the corresponding Cartesian point would be in the third quadrant. Since \(r = \cos(3\theta)\) is negative on \(\dfrac{\pi}{6} \le \theta < \dfrac{\pi}{2}\), this interval corresponds to the loop of the graph in the third quadrant.
Exercise \(\PageIndex{4}\)
Sketch a graph of the polar equation \(r = \sin(2\theta)\). Would you call this function a limaçon or a rose ?
- Answer
-
This is a 4-leaf rose.
Converting Equations
While many polar equations cannot be expressed nicely in Cartesian form (and vice versa), it can be beneficial to convert between the two forms, when possible. To do this we use the same relationships we used to convert points between coordinate systems.
Example \(\PageIndex{10}\)
Rewrite the Cartesian equation \(x^2 + y^2 = 6y\) as a polar equation.
Solution
We wish to eliminate \(x\) and \(y\) from the equation and introduce \(r\) and \(\theta\). Ideally, we would like to write the equation with \(r\) isolated, if possible, which represents \(r\) as a function of \(\theta\).
\[x^2 + y^2 = 6y\nonumber\]Remembering \(x^2 + y^2 = r^2\) we substitute
\[r^2 = 6y\nonumber\] \(y = r\sin(\theta)\) and so we substitute again
\[r^2 = 6r \sin(\theta)\nonumber\] Subtract \(6r\sin(\theta)\) from both sides
\[r^2 - 6r\sin(\theta) = 0\nonumber\]Factor
\[r (r - 6\sin(\theta)) = 0\nonumber\] Use the zero factor theorem
\[r = 6\sin(\theta)\text{ or }r = 0\nonumber\] Since \(r = 0\) is only a point, we reject that solution.
The solution \(r = 6\sin(\theta)\) is fairly similar to the one we graphed in Example 7. In fact, this equation describes a circle with bottom at the origin and top at the point (0, 6).
Example \(\PageIndex{11}\)
Rewrite the Cartesian equation \(y = 3x + 2\) as a polar equation.
Solution
\[y = 3x + 2\nonumber\]Use \(y = r\sin(\theta)\) and \(x = r\cos(\theta)\)
\[r\sin(\theta) = 3r\cos(\theta) + 2\nonumber\] Move all terms with \(r\) to one side
\[r\sin(\theta) - 3r\cos(\theta) = 2\nonumber\] Factor out \(r\)
\[r(\sin(\theta) - 3\cos(\theta)) = 2\nonumber\] Divide
\[r = \dfrac{2}{\sin(\theta) - 3\cos(\theta)}\nonumber\]
In this case, the polar equation is more unwieldy than the Cartesian equation, but there are still times when this equation might be useful.
Example \(\PageIndex{12}\)
Rewrite the polar equation \(r = \dfrac{3}{1- 2\cos(\theta)}\) as a Cartesian equation.
Solution
We want to eliminate \(\theta\) and \(r\) and introduce \(x\) and \(y\). It is usually easiest to start by clearing the fraction and looking to substitute values that will eliminate \(\theta\).
\[r = \dfrac{3}{1 - 2\cos(\theta)}\nonumber\]Clear the fraction
\[r(1 - 2\cos(\theta)) = 3\nonumber\]Use \(\cos(\theta) = \dfrac{x}{r}\) to eliminate \(\theta\)
\[r(1 - 2\dfrac{x}{r}) = 3\nonumber\] Distribute and simplify
\[r - 2x = 3\nonumber\] Isolate the \(r\)
\[r = 3 + 2x\nonumber\] Square both sides
\[r^2 = (3 + 2x)^2\nonumber\] Use \(x^2 + y^2 = r^2\)
\[x^2 + y^2 = (3+2x)^2\nonumber\]
When our entire equation has been charged from \(r\) and \(\theta\) to \(x\) and \(y\) we can stop unless asked to solve for \(y\) or simplify.
In this example, if desired, the right side of the equation could be expanded and the equation simplified further. However, the equation cannot be written as a function in Cartesian form.
Exercise \(\PageIndex{5}\)
a. Rewrite the Cartesian equation in polar form: \(y = \pm \sqrt{3 - x^2}\)
b. Rewrite the polar equation in Cartesian form: \(r = 2\sin(\theta)\)
- Answer
-
a. \(y = \pm \sqrt{3 - x^2}\) can be rewritten as \(x^2 + y^2 = 3\), and becomes \(r = \sqrt{3}\)
b. \[r = 2\sin(\theta)\nonumber\]
\[r= 2\dfrac{y}{r}\nonumber\]
\[r^2 = 2y\nonumber\]
\[x^2 + y^2 = 2y\nonumber\]
Example \(\PageIndex{13}\)
Rewrite the polar equation \(r = \sin(2\theta)\) in Cartesian form.
Solution
\[r = \sin(2\theta)\nonumber\] Use the double angle identity for sine
\[r = 2\sin(\theta)\cos(\theta)\nonumber\] Use \(\cos(\theta) = \dfrac{x}{r}\) and \(\sin(\theta) = \dfrac{y}{r}\)
\[r = 2 \cdot \dfrac{x}{r} \cdot \dfrac{y}{r}\nonumber\] Simplify
\[r = \dfrac{2xy}{r^2}\nonumber\]Multiply by \(r^2\)
\[r^3 = 2xy\nonumber\]Since \(x^2 + y^2 = r^2\), \(r = \sqrt{x^2 + y^2}\)
\[(\sqrt{x^2 + y^2})^3 = 2xy\nonumber\]
This equation could also be written as
\[(x^2 + y^2)^{3/2} = 2xy\text{ or }x^2 + y^2 = (2xy)^{2/3}\nonumber\]
Important Topics of This Section
- Cartesian coordinate system
- Polar coordinate system
- Plotting points in polar coordinates
- Converting coordinates between systems
- Polar equations: Spirals, circles, limaçons and roses Converting equations between systems


9.6: Integration and differentiation over angles
The vector equations from the previous sections are very useful for determining the kinematics of points in the rigid body at a certain time instant. However, to predict the motion at all times, the coordinate functions \(x_{A}(t), y_{A}(t)\) and
\(\phi_{B / A}(t)\) need to be determined. Using Equation 9.5 we can then describe the planar time-dependent motion of a point \(B\) as follows:
\[\overrightarrow{\boldsymbol{r}}_{B, 2 D}(t)=\overrightarrow{\boldsymbol{r}}_{A}(t)+\left|\overrightarrow{\boldsymbol{r}}_{B / A}\right| \cos \phi_{B / A}(t) \hat{\boldsymbol{\imath}}+\left|\overrightarrow{\boldsymbol{r}}_{B / A}\right| \sin \phi_{B / A}(t) \hat{\boldsymbol{\jmath}} \tag{9.66} \label{9.66}\]
The kinematics to describe the motion \(\overrightarrow{\boldsymbol{r}}_{A}(t)\) of point \(A\) is exactly the same as for the kinematics of point masses that were described in Ch. 5, using different coordinate systems and path curves. The new aspect is that for rigid bodies we also have to describe the time dependent angle \(\phi(t) \equiv \phi_{B / A}(t)\).
Luckily, for planar kinematics in 2D the required mathematics is completely the same as that for the path coordinate \(s_{A}\), such that the methods we discussed in Sec. 5.7 and Sec. 5.8 can be directly applied to the angle \(\phi(t)\). This thus just requires replacing \(s\) by \(\phi, v\) by \(\omega\) and \(a\) by \(\alpha\). For completeness we provide all the resulting equations for angles here. For time differentiation we get:
\[\begin{align} \phi(t) & \rightarrow \tag{9.67} \label{9.67}\\[4pt] \omega(t) & =\dot{\phi}(t) \rightarrow \tag{9.68} \label{9.68}\\[4pt] \alpha(t) & =\dot{\omega}(t)=\ddot{\phi}(t) \tag{9.69} \label{9.69}\end{align}\]
For time integration we obtain:
\[\begin{align} \alpha(t) & \rightarrow \tag{9.70} \label{9.70}\\[4pt] \omega\left(t_{2}\right) & =\omega\left(t_{1}\right)+\int_{t_{1}}^{t_{2}} \alpha(t) \mathrm{d} t \rightarrow \tag{9.71} \label{9.71}\\[4pt] \phi\left(t_{2}\right) & =\phi\left(t_{1}\right)+\int_{t_{1}}^{t_{2}} \omega(t) \mathrm{d} t \tag{9.72} \label{9.72}\end{align}\]
Similar to having functions along the path curve like \(a_{s}(s)\) and \(v_{s}(s)\), we can have functions for angle dependent angular velocity \(\omega_{\phi}(\phi)\) and angular acceleration \(\alpha_{\phi}(\phi)\). Differentiation and integration over these functions is similar to that over the path coordinate \(s\) :
\[\begin{align} \alpha_{\phi}(\phi)=\frac{\mathrm{d} \omega_{\phi}}{\mathrm{d} \phi} \omega_{\phi}(\phi) \tag{9.73} \label{9.73}\\[4pt] \omega_{\phi}^{2}\left(\phi_{2}\right)=\omega_{\phi}^{2}\left(\phi_{1}\right)+2 \int_{\phi_{1}}^{\phi_{2}} \alpha_{\phi}(\phi) \mathrm{d} \phi \tag{9.74} \label{9.74}\end{align}\]
Gearwheel \(A\) in Figure 9.8 has a constant acceleration \(\alpha_{A}\) and \(\omega_{A}=0, \phi_{A}=0\) at \(t=0\). Determine the time dependent angle \(\phi_{A}(t)\).
Solution
It is straightforward to obtain the solution by integration:
\[\begin{align} \omega_{A}(t) & =\omega_{A}(0)+\int_{0}^{t} \alpha_{A} \mathrm{~d} t=\alpha_{A} t \tag{9.75} \label{9.75}\\[4pt] \phi_{A}(t) & =\phi_{A}(0)+\int_{0}^{t} \alpha_{A} t \mathrm{~d} t=\frac{1}{2} \alpha_{A} t^{2} \tag{9.76} \label{9.76}\end{align}\]


5.2.7: Summary and Exercises
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Summary
The Constitution gives Congress the power to legislate on bankruptcy. The current law is the Bankruptcy Abuse Prevention and Consumer Protection Act of 2005, which provides for six types of proceedings: (1) liquidation, Chapter 7; (2) adjustment of debts of a municipality, Chapter 9; (3) reorganization, Chapter 11; (4) family farmers with regular income, Chapter 12; (5) individuals with regular income, Chapter 13; and (6) cross-border bankruptcies, Chapter 15.
