5.2: Drag Forces
Learning Objectives
By the end of this section, you will be able to:
- Express mathematically the drag force.
- Discuss the applications of drag force.
- Define terminal velocity.
- Determine the terminal velocity given mass.
Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid). You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air—you have decreased the area of your hand that faces the direction of motion. Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. For most large objects such as bicyclists, cars, and baseballs not moving too slowly, the magnitude of the drag force \(F_D\) is found to be proportional to the square of the speed of the object. We can write this relationship mathematically as \(F_D\propto v^2 \). When taking into account other factors, this relationship becomes
\[F_D = \dfrac{1}{2} C_{\rho} Av^2, \label{5.3.1}\]
where \(C\) is the drag coefficient, \(A\) is the area of the object facing the fluid, and \(\rho\) is the density of the fluid. (Recall that density is mass per unit volume.) This equation can also be written in a more generalized fashion as \(F_D = bv^2, \) where \(b\) is a constant equivalent to \(0.5 C\rho A.\) We have set the exponent for these equations as 2 because, when an object is moving at high velocity through air, the magnitude of the drag force is proportional to the square of the speed. As we shall see in a few pages on fluid dynamics, for small particles moving at low speeds in a fluid, the exponent is equal to 1.
Definition: DRAG FORCE
Drag Force \(F_D\) is found to be proportional to the square of the speed of the object. Mathematically \[F_D\propto v^2 \]
\[F_D = \dfrac{1}{2} C_{\rho} Av^2, \] where \(C\) is the drag coefficient \(A\) is the area of the object facing the fluid, and \(\rho\) is the density of the fluid.
Athletes as well as car designers seek to reduce the drag force to lower their race times. (See Figure). “Aerodynamic” shaping of an automobile can reduce the drag force and so increase a car’s gas mileage.
The value of the drag coefficient, \(C\) is determined empirically, usually with the use of a wind tunnel. (See Figure).
The drag coefficient can depend upon velocity, but we will assume that it is a constant here. Table lists some typical drag coefficients for a variety of objects. Notice that the drag coefficient is a dimensionless quantity. At highway speeds, over 50% of the power of a car is used to overcome air drag. The most fuel-efficient cruising speed is about 70–80 km/h (about 45–50 mi/h). For this reason, during the 1970s oil crisis in the United States, maximum speeds on highways were set at about 90 km/h (55 mi/h).
| Object | C |
|---|---|
| Airfoil | 0.05 |
| Toyota Camry | 0.28 |
| Ford Focus | 0.32 |
| Honda Civic | 0.36 |
| Ferrari Testarossa | 0.37 |
| Dodge Ram pickup | 0.43 |
| Sphere | 0.45 |
| Hummer H2 SUV | 0.64 |
| Skydiver (feet first) | 0.70 |
| Bicycle | 0.90 |
| Skydiver (horizontal) | 1.0 |
| Circular flat plate | 1.12 |
Substantial research is under way in the sporting world to minimize drag.  Bicycle racers and some swimmers and runners wear full bodysuits. Australian Cathy Freeman wore a full body suit in the 2000 Sydney Olympics, and won the gold medal for the 400 m race. Many swimmers in the 2008 Beijing Olympics wore (Speedo) body suits; it might have made a difference in breaking many world records (See Figure). Most elite swimmers (and cyclists) shave their body hair. Such innovations can have the effect of slicing away milliseconds in a race, sometimes making the difference between a gold and a silver medal. One consequence is that careful and precise guidelines must be continuously developed to maintain the integrity of the sport.
Some interesting situations connected to Newton’s second law occur when considering the effects of drag forces upon a moving object. For instance, consider a skydiver falling through air under the influence of gravity. The two forces acting on him are the force of gravity and the drag force (ignoring the buoyant force). The downward force of gravity remains constant regardless of the velocity at which the person is moving. However, as the person’s velocity increases, the magnitude of the drag force increases until the magnitude of the drag force is equal to the gravitational force, thus producing a net force of zero. A zero net force means that there is no acceleration, as given by Newton’s second law. At this point, the person’s velocity remains constant and we say that the person has reached his terminal velocity \((v_t)\). Since \(F_D\) is proportional to the speed, a heavier skydiver must go faster for \(F_D\) to equal his weight. Let’s see how this works out more quantitatively.