With some exceptions, any individual, partnership, or corporation seeking liquidation may file a voluntary petition in bankruptcy. An involuntary petition is also possible; creditors petitioning for that must meet certain criteria.
A petition operates as a stay against the debtor for lawsuits to recover claims or enforce judgments or liens. A judge will issue an order of relief and appoint a trustee, who takes over the debtor’s property and preserves security interests. To recover monies owed, creditors must file proof of claims. The trustee has certain powers to recover property for the estate that the debtor transferred before bankruptcy. These include the power to act as a hypothetical lien creditor, to avoid fraudulent transfers and voidable preferences.
The bankruptcy act sets out categories of claimants and establishes priority among them. After secured parties take their security, the priorities are (1) domestic support obligations, (2) administrative expenses, (3) gap creditor claims, (4) employees’ wages, salaries, commissions, (5) contributions to employee benefit plans, (6) grain or fish producers’ claims against a storage facility, (7) consumer deposits, (8) taxes owed to governments, (9) allowed claims for personal injury or death resulting from debtor’s driving or operating a vessel while intoxicated. After these priority claims are paid, the trustee must distribute the estate in this order: (a) unsecured creditors who filed timely, (b) unsecured creditors who filed late, (c) persons claiming fines and the like, (d) all other creditors, (e) the debtor. Most bankruptcies are no-asset, so creditors get nothing.
Under Chapter 7’s 2005 amendments, debtors must pass a means test to be eligible for relief; if they make too much money, they must file Chapter 13.
Certain property is exempt from the estate of an individual debtor. States may opt out of the federal list of exemptions and substitute their own; most have.
Once discharged, the debtor is no longer legally liable for most debts. However, some debts are not dischargeable, and bad faith by the debtor may preclude discharge. Under some circumstances, a debtor may reaffirm a discharged debt. A Chapter 7 case may be converted to Chapter 11 or 13 voluntarily, or to Chapter 11 involuntarily.
Chapter 11 provides for reorganization. Any person eligible for discharge in Chapter 7 is eligible for Chapter 11, except stockbrokers and commodity brokers; those who have too much debt to file Chapter 13 and surpass the means test for Chapter 7 file Chapter 11. Under Chapter 11, the debtor retains possession of the business and may continue to operate it with its own management unless the court appoints a trustee. The court may do so either for cause or if it is in the best interests of the creditors. The court must appoint a committee of unsecured creditors, who remain active throughout the proceeding. The debtor may file its own reorganization plan and has the exclusive right to do so within 120 days if it remains in possession. The plan must be accepted by certain proportions of each impaired class of claims and interests. It is binding on all creditors, and the debtor is discharged from all debts once the court confirms the plan.
Chapter 13 is for any individual with regular income who has difficulty paying debts; it is voluntary only; the debtor must get credit counseling. The debtor presents a payment plan to creditors, and the court appoints a trustee. The plan extends the time to pay and may reduce the size of the debt. If the creditors wind up with more in this proceeding than they would have in Chapter 7, the court is likely to approve the plan. The court may approve a stretch-out of five years. Some debts not dischargeable under Chapter 7 may be under Chapter 13.
Alternatives to bankruptcy are (1) composition (agreement by creditors to accept less than the face amount of the debt), (2) assignment for benefit of creditors (transfer of debtor’s property to a trustee, who uses it to pay debts), and (3) receivership (a disinterested person is appointed by the court to preserve assets and distribute them at the court’s direction). Because these are voluntary procedures, they are ineffective if all parties do not agree to them.
Exercises
- David has debts of $18,000 and few assets. Because his debts are less than $25,000, he decides to file for bankruptcy using the state court system rather than the federal system. Briefly describe the procedure he should follow to file for bankruptcy at the state level.
- Assume that David in Exercise 1 is irregularly employed and has developed a plan for paying off his creditors. What type of bankruptcy should he use, Chapter 7, 11, or 13? Why?
- Assume that David owns the following unsecured property: a $3,000 oboe, a $1,000 piano, a $2,000 car, and a life insurance policy with a cash surrender value of $8,000. How much of this property is available for distribution to his creditors in a bankruptcy? Explain.
- If David owes his ex-wife alimony (maintenance) payments and is obligated to pay $12,000 for an educational loan, what effect will his discharge have on these obligations?
- Assume that David owns a corporation that he wants to liquidate under Chapter 7. After the corporate assets are distributed to creditors, there is still money owing to many of them. What obstacle does David face in obtaining a discharge for the corporation?
- The famous retired professional football player—with a pension from the NFL—Orenthal James “O.J.” Simpson was convicted of wrongful death in a celebrated Santa Monica, California, trial in 1997 and ordered to pay $33.5 million in damages to the families of the deceased. Mr. Simpson sold his California house, moved to Florida, and, from occasional appearances in the press, seemed to be living a high-style life with a big house, nice cars, and sharp clothing. He has never declared bankruptcy. Why hasn’t he been forced into an involuntary Chapter 7 bankruptcy by his creditors?
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- A debtor has an automobile worth $5,000. The federal exemption applicable to her is $3,225. The trustee sells the car and gives the debtor the amount of the exemption. The debtor, exhausted by the bankruptcy proceedings, takes the $3,225 and spends it on a six-week vacation in Baja California. Is this an “abuse” of the bankruptcy system?
- A debtor has $500 in cash beyond what is exempt in bankruptcy. She takes the cash and buys new tires for her car, which is worth about $2,000. Is this an “abuse” of the bankruptcy system?
SELF CHECK QUESTIONS
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Alternatives to bankruptcy include
- an assignment
- a composition
- receivership
- all of the above
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A composition is
- a procedure where a receiver takes over the debtor’s property
- an agreement by creditors to take less than the face value of their debt
- basically the same as an assignment
- none of these
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The highest-priority class set out by the 2005 act is for
- employees’ wages
- administrative expenses
- property settlements arising from divorce
- domestic support obligations
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Darlene Debtor did the following within ninety days of filing for bankruptcy. Which could be set aside as a preferential payment?
- paid water and electricity bills
- made a gift to the Humane Society
- prepaid an installment loan on inventory
- borrowed money from a bank secured by a mortgage on business property
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Donald Debtor sold his 1957 Chevrolet to his brother for one-fifth its value sixty days before filing for bankruptcy. The trustee wishes to avoid the transaction on the basis that it was
- a hypothetical lien
- a lease disguised as a sale
- a preferential payment
- a voidable preference
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Acme Co. filed for bankruptcy with the following debts; which is their correct priority from highest to lowest?i. wages of $15,000 owed to employeesii. unpaid federal taxesiii. balance owed to a creditor who claimed its security with a $5,000 deficiency owing
- i, ii, iii
- ii, iii, i
- iii, ii, i
- i, iii, ii
AnswerS
- d
- b
- d
- c
- d
- a


6.4: “A South African Storm” by Allison Howard
As you read, look for the following:
- What “points for comparison” does the author use?
- How does the author go beyond the obvious similarities and differences to surface interesting ideas and insights?
By Allison Howard – Peace Corps Volunteer: South Africa (2003-2005)
It’s a Saturday afternoon in January in South Africa. When I begin the 45–minute walk to the shops for groceries, I can hear thunder cracking in the distance up the mountain in Mageobaskloof. But at 4 p.m. the sky is still light and bright and I am sure—famous last words—I will be fine without an umbrella.
Just the basics: eggs, bread, Diet Coke in a bag slung into the crook of my elbow. Halfway from town, two black South African women—domestic workers in the homes of white Afrikaner families—stop me with wide smiles. They know me; I’m the only white person in town who walks everywhere, as they do. They chatter quickly in northern Sotho: “Missus, you must go fast. Pula e tla na! The rain, it comes!” They like me, and it feels very important to me that they do. “Yebo, yebo, mma,” I say—Yes, it’s true—and I hurry along in flip-flops, quickening my pace, feeling good about our brief but neighborly conversation. These are Venda women.
My black South African friends tell me it’s easy to tell a Venda from a Shangaan from a Xhosa from a Pedi. “These ones from Venda , they have wide across the nose and high in the cheekbones,” they say. But I don’t see it; I’m years away from being able to distinguish the nuances of ethnicity. Today, I know these women are Vendas simply because of their clothing: bright stripes of green and yellow and black fabric tied at one shoulder and hanging quite like a sack around their bodies. They’ve already extended a kindness to me by speaking in northern Sotho. It’s not their language but they know I don’t speak a word of Afrikaans (though they don’t understand why; Afrikaans is the language of white people). They know I struggle with Sotho and they’re trying to help me learn. So they speak Sotho to me and they’re delighted and amused by my fumbling responses. And I am, quite simply, delighted by their delight.
The Venda ladies are right: the rain, it comes. Lightly at first, and by habit I begin trotting to hurry my way home. Just a little rain at first and there are plenty of us out in it. I can see others up ahead on the street and others still just leaving the shops to get back before the real rain begins.
The people who are walking along this swath of tar road are black. Black people don’t live in this neighborhood—or in my town at all, for the most part. They work and board here as domestic workers, nannies, gardeners. Their families live in black townships and rural villages—some just outside of my town; others far away, in places like Venda.
Today, we’re walking together in the rain, and I’m quickening my pace because—after all , it’s raining . That’s what you do in the rain. And even though it’s coming down noticeably harder, it’s 80 degrees and I’m not cold, I’m just wet. My hair is stuck to my forehead and my T-shirt is soaked … and I’m the only one running for cover. And I think: So what? It’s just water and in the middle of the January summer, it’s warm, refreshing water. Why run? Why do we run from the rain?
In my life back in the United States, I might run because I was carrying a leather handbag, or because I wore an outfit that shouldn’t get wet. I would run because rain dishevels and messes things up. Mostly though, we run because we just do; it’s a habit. I’ve done it a hundred times: running to my car or the subway station with a newspaper sheltering my head. I have never not quickened my pace in the rain until today.
It took all of my 27 years and a move to Africa , where I don’t have a leather handbag to shelter or a pretty outfit to protect. I’m wearing an old cotton skirt and a T-shirt, and I’m drenched, and I love it. I learn things here in the most ordinary circumstances. And I feel like a smarter, better woman today because I got groceries in the rain.
But on the long walk home, positively soaked and smiling like a fool, I notice a car pulling over and a man yelling in Afrikaans to get in, get in.  On the road ahead, a woman about my age carries a parcel wrapped in plastic, balanced precariously on her head.  And the man in the car is white and he’s gesturing frantically for me to get in. Why me? Why not the others? Because I’m white and it’s about race. Everything is about race here.
This man in the car is trying to do something kind and neighborly. He wants to help me and his gesture is right, but his instincts are so wrong. How do you resent someone who is, for no benefit of his own, trying to help? But I do.  This whole event unravels in a few seconds’ time. He’s leaned over and opened the car door, urging me in … and I get in. And we speed past my fellow walkers and he drops me at my doorstep before I have time to think of anything besides giving him directions.
It feels like a mistake because I’m ashamed to think what the Venda women would have felt if he’d ignored them and they had watched me climb into that car. In some ways, the whole episode seems absurd. I’m not going to atone for 400 years of South African history by walking with black people in the rain. If I’d refused his ride, he wouldn’t have thought anything besides the fact that I was certifiably crazy. That’s the thing about being here: I’m not going to change anything. But I believe it matters in some infinitesimal way that people like the Venda women, and the dozens of people who may walk alongside me on any given day, know that I’m there. In black South African culture it is polite to greet every person you pass. That’s what they do, so I do it, too. On the occasional morning, someone might greet me as “sesi,” sister. I have to believe that matters; I know it matters to me.
I was disappointed in myself for getting into the car because I acted according to the same habit that makes us think rain an inconvenience. Just as we run from the rain, I hopped into that car because I’m supposed to. Conventionally, it makes sense. But convention compels us to do so many things that don’t make any sense at all. Convention misinforms our instincts. And in a larger sense, it is convention that propels Afrikaner culture anachronistically into the future. Ten years after the supposed end of apartheid, I’m living in a world of institutionalized racism. Convention becomes institution—and it’s oppressive and it’s unjust. I know that if I’m going to make it here for two more years, I need to walk in the rain. It’s a small, wasted gesture, but it’s an uncorrupted instinct that makes me feel human.
So much about living here feels like that fraction of a second when the Afrikaner man was appealing to my conventional sensibilities and the people on the street were appealing to my human instincts. It may feel unnatural to reject those sensibilities just as, at first, it feels unnatural to walk in the rain. But if I lose a hold on my instincts here, I’ll fail myself and I’ll fail to achieve those tiny things that matter so much. It’s simple and it’s small; and it’s everything. Gandhi said, “Be the change you wish to see in the world.” Indeed. Let it rain.


3.3: Independent and Mutually Exclusive Events
Independent and mutually exclusive do not mean the same thing.
Independent Events
Two events are independent if one of the following are true:
- \(P(A|B)=P(A)\)
- \(P(B|A)=P(B)\)
- \(P(A \cap B)=P(A) P(B)\)
Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent .
Sampling may be done with replacement or without replacement .
- With replacement : If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.
- Without replacement : When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.
If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise .
Example \(\PageIndex{1}\)
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit.
a. Sampling with replacement:
Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is the
Q
of spades. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the
Q
of spades again. Your picks are {
Q
of spades, ten of clubs,
Q
of spades}. You have picked the
Q
of spades twice. You pick each card from the 52-card deck.
b. Sampling without replacement:
Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the
K
of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the
J
of spades. Your picks are {
K
of hearts, three of diamonds,
J
of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice. The probability of picking the three of diamonds is called a conditional probability because it is conditioned on what was picked first. This is true also of the probability of picking the J of spades. The probability of picking the J of spades is actually conditioned on
both
the previous picks.
Exercise \(\PageIndex{1}\)
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit. Three cards are picked at random.
- Suppose you know that the picked cards are Q of spades, K of hearts and Q of spades. Can you decide if the sampling was with or without replacement?
- Suppose you know that the picked cards are Q of spades, K of hearts, and J of spades. Can you decide if the sampling was with or without replacement?
Example \(\PageIndex{2}\)
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs.
- Suppose you pick four cards, but do not put any cards back into the deck. Your cards are QS , 1 D , 1 C , QD .
- Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH , 7 D , 6 D , KH .
Which of a. or b. did you sample with replacement and which did you sample without replacement?
Answer
a. Without replacement; b. With replacement
Exercise \(\PageIndex{2}\)
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement.
- QS , 1 D , 1 C , QD
- KH , 7 D , 6 D , KH
- QS , 7 D , 6 D , KS
Mutually Exclusive Events
\(A\) and \(B\) are mutually exclusive events if they cannot occur at the same time. Said another way, if \(A\) occurred then \(B\) cannot occur and vice versa. This means that \(A\) and \(B\) do not share any outcomes and \(A \cap B = 0\).
For example, suppose the sample space \(S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\). Let \(A = \{1, 2, 3, 4, 5\}\), \(B = \{4, 5, 6, 7, 8\}\), and \(C = \{7, 9\}\). \(A \cap B = \{4, 5\}\). \(P(A \cap B)= \frac{2}{10}\) and is not equal to zero. Therefore, \(A\) and \(B\) are not mutually exclusive. On the other hand, \(A\) and \(C\) do not have any numbers in common, so \(P(A \cap C) = 0\). Therefore, \(A\) and \(C\) are mutually exclusive.
If it is not known whether \(A\) and \(B\) are mutually exclusive, assume they are not until you can show otherwise . The following examples illustrate these definitions and terms.
Example \(\PageIndex{3}\)
Flip two fair coins. (This is an experiment.)
The sample space is \(\{HH, HT, TH, TT\}\) where \(T\) = tails and \(H\) = heads. The outcomes are \(HH\), \(HT\), \(TH\), and \(TT\). The outcomes \(HT\) and \(TH\) are different. The \(HT\) means that the first coin showed heads and the second coin showed tails. The \(TH\) means that the first coin showed tails and the second coin showed heads.
- Let \(A\) = the event of getting at most one tail . (At most one tail means zero or one tail.) Then \(A\) can be written as \(\{HH, HT, TH\}\). The outcome \(HH\) shows zero tails. \(HT\) and \(TH\) each show one tail.
- Let \(B\) = the event of getting all tails. \(B\) can be written as \(\{TT\}\). \(B\) is the complement of \(A\), so \(B\) = \(A'\). Also, \(P(A) + P(B) = P(A) + P(A′) = 1\).
- The probabilities for \(A\) and for \(B\) are \(P(A) = \frac{3}{4}\) and \(P(B) = \frac{1}{4}\).
- Let \(C\) = the event of getting all heads. \(C = \{HH\}\). Since \(B = \{TT\}\), \(P(B \cap C) = 0\). \(B\) and \(C\) are mutually exclusive. (\(B\) and \(C\) have no members in common because you cannot have all tails and all heads at the same time.)
- Let \(D\) = event of getting more than one tail. \(D = \{TT\}\). \(P(D) = \frac{1}{4}\).
- Let \(E\) = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) \(E = \{HT, HH\}\). \(P(E) = \frac{2}{4}\).
- Find the probability of getting at least one (one or two) tail in two flips. Let \(F\) = event of getting at least one tail in two flips. \(F = \{HT, TH, TT\}\). \(P(F) = \frac{3}{4}\).
Exercise \(\PageIndex{3}\)
Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card.
Example \(\PageIndex{4}\)
Flip two fair coins. Find the probabilities of the events.
- Let \(F\) = the event of getting at most one tail (zero or one tail).
- Let \(G\) = the event of getting two faces that are the same.
- Let \(H\) = the event of getting a head on the first flip followed by a head or tail on the second flip.
- Are \(F\) and \(G\) mutually exclusive?
- Let \(J\) = the event of getting all tails. Are \(J\) and \(H\) mutually exclusive?
Answer
Look at the sample space in Example \(\PageIndex{3}\) .
-
Zero (0) or one (1) tails occur when the outcomes \(HH, TH, HT\) show up. \(P(F) = \frac{3}{4}\).
-
Two faces are the same if \(HH\) or \(TT\) show up. \(P(G) = \frac{2}{4}\).
-
A head on the first flip followed by a head or tail on the second flip occurs when \(HH\) or \(HT\) show up. \(P(H) = \frac{2}{4}\).
-
\(F\) and \(G\) share \(HH\) so \(P(F \cap G)\) is not equal to zero (0). \(F\) and \(G\) are not mutually exclusive.
-
Getting all tails occurs when tails shows up on both coins \(TT\). \(H\)'s outcomes are \(HH\) and \(HT\). \(J\) and \(H\) have nothing in common, so \(P(J \cap H) = 0\). \(J\) and \(H\) are mutually exclusive.
Exercise \(\PageIndex{4}\)
A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:
- Let \(F\) = the event of getting the white ball twice.
- Let \(G\) = the event of getting two balls of different colors.
- Let \(H\) = the event of getting white on the first pick.
- Are \(F\) and \(G\) mutually exclusive?
- Are \(G\) and \(H\) mutually exclusive?
Example \(\PageIndex{5}\)
Roll one fair, six-sided die. The sample space is \(\{1, 2, 3, 4, 5, 6\}\). Let event \(A\) = a face is odd. Then \(A = \{1, 3, 5\}\). Let event \(B\) = a face is even. Then \(B = \{2, 4, 6\}\).
- Find \(A'\), the complement of \(A\). The complement of \(A\) is \(B\) because \(A\) and \(B\) together make up the sample space. \(P(A) + P(B) = P(A) + P(A′) = 1\). Also, \(P(A) = \frac{3}{6}\) and \(P(B) = \frac{3}{6}\).
- Let event \(C\) = odd faces larger than two. Then \(C = \{3, 5\}\). Let event \(D\) = all even faces smaller than five. Then \(D = \{2, 4\}\). \(P(C \cap D) = 0\) because you cannot have an odd and even face at the same time. Therefore, \(C\) and \(D\) are mutually exclusive events.
- Let event \(E\) = all faces less than five. \(E = \{1, 2, 3, 4\}\). Are \(C\) and \(E\) mutually exclusive events? Why or why not?