At the terminal velocity,
\[ F_{net} = mg - F_D = ma = 0\]
Thus,
\[ mg = F_D. \]
Using the equation for drag force, we have \[ mg = \dfrac{1}{2}\rho C Av^2. \]
Solving for the velocity, we obtain \[ v = \sqrt{\dfrac{2mg}{\rho C A}} \]
Assume the density of air is \(\rho = 1.21 \, kg/m^3. \) A 75-kg skydiver descending head first will have an area approximately \(A = 0.18 \, m^2 \) and a drag coefficient of approximately \(C = 0.70. \) We find that
\[v = \sqrt{\dfrac{2(75 \, kg)(9.80 \, m/s^2)}{(1.21 \, kg/m^3)(0.70)(0.18 \, m^2)}} \]
\[ = 98 \, m/s \]
\[ = 350 \, km/h. \]
This means a skydiver with a mass of 75 kg achieves a maximum terminal velocity of about 350 km/h while traveling in a pike (head first) position, minimizing the area and his drag. In a spread-eagle position, that terminal velocity may decrease to about 200 km/h as the area increases. This terminal velocity becomes much smaller after the parachute opens.
TAKE-HOME EXPERIMENT
This interesting activity examines the effect of weight upon terminal velocity. Gather together some nested coffee filters. Leaving them in their original shape, measure the time it takes for one, two, three, four, and five nested filters to fall to the floor from the same height (roughly 2 m). (Note that, due to the way the filters are nested, drag is constant and only mass varies.) They obtain terminal velocity quite quickly, so find this velocity as a function of mass. Plot the terminal velocity \(v\) versus mass. Also plot \(v^2\) versus mass. Which of these relationships is more linear? What can you conclude from these graphs?
Example \(\PageIndex{1}\): The terminal Velocity
Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position.
Strategy
At terminal velocity, \(F_{net} = 0 \) Thus the drag force on the skydiver must equal the force of gravity (the person’s weight). Using the equation of drag force, we find \(mg = \frac{1}{2} \rho C A v^2\).
Thus the terminal velocity \(v_t\) can be written as
\[v_t = \sqrt{\dfrac{2mg}{\rho CA}} \]
Solution
All quantities are known except the person’s projected area. This is an adult (85 kg) falling spread eagle. We can estimate the frontal area as \[ A = (2 \, m)(0.35 \, m) = 0.70 \, m^2 \]
Using our equation for \(v_t,\) we find that
\[ v_t = \sqrt{\dfrac{2(85\space kg)(9.80 \, m/s^2)}{91.21 \, kg/m^3)(1.0)(0.70 \, m^2)}} \]
\[ = 44 \, m/s\]
Discussion
This result is consistent with the value for \(v_t\) mentioned earlier. The 75-kg skydiver going feet first had a \(v = 98 \, m/s.\) He weighed less but had a smaller frontal area and so a smaller drag due to the air.
The size of the object that is falling through air presents another interesting application of air drag. If you fall from a 5-m high branch of a tree, you will likely get hurt—possibly fracturing a bone. However, a small squirrel does this all the time, without getting hurt. You don’t reach a terminal velocity in such a short distance, but the squirrel does.
The following interesting quote on animal size and terminal velocity is from a 1928 essay by a British biologist, J.B.S. Haldane, titled “On Being the Right Size.”
To the mouse and any smaller animal, [gravity] presents practically no dangers. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, and a horse splashes. For the resistance presented to movement by the air is proportional to the surface of the moving object. Divide an animal’s length, breadth, and height each by ten; its weight is reduced to a thousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively ten times greater than the driving force.
The above quadratic dependence of air drag upon velocity does not hold if the object is very small, is going very slow, or is in a denser medium than air. Then we find that the drag force is proportional just to the velocity. This relationship is given by Stokes’ law , which states that \[ F_S = 6 \pi r \eta v \] where \(r\) is the radius of the object, \(\eta\) is the viscosity of the fluid and \(v\) is the object's velocity.
Definition: STOKES LAW
\[ F_S = 6 \pi r \eta v \]
where \(r\) is the radius of the object, \(\eta\) is the viscosity of the fluid and \(v\) is the object's velocity.