No. \(C = \{3, 5\}\) and \(E = \{1, 2, 3, 4\}\). \(P(C\cap E) = \frac{1}{6}\). To be mutually exclusive, \(P(C\cap E)\) must be zero.
- Find \(P(C|A)\). This is a conditional probability. Recall that the event \(C = \{3, 5\}\) and event \(A = \{1, 3, 5\}\). To find \(P(C|A)\), find the probability of \(C\) using the sample space \(A\). You have reduced the sample space from the original sample space \(\{1, 2, 3, 4, 5, 6\}\) to \(\{1, 3, 5\}\). So, \(P(C|A) = \frac{2}{3}\).
Exercise \(\PageIndex{5}\)
Let event \(A\) = learning Spanish. Let event \(B\) = learning German. Then \(A \cap B\) = learning Spanish and German. Suppose \(P(A) = 0.4\) and \(P(B) = 0.2\). \(P(A\cap B) = 0.08\). Are events \(A\) and \(B\) independent? Hint: You must show ONE of the following:
- \(P(A|B) = P(A)\)
- \(P(B|A) = P(B)\)
- \(P(A\cap B) = P(A)*P(B)\)
Example \(\PageIndex{6}\)
Let event \(G\) = taking a math class. Let event \(H\) = taking a science class. Then, \(G \cap H\) = taking a math class and a science class. Suppose \(P(G) = 0.6\), \(P(H) = 0.5\), and \(P(G\cap H) = 0.3\). Are \(G\) and \(H\) independent?
If \(G\) and \(H\) are independent, then you must show ONE of the following:
- \(P(G|H) = P(G)\)
- \(P(H|G) = P(H)\)
- \(P(G\cap H) = P(G)*P(H)\)
Note
The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information.
a. Show that \(P(G|H) = P(G)\) .
Answer 1
\(P(G|H) = \frac{P(G \cap H)}{P(H)} = \frac{0.3}{0.5} = 0.6 = P(G)\)
b. Show \(P(G\cap H) = P(G)*P(H)\).
Answer 2
\(P(G)*P(H) = (0.6)*(0.5) = 0.3 = P(G\cap H)\).
Since \(G\) and \(H\) are independent, knowing that a person is taking a science class does not change the chance that he or she is taking a math class. If the two events had not been independent (that is, they are dependent), then knowing that a person is taking a science class would change the chance they are taking math. For practice, show that \(P(H|G) = P(H)\) to show that \(G\) and \(H\) are independent events.
Exercise \(\PageIndex{6}\)
In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4.
-
\(R\) = a red marble
-
\(G\) = a green marble
-
\(O\) = an odd-numbered marble
-
The sample space is \(S = \{R1, R2, R3, R4, R5, R6, G1, G2, G3, G4\}\).
\(S\) has ten outcomes. What is \(P(G\cap O)\)?
Example \(\PageIndex{7}\)
Let event \(C\)= taking an English class. Let event \(D\) = taking a speech class. Suppose \(P(C) = 0.75\), \(P(D) = 0.3\), \(P(C|D) = 0.75\) and \(P(C\cap D) = 0.225\). Justify your answers to the following questions numerically.
-
Are \(C\) and \(D\) independent?
-
Are \(C\) and \(D\) mutually exclusive?
-
What is \(P(D|C)\)?
Answer
-
Yes, because \(P(C|D) = P(C)\).
-
No, because \(P(C\cap D)\) is not equal to zero.
-
\(P(D|C) = \frac{P(C \cap D)}{P(C)} = \frac{0.225}{0.75} = 0.3\)
Exercise \(\PageIndex{7}\)
A student goes to the library. Let events \(B\) = the student checks out a book and \(D\) = the student checks out a DVD. Suppose that \(P(B) = 0.4\), \(P(D) = 0.3\), and \(P(B\cap D) = 0.20\).
- Find \(P(B|D)\).
- Find \(P(D|B)\).
- Are \(B\) and \(D\) independent?
- Are \(B\) and \(D\) mutually exclusive?
Example \(\PageIndex{8}\)
In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.
Let \(R\) = red card is drawn, \(B\) = blue card is drawn, \(E\) = even-numbered card is drawn.
The sample space \(S = \{R1, R2, R3, B1, B2, B3, B4, B5\}\). \(S\) has eight outcomes.
- \(P(R) = \frac{3}{8}\). \(P(B) = \frac{5}{8}\). \(P(R\cap B) = 0\). (You cannot draw one card that is both red and blue.)
- \(P(E) = \frac{3}{8}\). (There are three even-numbered cards, \(R2\), \(B2\), and \(B4\).)
- \(P(E|B) = \frac{2}{5}\). (There are five blue cards: \(B1\), \(B2\), \(B3\), \(B4\), and \(B5\). Out of the blue cards, there are two even cards, \(B2\) and \(B4\).)
- \(P(B|E) = \frac{2}{3}\). (There are three even-numbered cards: \(R2\), \(B2\), and \(B4\). Out of the even-numbered cards, two are blue, \(B2\), and \(B4\).)
- The events \(R\) and \(B\) are mutually exclusive because \(P(R\cap B) = 0\).
- Let \(G\)= card with a number greater than 3. \(G = \{B4, B5\}\). \(P(G) = \frac{2}{8}\). Let \(H\) = blue card numbered between one and four, inclusive. \(H = \{B1, B2, B3, B4\}\). \(P(G|H) = \frac{1}{4}\). (The only card in \(H\) that has a number greater than three is \(B4\).) Since \(\frac{2}{8} = \frac{1}{4}\), \(P(G) = P(G|H)\), which means that \(G\) and \(H\) are independent.
Exercise \(\PageIndex{8}\)
In a basketball arena,
- 70% of the fans are rooting for the home team.
- 25% of the fans are wearing blue.
- 20% of the fans are wearing blue and are rooting for the away team.
- Of the fans rooting for the away team, 67% are wearing blue.
Let \(A\) be the event that a fan is rooting for the away team. Let \(B\) be the event that a fan is wearing blue.
Are the events of rooting for the away team and wearing blue independent? Are they mutually exclusive?
Example \(\PageIndex{9}\)
In a particular college class, 60% of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75% have long hair. Let \(F\) be the event that a student is female. Let \(L\) be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent?
The following probabilities are given in this example:
- \(P(F) = 0.6\), \(P(L) = 0.5\)
- \(P(F\cap L) = 0.45\)
- \(P(L|F) = 0.75\)
Note
The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know \(P(F|L)\) yet, so you cannot use the second condition.
Check whether \(P(F\cap L) = P(F)*P(L)\). We are given that \(P(F\cap L) = 0.45\), but \(P(F)*P(L) = (0.6)*(0.5) = 0.3\). The events of being female and having long hair are not independent because \(P(F\cap L)\) does not equal \(P(F)*P(L)\).
Check whether \(P(L|F)\) equals \(P(L)\). Since we are given that \(P(L|F) = 0.75\) and \(P(L) = 0.5\), they are not equal. The events of being female and having long hair are not independent.
Interpretation of Results
The events of being female and having long hair are not independent. Knowing that a student is female changes the probability that a student has long hair.
Exercise \(\PageIndex{9}\)
Mark is deciding which route to take to work. His choices are \(I\) = the Interstate and \(F\) = Fifth Street.
- \(P(I) = 0.44\), \(P(F) = 0.56\)
- \(P(I\cap F) = 0\) because Mark will take only one route to work.
What is the probability of \(P(I\cup F)\)?
Example \(\PageIndex{10}\)
- Toss one fair coin (the coin has two sides, \(H\) and \(T\)). The outcomes are ________. Count the outcomes. There are ____ outcomes.
- Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5 or 6 dots on a side). The outcomes are ________________. Count the outcomes. There are ___ outcomes.
- Multiply the two numbers of outcomes. The answer is _______.
- If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer to c is the number of outcomes (size of the sample space). What are the outcomes? (Hint: Two of the outcomes are \(H1\) and \(T6\).)
-
Event \(A\) = heads \((H)\) on the coin followed by an even number (2, 4, 6) on the die.
\(A\) = {_________________}. Find \(P(A)\). - Event \(B\) = heads on the coin followed by a three on the die. \(B\) = {________}. Find \(P(B)\).
- Are \(A\) and \(B\) mutually exclusive? (Hint: What is \(P(A\cap B)\)? If \(P(A\cap B) = 0\), then \(A\) and \(B\) are mutually exclusive.)
- Are \(A\) and \(B\) independent? (Hint: Is \(P(A\cap B) = P(A)*P(B)\)? If \(P(A\cap B) = P(A)*P(B)\), then \(A\) and \(B\) are independent. If not, then they are dependent).
Answer
- \(H\) and \(T\); 2
- 1, 2, 3, 4, 5, 6; 6
- 2(6) = 12
- \(T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6\)
- \(A = \{H2, H4, H6\}\); \(P(A) = \frac{3}{12}\)
- \(B = \{H3\}\); \(P(B) = \frac{1}{12}\)
- Yes, because \(P(A\cap B) = 0\).
- \(P(A\cap B) = 0\). \(P(A)*P(B) = \frac{3}{12}\). \(P(A\cap B)\) does not equal \(P(A)*P(B)\), so \(A\) and \(B\) are dependent.
Exercise \(\PageIndex{10}\)
A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement).
- Compute \(P(T)\).
- Compute \(P(T|F)\).
- Are \(T\) and \(F\) independent?.
- Are \(F\) and \(S\) mutually exclusive?
- Are \(F\) and \(S\) independent?


9.6: Oxidative Cleavage of Alkynes
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- Steven Farmer & Dietmar Kennepohl
- University of Illinois Springfield
Objectives
After completing this section, you should be able to
- write an equation to represent the oxidative cleavage of an alkyne with potassium permanganate or ozone.
- identify the products that result from the oxidative cleavage of a given alkyne.
- identify the reagents needed to carry out the oxidative cleavage of an alkyne.
- use the results of an oxidative cleavage to determine the identity of an alkyne of unknown structure.
Study Notes
Compare the oxidative cleavage of alkynes with the oxidative cleavage of alkenes, discussed in Section 8.8.
Alkynes, similar to alkenes, can be oxidized gently or strongly depending on the reaction environment. Since alkynes are less stable than alkenes, the reaction conditions can be gentler. For example, alkynes form vicinal dicarbonyls in neutral permanganate solution.
Gentle Alkyne Oxidation
Strong Alkyne Oxidation - Oxidative Cleavage
During strong oxidation with ozone or basic potassium permanganate, the alkyne is cleaved into two products. Because at least one of the reaction products is a carboxylic acid, it is important to consider the acid-base chemistry of the product in the reaction solution. Carboxylic acids are deprotonated in basic solutions to carboxylates. A second reaction step is required to protonate the carboxylate to the neutral form of the carboxylic acid.
Exercise \(\PageIndex{1}\)
Draw the bond-line structures for the product(s) of the following reactions.
- Answer
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Oxidative cleavage of alkynes produces carboxylic acids and/or carbon dioxide. Aldehydes are not produced.


9.3: Independent Samples for Two Means
This section will look at how to analyze when two samples are collected that are independent. As with all other hypothesis tests and confidence intervals, the process is the same though the formulas and assumptions are different. The only difference with the independent t-test, as opposed to the other tests that have been done, is that there are actually two different formulas to use depending on if a particular assumption is met or not.
Hypothesis Test for Independent t-Test (2-Sample t-Test)
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State the random variables and the parameters in words.
\(x_{1}\) = random variable 1
\(x_{2}\) = random variable 2
\(\mu_{1}\) = mean of random variable 1
\(\mu_{2}\)= mean of random variable 2 -
State the null and alternative hypotheses and the level of significance The normal hypotheses would be
\(\begin{array}{ll}{H_{o} : \mu_{1}=\mu_{2}} & {\text { or } \quad H_{o} : \mu_{1}-\mu_{2}=0} \\ {H_{A} : \mu_{1}<\mu_{2}} & \quad\quad\: {H_{A} : \mu_{1}-\mu_{2}<0} \\ {H_{A} : \mu_{1}>\mu_{2}} & \quad\quad\: {H_{A} : \mu_{1}-\mu_{2}>0} \\ {H_{A} : \mu_{1} \neq \mu_{2}} & \quad\quad\: {H_{A} : \mu_{1}-\mu_{2} \neq 0}\end{array}\)
Also, state your \(\alpha\) level here. -
State and check the assumptions for the hypothesis test
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A random sample of size \(n_{1}\) is taken from population 1. A random sample of size \(n_{2}\) is taken from population 2.
Note
The samples do not need to be the same size, but the test is more robust if they are.
- The two samples are independent.
- Population 1 is normally distributed. Population 2 is normally distributed. Just as before, the t-test is fairly robust to the assumption if the sample size is large. This means that if this assumption isn’t met, but your sample sizes are quite large (over 30), then the results of the t-test are valid.
- The population variances are unknown and not assumed to be equal. The old assumption is that the variances are equal. However, this assumption is no longer an assumption that most statisticians use. This is because it isn’t really realistic to assume that the variances are equal. So we will just assume the assumption of the variances being unknown and not assumed to be equal is true, and it will not be checked.
-
A random sample of size \(n_{1}\) is taken from population 1. A random sample of size \(n_{2}\) is taken from population 2.
-
Find the sample statistic, test statistic, and p-value
Sample Statistic:
Calculate \(\overline{x}_{1}, \overline{x}_{2}, s_{1}, s_{2}, n_{1}, n_{2}\)
Test Statistic:
Since the assumption that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) isn’t being satisfied, then
\(t=\dfrac{\left(\overline{x}_{1}-\overline{x}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{\sqrt{\stackrel{s_{1}^{2}}{n_{1}}+\stackrel{s_{2}^{2}}{n_{2}}}}\)
Usually \(\mu_{1}-\mu_{2}=0\), since \(H_{o} : \mu_{1}-\mu_{2}=0\)
Degrees of freedom: (the Welch–Satterthwaite equation)
\(d f=\dfrac{(A+B)^{2}}{\dfrac{A^{2}}{n_{1}-1}+\dfrac{B^{2}}{n_{2}-1}}\)
where \(A=\dfrac{s_{1}^{2}}{n_{1}} \text { and } B=\dfrac{s_{2}^{2}}{n_{2}}\)
p-value:
Using the TI-83/84: tcdf(lower limit, upper limit, df )Using R: pt(t, df )Note
If \(H_{A} : \mu_{1}-\mu_{2}<0\), then lower limit is \(-1 E 99\) and upper limit is your test statistic. If \(H_{A} : \mu_{1}-\mu_{2}>0\), then lower limit is your test statistic and the upper limit is \(1 E 99\). If \(H_{A} : \mu_{1}-\mu_{2} \neq 0\), then find the p-value for \(H_{A} : \mu_{1}-\mu_{2}<0\), and multiply by 2.
Note
If \(H_{A} : \mu_{1}-\mu_{2}<0\), then use pt(t, df ). If \(H_{A} : \mu_{1}-\mu_{2}>0\), then use 1 - pt(t, df ). If \(H_{A} : \mu_{1}-\mu_{2} \neq 0\), then find the p-value for \(H_{A} : \mu_{1}-\mu_{2}<0\), and multiply by 2.
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Conclusion
This is where you write reject \(H_{o}\) or fail to reject \(H_{o}\). The rule is: if the p-value < \(\alpha\), then reject \(H_{o}\). If the p-value \(\geq \alpha\), then fail to reject \(H_{o}\) - Interpretation This is where you interpret in real world terms the conclusion to the test. The conclusion for a hypothesis test is that you either have enough evidence to show \(H_{A}\) is true, or you do not have enough evidence to show \(H_{A}\) is true.
Confidence Interval for the Difference in Means from Two Independent Samples (2 Samp T-Int)
The confidence interval for the difference in means has the same random variables and means and the same assumptions as the hypothesis test for independent samples. If you have already completed the hypothesis test, then you do not need to state them again. If you haven’t completed the hypothesis test, then state the random variables and means and state and check the assumptions before completing the confidence interval step.
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Find the sample statistic and confidence interval
Sample Statistic:
Calculate Confidence Interval: \(\overline{x}_{1}, \overline{x}_{2}, s_{1}, s_{2}, n_{1}, n_{2}\)
The confidence interval estimate of the difference is \(\mu_{1}-\mu_{2}\)
Since the assumption that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) isn’t being satisfied, then
\(\left(\overline{x}_{1}-\overline{x}_{2}\right)-E<\mu_{1}-\mu_{2}<\left(\overline{x}_{1}-\overline{x}_{2}\right)+E\)
where \(E=t_{c} \sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}\)
where \(t_{c}\) is the critical value with degrees of freedom:
Degrees of freedom: (the Welch–Satterthwaite equation)
\(d f=\dfrac{(A+B)^{2}}{\dfrac{A^{2}}{n_{1}-1}+\dfrac{B^{2}}{n_{2}-1}}\)
where \(A=\dfrac{s_{1}^{2}}{n_{1}} \text { and } B=\dfrac{s_{2}^{2}}{n_{2}}\) - Statistical Interpretation: In general this looks like, “there is a C% chance that \(\left(\overline{x}_{1}-\overline{x}_{2}\right)-E<\mu_{1}-\mu_{2}<\left(\overline{x}_{1}-\overline{x}_{2}\right)+E\) contains the true mean difference.”
- Real World Interpretation: This is where you state what interval contains the true difference in means, though often you state how much more (or less) the first mean is from the second mean.
The critical value is a value from the Student’s t-distribution. Since a confidence interval is found by adding and subtracting a margin of error amount from the difference in sample means, and the interval has a probability of containing the true difference in means, then you can think of this as the statement \(P\left(\left(\overline{x}_{1}-\overline{x}_{2}\right)-E<\mu_{1}-\mu_{2}<\left(\overline{x}_{1}-\overline{x}_{2}\right)+E\right)=C\). To find the critical value you use table A.2 in the Appendix.
How to check the assumptions of two sample t-test and confidence interval:
In order for the t-test or confidence interval to be valid, the assumptions of the test must be true. So whenever you run a t-test or confidence interval, you must make sure the assumptions are true. So you need to check them. Here is how you do this:
- For the random sample assumption, describe how you took the two samples. Make sure your sampling technique is random for both samples.
- For the independent assumption, describe how they are independent samples.
- For the assumption about each population being normally distributed, remember the process of assessing normality from chapter 6. Make sure you assess each sample separately.
- You do not need to check the equal variance assumption since it is not being assumed.
Example \(\PageIndex{1}\) hypothesis test for two means
The cholesterol level of patients who had heart attacks was measured two days after the heart attack. The researchers want to see if patients who have heart attacks have higher cholesterol levels over healthy people, so they also measured the cholesterol level of healthy adults who show no signs of heart disease. The data is in Table \(\PageIndex{1}\) ("Cholesterol levels after," 2013). Do the data show that people who have had heart attacks have higher cholesterol levels over patients that have not had heart attacks? Test at the 1% level.
| Cholesterol Level of Heart Attack Patients | Cholesterol Level of Healthy Individual |
|---|---|
| 270 | 196 |
| 236 | 232 |
| 210 | 200 |
| 142 | 242 |
| 280 | 206 |
| 272 | 178 |
| 160 | 184 |
| 220 | 198 |
| 226 | 160 |
| 242 | 182 |
| 186 | 182 |
| 266 | 198 |
| 206 | 182 |
| 318 | 238 |
| 294 | 198 |
| 282 | 188 |
| 234 | 166 |
| 224 | 204 |
| 276 | 182 |
| 282 | 178 |
| 360 | 212 |
| 310 | 164 |
| 280 | 230 |
| 278 | 186 |
| 288 | 162 |
| 288 | 182 |
| 244 | 218 |
| 236 | 170 |
| 200 | |
| 176 |
- State the random variables and the parameters in words.
- State the null and alternative hypotheses and the level of significance.
- State and check the assumptions for the hypothesis test.
- Find the sample statistic, test statistic, and p-value.