Good examples of this law are provided by microorganisms, pollen, and dust particles. Because each of these objects is so small, we find that many of these objects travel unaided only at a constant (terminal) velocity. Terminal velocities for bacteria (size about \(1 \, \mu m\)) can be about \(2 \, m/s.\) To move at a greater speed, many bacteria swim using flagella (organelles shaped like little tails) that are powered by little motors embedded in the cell. Sediment in a lake can move at a greater terminal velocity (about \(5 \, \mu m/s,\)) so it can take days to reach the bottom of the lake after being deposited on the surface.
If we compare animals living on land with those in water, you can see how drag has influenced evolution. Fishes, dolphins, and even massive whales are streamlined in shape to reduce drag forces. Birds are streamlined and migratory species that fly large distances often have particular features such as long necks. Flocks of birds fly in the shape of a spear head as the flock forms a streamlined pattern (see Figure). In humans, one important example of streamlining is the shape of sperm, which need to be efficient in their use of energy.
GALILEO’S EXPERIMENT
Galileo is said to have dropped two objects of different masses from the Tower of Pisa. He measured how long it took each to reach the ground. Since stopwatches weren’t readily available, how do you think he measured their fall time? If the objects were the same size, but with different masses, what do you think he should have observed? Would this result be different if done on the Moon?
Summary
- Drag forces acting on an object moving in a fluid oppose the motion. For larger objects (such as a baseball) moving at a velocity \(v\) in air, the drag force is given by \[ F_D = \dfrac{1}{2}C \rho A v^2,\] where \(C\) is the drag coefficient (typical values are given in Table), \(A\) is the area of the object facing the fluid, and \(\rho\) is the fluid density.
- For small objects (such as a bacterium) moving in a denser medium (such as water), the drag force is given by Stokes’ law,


9.3: The F-Distribution and the F-Ratio
The distribution used for the hypothesis test is a new one. It is called the F -distribution , invented by George Snedecor but named in honor of Sir Ronald Fisher, an English statistician. The \(F\)-statistic is a ratio (a fraction). There are two sets of degrees of freedom; one for the numerator and one for the denominator.
Here are some facts about the F -distribution.
- The curve is not symmetrical but skewed to the right.
- There is a different curve for each set of degrees of freedom.
- The \(F\)-statistic is greater than or equal to zero.
- As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal as can be seen in the two figures below. Notice that with more degrees of freedom as shown in the figure below, the curve is more closely approaching the normal distribution, but remember that the \(F\) cannot ever be less than zero so the distribution does not have a tail that goes to infinity on the left as the normal distribution does.
- Other uses for the \(F\)-distribution include comparing two variances and two-way Analysis of Variance. Two-Way Analysis is beyond the scope of this chapter.
For example, if \(F\) follows an \(F\) distribution and the number of degrees of freedom for the numerator is four, and the number of degrees of freedom for the denominator is ten, then \(F \sim F_{4,10}\).
To calculate the F -ratio , two estimates of the variance are made.
- Variance between samples : An estimate of \(\sigma^2\) that is the variance of the sample means multiplied by n . (If the samples are different sizes, the variance between samples is thus weighted to account for the different sample sizes.) This variance is also called variation due to treatment or group, or explained variation. It may also be called "factor" variance . \[M S_{\text {Between }}=\frac{S S_{\text {Between }}}{d f_{\text {Between }}}=\frac{n_{1}\left(\bar{x}_{1}-\bar{x}\right)^{2}+\cdots+n_{g}\left(\bar{x}_{g}-\bar{x}\right)^{2}}{g-1}\nonumber\]
- Variance within samples : An estimate of \(\sigma^2\) that is the average of the sample variances. When the sample sizes are different, the variance within samples is weighted. This variance is also called the variation due to error, or unexplained variation. \[M S_{\text {Within }}=\frac{S S_{\text {Within }}}{d f_{\text {Within }}}=\frac{\left(n_{1}-1\right) s_{1}^{2}+\cdots+\left(n_{g}-1\right) s_{g}^{2}}{n-g}\nonumber\]
- \(SS_{between}\) = the sum of squares that represents the variation among the different samples
- \(SS_{within}\) = the sum of squares that represents the variation within samples that is due to chance.