- Conclusion
- Interpretation
Solution
1. \(x_{1}\) = Cholesterol level of patients who had a heart attack
\(x_{2}\) = Cholesterol level of healthy individuals
\(\mu_{1}\) = mean cholesterol level of patients who had a heart attack
\(\mu_{2}\) = mean cholesterol level of healthy individuals
2. The normal hypotheses would be
\(\begin{array}{ll}{H_{o} : \mu_{1}=\mu_{2}} & {\text { or } \quad H_{o} : \mu_{1}-\mu_{2}=0} \\ {H_{A} : \mu_{1}>\mu_{2}} & \quad\quad\:{H_{A} : \mu_{1}-\mu_{2}>0} \\ {\alpha=0.01}\end{array}\)
3.
- A random sample of 28 cholesterol levels of patients who had a heart attack is taken. A random sample of 30 cholesterol levels of healthy individuals is taken. The problem does not state if either sample was randomly selected. So this assumption may not be valid.
- The two samples are independent. This is because either they were dealing with patients who had heart attacks or healthy individuals.
- Population of all cholesterol levels of patients who had a heart attack is normally distributed. Population of all cholesterol levels of healthy individuals is normally distributed.
Patients who had heart attacks:
Figure \(\PageIndex{1}\): Histogram of Cholesterol Levels of Patients who had Heart AttacksThis looks somewhat bell shaped.
Figure \(\PageIndex{2}\): Modified Box Plot of Cholesterol Levels of Patients who had Heart AttacksThere are no outliers
Figure \(\PageIndex{3}\): Normal Quantile Plot of Cholesterol Levels of Patients who had Heart AttacksThis looks somewhat linear.
So, the population of all cholesterol levels of patients who had heart attacks is probably somewhat normally distributed.
Healthy individuals:
Figure \(\PageIndex{4}\): Histogram of Cholesterol Levels of Healthy IndividualsThis does not look bell shaped.
Figure \(\PageIndex{5}\): Modified Box Plot of Cholesterol Levels of Healthy IndividualsThere are no outliers.
Figure \(\PageIndex{6}\): Normal Quantile Plot of Cholesterol Levels of Healthy IndividualsThis doesn't look linear.
So, the population of all cholesterol levels of healthy individuals is probably not normally distributed.
This assumption is not valid for the second sample. Since the sample is fairly large, and the t-test is robust, it may not be an issue. However, just realize that the conclusions of the test may not be valid.
4. Sample Statistic:
\(\overline{x}_{1} \approx 252.32, \overline{x}_{2} \approx 193.13, s_{1} \approx 47.0642, s_{2} \approx 22.3000, n_{1}=28, n_{2}=30\)
Test Statistic:
\(t=\dfrac{\left(\overline{x}_{1}-\overline{x}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}}\)
\(=\dfrac{(252.32-193.13)-0}{\sqrt{\dfrac{47.0642^{2}}{28}+\dfrac{22.3000^{2}}{30}}}\)
\(\approx 6.051\)
Degrees of freedom: (the Welch-Satterthwaite equation)
\(A=\dfrac{s_{1}^{2}}{n_{1}}=\dfrac{47.0642^{2}}{28} \approx 79.1085\)
\(B=\dfrac{s_{2}^{2}}{n_{2}}=\dfrac{22.3000^{2}}{30} \approx 16.5763\)
\(d f=\dfrac{(A+B)^{2}}{\dfrac{A^{2}}{n_{1}-1}+\dfrac{B^{2}}{n_{2}-1}}=\dfrac{(79.1085+16.5763)^{2}}{\dfrac{79.1085^{2}}{28-1}+\dfrac{16.5763^{2}}{30-1}} \approx 37.9493\)
p-value:
Using TI-83/84: \(\operatorname{tcdf}(6.051,1 E 99,37.9493) \approx 2.44 \times 10^{-7}\)
Using R: \(1-\mathrm{pt}(6.051,37.9493) \approx 2.44 \times 10^{-7}\)
Using Technology: Using the TI-83/84:
Figure \(\PageIndex{7}\): Setup for 2-SampTTest on TI-83/84 CalculatorNote
The Pooled question on the calculator is for whether you are assuming the variances are equal. Since this assumption is not being made, then the answer to this question is no. Pooled means that you assume the variances are equal and can pool the sample variances together.
Using R: command in general: t.test(variable1, variable2, alternative = "less" or "greater")
For this example, the R command is:
t.test(heartattack, healthy, alternative="greater")
Welch Two Sample t-test
data: heartattack and healthy
t = 6.1452, df = 37.675, p-value = 1.86e-07
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
44.1124 Inf
sample estimates:
mean of x mean of y
253.9286 193.1333
The test statistic is t = 6.1452. The p-value is \(1.86 \times 10^{-7}\)
5. Reject \(H_{o}\) since the p-value < \(\alpha\).
6. This is enough evidence to show that patients who have had heart attacks have higher cholesterol level on average from healthy individuals. (Though do realize that some of assumptions are not valid, so this interpretation may be invalid.)
Example \(\PageIndex{2}\) confidence interval for \(\mu_{1}-\mu_{2}\)
The cholesterol level of patients who had heart attacks was measured two days after the heart attack. The researchers want to see if patients who have heart attacks have higher cholesterol levels over healthy people, so they also measured the cholesterol level of healthy adults who show no signs of heart disease. The data is in Example \(\PageIndex{1}\) ("Cholesterol levels after," 2013). Find a 99% confidence interval for the mean difference in cholesterol levels between heart attack patients and healthy individuals.
- State the random variables and the parameters in words.
- State and check the assumptions for the hypothesis test.
- Find the sample statistic and confidence interval.
- Statistical Interpretation
- Real World Interpretation
Solution
1. These were stated in Example \(\PageIndex{1}\), but are reproduced here for reference.
\(x_{1}\) = Cholesterol level of patients who had a heart attack
\(x_{2}\) = Cholesterol level of healthy individuals
\(\mu_{1}\) = mean cholesterol level of patients who had a heart attack
\(\mu_{2}\) = mean cholesterol level of healthy individuals
2. The assumptions were stated and checked in Example \(\PageIndex{1}\).
3. Sample Statistic:
\(\overline{x}_{1} \approx 252.32, \overline{x_{2}} \approx 193.13, s_{1} \approx 47.0642, s_{2} \approx 22.3000, n_{1}=28, n_{2}=30\)
Test Statistic:
Degrees of freedom: (the Welch–Satterthwaite equation)
\(A=\dfrac{s_{1}^{2}}{n_{1}}=\dfrac{47.0642^{2}}{28} \approx 79.1085\)
\(B=\dfrac{s_{2}^{2}}{n_{2}}=\dfrac{22.3000^{2}}{30} \approx 16.5763\)
\(d f=\dfrac{(A+B)^{2}}{\dfrac{A^{2}}{n_{1}-1}+\dfrac{B^{2}}{n_{2}-1}}=\dfrac{(79.1085+16.5763)^{2}}{\dfrac{79.1085^{2}}{28-1}+\dfrac{16.5763^{2}}{30-1}} \approx 37.9493\)
Since this df is not in the table, round to the nearest whole number.
\(t_{c}=2.712\)
\(E=t_{c} \sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}=2.712 \sqrt{\dfrac{47.0642^{2}}{28}+\dfrac{22.3000^{2}}{30}} \approx 26.53\)
\(\left(\overline{x}_{1}-\overline{x}_{2}\right)-E<\mu_{1}-\mu_{2}<\left(\overline{x}_{1}-\overline{x}_{2}\right)+E\)
\((252.32-193.13)-26.53<\mu_{1}-\mu_{2}<(252.32-193.13)+26.53\)
\(32.66 \mathrm{mg} / \mathrm{dL}<\mu_{1}-\mu_{2}<85.72 \mathrm{mg} / \mathrm{dL}\)
Using Technology:
Using TI-83/84:
Note
The Pooled question on the calculator is for whether you are assuming the variances are equal. Since this assumption is not being made, then the answer to this question is no. Pooled means that you assume the variances are equal and can pool the sample variances together.
Using R: the commands is t.test(variable1, variable2, conf.level=C), where C is in decimal form.
For this example, the command is
t.test(heartattack, healthy, conf.level=.99)
Output:
Welch Two Sample t-test
data: heartattack and healthy
t = 6.1452, df = 37.675, p-value = 3.721e-07
alternative hypothesis: true difference in means is not equal to 0
99 percent confidence interval:
33.95750 87.63298
sample estimates:
mean of x mean of y
253.9286 193.1333
The confidence interval is \(33.96<\mu_{1}-\mu_{2}<87.63\)
4. There is a 99% chance that \(33.96<\mu_{1}-\mu_{2}<87.63\) contains the true difference in means.
5. The mean cholesterol level for patients who had heart attacks is anywhere from 32.66 mg/dL to 85,72 mg/dL more than the mean cholesterol level for healthy patients. (Though do realize that many of assumptions are not valid, so this interpretation may be invalid.)
If you do assume that the variances are equal, that is \(\sigma_{1}^{2}=\sigma_{2}^{2}\), then the test statistic is:
\(t=\dfrac{\left(\overline{x}_{1}-\overline{x}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{s_{p} \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}}\)
where \(s_{p}=\sqrt{\dfrac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{\left(n_{1}-1\right)+\left(n_{2}-1\right)}}\)
\(s_{p}\) = pooled standard deviation
The Degrees of Freedom is: df = \(n_{1}+n_{2}-2\)
The confidence interval if you do assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) has been met, is
\(\left(\overline{x}_{1}-\overline{x}_{2}\right)-E<\mu_{1}-\mu_{2}<\left(\overline{x}_{1}-\overline{x}_{2}\right)+E\)
where \(E=t_{c} s_{p} \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}\)
and \(s_{p}=\sqrt{\dfrac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{\left(n_{1}-1\right)+\left(n_{2}-1\right)}}\)
Degrees of Freedom: df = \(n_{1}+n_{2}-2\)
\(t_{c}\) is the critical value where C = 1 - \(\alpha\)
To show that the variances are equal, just show that the ratio of your sample variances is not unusual (probability is greater than 0.05). In other words, make sure the following is true.
\(P\left(F>s_{1}^{2} / s_{2}^{2}\right) \geq 0.05\left(\text { or } P\left(F>s_{2}^{2} / s_{1}^{2}\right) \geq 0.05\right.\) so that the larger variance is in the numerator). This probability is from an F-distribution. To find the probability on the TI-83/84 calculator use \(\operatorname{Fcdf}\left(s_{1}^{2} / s_{2}^{2}, 1 E 99, n_{1}-1, n_{2}-1\right)\). To find the probability on R, use \(1-\operatorname{pf}\left(s_{1}^{2} / s_{2}^{2}, n_{1}-1, n_{2}-1\right)\).
Note
The F-distribution is very sensitive to the normal distribution. A better test for equal variances is Levene's test, though it is more complicated. It is best to do Levene’s test when using statistical software (such as SPSS or Minitab) to perform the two-sample independent t-test.
Example \(\PageIndex{3}\) hypothesis test for two means
The amount of sodium in beef hotdogs was measured. In addition, the amount of sodium in poultry hotdogs was also measured ("SOCR 012708 id," 2013). The data is in Example \(\PageIndex{2}\). Is there enough evidence to show that beef has less sodium on average than poultry hotdogs? Use a 5% level of significance.
| Sodium in Beef Hotdogs | Sodium in Poultry Hotdogs |
|---|---|
| 495 | 430 |
| 477 | 375 |
| 425 | 396 |
| 322 | 383 |
| 482 | 387 |
| 587 | 542 |
| 370 | 359 |
| 322 | 357 |
| 479 | 528 |
| 375 | 513 |
| 330 | 426 |
| 300 | 513 |
| 386 | 358 |
| 401 | 581 |
| 645 | 588 |
| 440 | 522 |
| 317 | 545 |
| 319 | 430 |
| 298 | 375 |
| 253 | 396 |
- State the random variables and the parameters in words.
- State the null and alternative hypotheses and the level of significance.
- State and check the assumptions for the hypothesis test.
- Find the sample statistic, test statistic, and p-value.
- Conclusion
- Interpretation
Solution
1. \(x_{1}\) = sodium level in beef hotdogs
\(x_{2}\) = sodium level in poultry hotdogs
\(\mu_{1}\) = mean sodium level in beef hotdogs
\(\mu_{2}\) = mean sodium level in poultry hotdogs
2. The normal hypotheses would be
\(\begin{array}{ll}{H_{o} : \mu_{1}=\mu_{2}} & {\text { or } \quad H_{o} : \mu_{1}-\mu_{2}=0} \\ {H_{A} : \mu_{1}<\mu_{2}} & \quad\quad\: {H_{A} : \mu_{1}-\mu_{2}<0} \\ {\alpha=0.05}\end{array}\)
3.
- A random sample of 20 sodium levels in beef hotdogs is taken. A random sample of 20 sodium levels in poultry hotdogs. The problem does not state if either sample was randomly selected. So this assumption may not be valid.
- The two samples are independent since these are different types of hotdogs.
- Population of all sodium levels in beef hotdogs is normally distributed. Population of all sodium levels in poultry hotdogs is normally distributed. Beef Hotdogs:
This looks somewhat bell shaped.
Figure \(\PageIndex{12}\): Modified Box Plot of Sodium Levels in Beef HotdogsThere are no outliers.
Figure \(\PageIndex{13}\): Normal Quantile Plot of Sodium Levels in Beef HotdogsThis looks somewhat linear.
So, the population of all sodium levels in beef hotdogs may be normally distributed.
Poultry Hotdogs:
Figure \(\PageIndex{14}\): Histogram of Sodium Levels in Poultry HotdogsThis does not look bell shaped.
Figure \(\PageIndex{15}\): Modified Box Plot of Sodium Levels in Poultry HotdogsThere are no outliers.
Figure \(\PageIndex{16}\): Normal Quantile Plot of Sodium Levels in Poultry HotdogsThis does not look linear.
So, the population of all sodium levels in poultry hotdogs is probably not normally distributed.
This assumption is not valid. Since the samples are fairly large, and the t-test is robust, it may not be a large issue. However, just realize that the conclusions of the test may not be valid.
d. The population variances are equal, i.e. \(\sigma_{1}^{2}=\sigma_{2}^{2}\).
\(\begin{array}{l}{s_{1} \approx 102.4347} \\ {s_{2} \approx 81.1786} \\ {\dfrac{s_{1}^{2}}{s_{2}^{2}}=\dfrac{102.4347^{2}}{81.1786^{2}} \approx 1.592}\end{array}\)
Using TI-83/84: Fcdf \((1.592,1 E 99,19,19) \approx 0.1597 \geq 0.05\)
Using R: 1 - pf \((1.592,19,19) \approx 0.1597 \geq 0.05\)
So you can say that these variances are equal.
4. Find the sample statistic, test statistic, and p-value
Sample Statistic:
\(\overline{x}_{1}=401.15, \overline{x}_{2}=450.2, s_{1} \approx 102.4347, s_{2} \approx 81.1786, n_{1}=20, n_{2}=20\)
Test Statistic:
The assumption \(\sigma_{1}^{2}=\sigma_{2}^{2}\) has been met, so
\(s_{p}=\sqrt{\dfrac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{\left(n_{1}-1\right)+\left(n_{2}-1\right)}}\)
\(=\sqrt{\dfrac{102.4347^{2} * 19+81.1786^{2} * 19}{(20-1)+(20-1)}}\)
\(\approx 92.4198\)
Though you should try to do the calculations in the problem so you don’t create round off error.
\(t=\dfrac{\left(\overline{x}_{1}-\overline{x}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{s_{P} \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}}\)
\(=\dfrac{(401.15-450.2)-0}{92.4198 \sqrt{\dfrac{1}{20}+\dfrac{1}{20}}}\)
\(\approx-1.678\)
df = 20 + 20 - 2 = 38
p-value:
Using TI-83/84: tcdf \((-1 E 99,-1.678,38) \approx 0.0508\)
Using R: pt \((-1.678,38) \approx 0.0508\)
Using technology to find the t and p-value:
Using TI-83/84:
Figure \(\PageIndex{17}\): Setup for 2-SampTTest on TI-83/84 CalculatorNote
The Pooled question on the calculator is for whether you are using the pooled standard deviation or not. In this example, the pooled standard deviation was used since you are assuming the variances are equal. That is why the answer to the question is Yes.
Using R: the command is t.test(variable1, variable2, alternative="less" or "greater")
For this example, the command is
t.test(beef, poultry, alternative="less", equalvar=TRUE)
Welch Two Sample t-test
data: beef and poultry
t = -1.6783, df = 36.115, p-value = 0.05096
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
-Inf 0.2875363
sample estimates:
mean of x mean of y
401.15 450.20
The t = -1.6783 and the p-value = 0.05096.
5. Fail to reject \(H_{o}\) since the p-value > \(\alpha\).
6. This is not enough evidence to show that beef hotdogs have less sodium than poultry hotdogs. (Though do realize that many of assumptions are not valid, so this interpretation may be invalid.)
Example \(\PageIndex{4}\) confidence interval for \(\mu_{1}-\mu_{2}\)
The amount of sodium in beef hotdogs was measured. In addition, the amount of sodium in poultry hotdogs was also measured ("SOCR 012708 id," 2013). The data is in Example \(\PageIndex{2}\). Find a 95% confidence interval for the mean difference in sodium levels between beef and poultry hotdogs.
- State the random variables and the parameters in words.
- State and check the assumptions for the hypothesis test.
- Find the sample statistic and confidence interval.
- Statistical Interpretation
- Real World Interpretation
Solution
1. These were stated in Example \(\PageIndex{1}\), but are reproduced here for reference.
\(x_{1}\) = sodium level in beef hotdogs
\(x_{2}\) = sodium level in poultry hotdogs
\(\mu_{1}\) = mean sodium level in beef hotdogs
\(\mu_{2}\) = mean sodium level in poultry hotdogs
2. The assumptions were stated and checked in Example \(\PageIndex{3}\).
3. Sample Statistic:
\(\overline{x}_{1}=401.15, \overline{x}_{2}=450.2, s_{1} \approx 102.4347, s_{2} \approx 81.1786, n_{1}=20, n_{2}=20\)
Confidence Interval:
The confidence interval estimate of the difference \(\mu_{1}-\mu_{2}\) is
The assumption \(\sigma_{1}^{2}=\sigma_{2}^{2}\) has been met, so
\(s_{p}=\sqrt{\dfrac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{\left(n_{1}-1\right)+\left(n_{2}-1\right)}}\)
\(=\sqrt{\dfrac{102.4347^{2} * 19+81.1786^{2} * 19}{(20-1)+(20-1)}}\)
\(\approx 92.4198\)
Though you should try to do the calculations in the formula for E so you don’t create round off error.
df = \(=n_{1}+n_{2}-2=20+20-2=38\)
\(t_{c} = 2.024\)
\(E=t_{c} s_{p} \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}\)
\(=2.024(92.4198) \sqrt{\dfrac{1}{20}+\dfrac{1}{20}}\)
\(\approx 59.15\)
\(\left(\overline{x}_{1}-\overline{x}_{2}\right)-E<\mu_{1}-\mu_{2}<\left(\overline{x}_{1}-\overline{x}_{2}\right)+E\)
\((401.15-450.2)-59.15<\mu_{1}-\mu_{2}<(401.15-450.2)+59.15\)
\(-108.20 \mathrm{g}<\mu_{1}-\mu_{2}<10.10 \mathrm{g}\)
Using technology:
Using the TI-83/84:
Figure \(\PageIndex{19}\): Setup for 2-SampTInt on Ti-83/84 CalculatorNote
The Pooled question on the calculator is for whether you are using the pooled standard deviation or not. In this example, the pooled standard deviation was used since you are assuming the variances are equal. That is why the answer to the question is Yes.
Using R: the command is t.test(variable1, variable2, equalvar=TRUE, conf.level=C), where C is in decimal form.
For this example, the command is
t.test(beef, poultry, conf.level=.95, equalvar=TRUE)
Welch Two Sample t-test
data: beef and poultry
t = -1.6783, df = 36.115, p-value = 0.1019
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-108.31592 10.21592
sample estimates:
mean of x mean of y
401.15 450.20
The confidence interval is \(-108.32<\mu_{1}-\mu_{2}<10.22\).
4. There is a 95% chance that \(-108.20 \mathrm{g}<\mu_{1}-\mu_{2}<10.10 \mathrm{g}\) contains the true difference in means.
5. The mean sodium level of beef hotdogs is anywhere from 108.20 g less than the mean sodium level of poultry hotdogs to 10.10 g more. (The negative sign on the lower limit implies that the first mean is less than the second mean. The positive sign on the upper limit implies that the first mean is greater than the second mean.)
Realize that many of assumptions are not valid in this example, so the interpretation may be invalid.
Homework
Exercise \(\PageIndex{1}\)
In each problem show all steps of the hypothesis test or confidence interval. If some of the assumptions are not met, note that the results of the test or interval may not be correct and then continue the process of the hypothesis test or confidence interval. Unless directed by your instructor, do not assume the variances are equal (except in problems 11 through 16).
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The income of males in each state of the United States, including the District of Columbia and Puerto Rico, are given in Example \(\PageIndex{3}\), and the income of females is given in table #9.3.4 ("Median income of," 2013). Is there enough evidence to show that the mean income of males is more than of females? Test at the 1% level.
$42,951 $52,379 $42,544 $37,488 $49,281 $50,987 $60,705 $50,411 $66,760 $40,951 $43,902 $45,494 $41,528 $50,746 $45,183 $43,624 $43,993 $41,612 $46,313 $43,944 $56,708 $60,264 $50,053 $50,580 $40,202 $43,146 $41,635 $42,182 $41,803 $53,033 $60,568 $41,037 $50,388 $41,950 $44,660 $46,176 $41,420 $45,976 $47,956 $22,529 $48,842 $41,464 $40,285 $41,309 $43,160 $47,573 $44,057 $52,805 $53,046 $42,125 $46,214 $51,630 Table \(\PageIndex{3}\): Data of Income for Males $31,862 $40,550 $36,048 $30,752 $41,817 $40,236 $47,476 $40,500 $60,332 $33,823 $35,438 $37,242 $31,238 $39,150 $34,023 $33,745 $33,269 $32,684 $31,844 $34,599 $48,748 $46,185 $36,931 $40,416 $29,548 $33,865 $31,067 $33,424 $35,484 $41,021 $47,155 $32,316 $42,113 $33,459 $32,462 $35,746 $31,274 $36,027 $37,089 $22,117 $41,412 $31,330 $31,329 $33,184 $35,301 $32,843 $38,177 $40,969 $40,993 $29,688 $35,890 $34,381 Table \(\PageIndex{4}\): Data of Income for Females - The income of males in each state of the United States, including the District of Columbia and Puerto Rico, are given in Example \(\PageIndex{3}\), and the income of females is given in Example \(\PageIndex{4}\) ("Median income of," 2013). Compute a 99% confidence interval for the difference in incomes between males and females in the U.S.
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A study was conducted that measured the total brain volume (TBV) (in \(m m^{3}\)) of patients that had schizophrenia and patients that are considered normal. Example \(\PageIndex{5}\) contains the TBV of the normal patients and Example \(\PageIndex{6}\) contains the TBV of schizophrenia patients ("SOCR data oct2009," 2013). Is there enough evidence to show that the patients with schizophrenia have less TBV on average than a patient that is considered normal? Test at the 10% level.
1663407 1583940 1299470 1535137 1431890 1578698 1453510 1650348 1288971 1366346 1326402 1503005 1474790 1317156 1441045 1463498 1650207 1523045 1441636 1432033 1420416 1480171 1360810 1410213 1574808 1502702 1203344 1319737 1688990 1292641 1512571 1635918 Table \(\PageIndex{5}\): Total Brain Volume (in \(\mathrm{mm}^{3}\)) of Normal Patients 1331777 1487886 1066075 1297327 1499983 1861991 1368378 1476891 1443775 1337827 1658258 1588132 1690182 1569413 1177002 1387893 1483763 1688950 1563593 1317885 1420249 1363859 1238979 1286638 1325525 1588573 1476254 1648209 1354054 1354649 1636119 Table \(\PageIndex{6}\): Total Brain Volume (in \(\mathrm{mm}^{3}\)) of Schizophrenia Patients - A study was conducted that measured the total brain volume (TBV) (in \(m m^{3}\)) of patients that had schizophrenia and patients that are considered normal. Example \(\PageIndex{5}\) contains the TBV of the normal patients and Example \(\PageIndex{6}\) contains the TBV of schizophrenia patients ("SOCR data oct2009," 2013). Compute a 90% confidence interval for the difference in TBV of normal patients and patients with Schizophrenia.
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The length of New Zealand (NZ) rivers that travel to the Pacific Ocean are given in Example \(\PageIndex{7}\) and the lengths of NZ rivers that travel to the Tasman Sea are given in Example \(\PageIndex{8}\) ("Length of NZ," 2013). Do the data provide enough evidence to show on average that the rivers that travel to the Pacific Ocean are longer than the rivers that travel to the Tasman Sea? Use a 5% level of significance.
209 48 169 138 64 97 161 95 145 90 121 80 56 64 209 64 72 288 322 Table \(\PageIndex{7}\): Lengths (in km) of NZ Rivers that Flow into the Pacific Ocean 76 64 68 64 37 32 32 51 56 40 64 56 80 121 177 56 80 35 72 72 108 48 Table \(\PageIndex{8}\): Lengths (in km) of NZ Rivers that Flow into the Tasman Sea - The length of New Zealand (NZ) rivers that travel to the Pacific Ocean are given in Example \(\PageIndex{7}\) and the lengths of NZ rivers that travel to the Tasman Sea are given in Example \(\PageIndex{8}\) ("Length of NZ," 2013). Estimate the difference in mean lengths of rivers between rivers in NZ that travel to the Pacific Ocean and ones that travel to the Tasman Sea. Use a 95% confidence level.
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The number of cell phones per 100 residents in countries in Europe is given in Example \(\PageIndex{9}\) for the year 2010. The number of cell phones per 100 residents in countries of the Americas is given in Example \(\PageIndex{10}\) also for the year 2010 ("Population reference bureau," 2013). Is there enough evidence to show that the mean number of cell phones in countries of Europe is more than in countries of the Americas? Test at the 1% level.
100 76 100 130 75 84 112 84 138 133 118 134 126 188 129 93 64 128 124 122 109 121 127 152 96 63 99 95 151 147 123 95 67 67 118 125 110 115 140 115 141 77 98 102 102 112 118 118 54 23 121 126 47 Table \(\PageIndex{9}\): Number of Cell Phones per 100 Residents in Europe 158 117 106 159 53 50 78 66 88 92 42 3 150 72 86 113 50 58 70 109 37 32 85 101 75 69 55 115 95 73 86 157 100 119 81 113 87 105 96 Table \(\PageIndex{10}\): Number of Cell Phones per 100 Residents in the America -
The number of cell phones per 100 residents in countries in Europe is given in Example \(\PageIndex{9}\) for the year 2010. The number of cell phones per 100 residents in countries of the Americas is given in Example \(\PageIndex{10}\) also for the year 2010 ("Population reference bureau," 2013). Find the 98% confidence interval for the difference in mean number of cell phones per 100 residents in Europe and the Americas.
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A vitamin K shot is given to infants soon after birth. Nurses at Northbay Healthcare were involved in a study to see if how they handle the infants could reduce the pain the infants feel ("SOCR data nips," 2013). One of the measurements taken was how long, in seconds, the infant cried after being given the shot. A random sample was taken from the group that was given the shot using conventional methods (Example \(\PageIndex{11}\)), and a random sample was taken from the group that was given the shot where the mother held the infant prior to and during the shot (Example \(\PageIndex{12}\)). Is there enough evidence to show that infants cried less on average when they are held by their mothers than if held using conventional methods? Test at the 5% level.
63 0 2 46 33 33 29 23 11 12 48 15 33 14 51 37 24 70 63 0 73 39 54 52 39 34 30 55 58 18 Table \(\PageIndex{11}\): Crying Time of Infants Given Shots Using Conventional Methods 0 32 20 23 14 19 60 59 64 64 72 50 44 14 10 58 19 41 17 5 36 73 19 46 9 43 73 27 25 18 Table \(\PageIndex{12}\): Crying Time of Infants Given Shots Using New Methods -
A vitamin K shot is given to infants soon after birth. Nurses at Northbay Healthcare were involved in a study to see if how they handle the infants could reduce the pain the infants feel ("SOCR data nips," 2013). One of the measurements taken was how long, in seconds, the infant cried after being given the shot. A random sample was taken from the group that was given the shot using conventional methods (Example \(\PageIndex{11}\)), and a random sample was taken from the group that was given the shot where the mother held the infant prior to and during the shot (Example \(\PageIndex{12}\)). Calculate a 95% confidence interval for the mean difference in mean crying time after being given a vitamin K shot between infants held using conventional methods and infants held by their mothers.
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Redo problem 1 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in Example \(\PageIndex{3}\)).
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Redo problem 2 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in Example \(\PageIndex{3}\)).
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Redo problem 7 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in Example \(\PageIndex{3}\)).
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Redo problem 8 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in Example \(\PageIndex{3}\)).
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Redo problem 9 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in Example \(\PageIndex{3}\)).
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Redo problem 10 testing for the assumption of equal variances and then use the formula that utilizes the assumption of equal variances (follow the procedure in Example \(\PageIndex{3}\)).
- Answer
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For all hypothesis tests, just the conclusion is given. For all confidence intervals, just the interval using technology is given. See solution for the entire answer.
- Reject Ho
- \(\$ 65443.80<\mu_{1}-\mu_{2}<\$ 13340.80\)
- Fail to reject Ho
- \(-51564.6 \mathrm{mm}^{3}<\mu_{1}-\mu_{2}<75656.6 \mathrm{mm}^{3}\)
- Reject Ho
- \(23.2818 \mathrm{km}<\mu_{1}-\mu_{2}<103.67 \mathrm{km}\)
- Reject Ho
- \(4.3641<\mu_{1}-\mu_{2}<37.5276\)
- Fail to reject Ho
- \(-10.9726 \mathrm{s}<\mu_{1}-\mu_{2}<11.3059 \mathrm{s}\)
- Reject Ho
- \(\$ 6544.98<\mu_{1}-\mu_{2}<\$ 13339.60\)
- Reject Ho
- \(4.8267<\mu_{1}-\mu_{2}<37.0649\)
- Fail to reject Ho
- \(-10.9713 \mathrm{s}<\mu_{1}-\mu_{2}<11.3047 \mathrm{s}\)