To find a "sum of squares" means to add together squared quantities that, in some cases, may be weighted. We used sum of squares to calculate the sample variance and the sample standard deviation.
MS in these equations means " mean square ". \(MS_{between}\) is the variance between groups, and \(MS_{within}\) is the variance within groups.
The one-way ANOVA test depends on the fact that \(M S_{between}\) can be influenced by population differences among means of the several groups. Since \(M S_{w i t h i n}\) compares values of each group to its own group mean, the fact that group means might be different does not affect \(M S_{w i t h i n}\).
The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions.
Note
The null hypothesis says that all the group population means are equal. The hypothesis of equal means implies that the populations have the same normal distribution, because it is assumed that the populations are normal and that they have equal variances.
F -Ratio or F -Statistic
\[F_{obs}=\frac{M S_{\text { between }}}{M S_{\text { within }}}\nonumber\]
If \(M S_{between}\) and \(M S_{w i t h i n}\) estimate the same value (following the belief that \(H_0\) is true), then the \(F\)-ratio should be approximately equal to one. Mostly, just sampling errors would contribute to variations away from one. As it turns out, \(M S_{between}\) consists of the population variance plus a variance produced from the differences between the samples. \(M S_{w i t h i n}\) is an estimate of the population variance. If the null hypothesis is false, \(M S_{between}\) will generally be larger than \(M S_{w i t h i n}\), and the \(F\)-ratio will be larger than one. However, if the population effect is small, it is not unlikely that \(M S_{w i t h i n}\) will be larger in a given sample.
To determine the critical value, we have to find \(F_{\alpha,df1,df2}\). See Appendix A for the \(F\) table. This table has \(F\) values for various levels of significance on different pages, as indicated in the first row of each page's table. To find the critical value, choose the page/table with the desired significance level, and follow down and across to find the critical value at the intersection of the two different degrees of freedom. The \(F\) distribution has two different degrees of freedom, one associated with the numerator, \(df_1\), and one associated with the denominator, \(df_2\). The degrees of freedom in the numerator is \(g-1\), where \(g\) is the number of groups; and the degrees of freedom in the denominator is \(n-g\), where \(n\) is the total sample size across all groups. \(F_{\alpha,df1,df2}\) will give the critical value on the upper end of the \(F\) distribution.
Data are typically put into a table for easy viewing. One-way ANOVA results are often displayed in this manner by computer software.
|
Source of variation |
Sum of squares (\(SS\)) |
Degrees of freedom (\(df\)) |
Mean square (\(MS\)) |
\(F\) |
|---|---|---|---|---|
|
Factor
|
\(SS\)(Factor) |
\(g – 1\) |
\(MS(Factor) =
|
\(F = MS(Factor)/MS(Error)\) |
|
Error
|
\(SS\)(Error) |
\(n – g\) |
\(MS(Error) =
|
|
|
Total |
\(SS\)(Total) |
\(n – 1\) |
Example \(\PageIndex{1}\)
Three different diet plans are to be tested for mean weight loss. The entries in the table are the weight losses for the different plans. The one-way ANOVA results are shown in the table below.
|
Plan 1: \(n_1 = 4\) |
Plan 2: \(n_2 = 3\) |
Plan 3: \(n_3 = 3\) |
|---|---|---|
|
5 |
3.5 |
8 |
|
4 |
7 |
4 |
|
4 |
4.5 |
3 |
|
3 |
\(H_{0} : \mu_{1}=\mu_{2}=\mu_{3}\)
\(H_a\): At least two of the means (\(\mu_{1}, \mu_{2}, \mu_{3}\)) are not equal.
Means: \(\bar{x}_{1}=\frac{16}{4}=4, \bar{x}_{2}=\frac{15}{3}=5, \bar{x}_{3}=\frac{15}{3}=5, \bar{x}=\frac{46}{10}=4.6\)
Variances: \(s_{1}^2=0.67, s_{2}^2=3.16, s_{3}^2=7\)
\(SS_{Between}=4(4-4.6)^2+3(5-4.6)^2+3(5-4.6)^2=2.4\)
\(SS_{Within}=(4-1)0.67+(3-1)3.16+(3-1)7=22.33\)
\(df_{Between}=3-1=2\)
\(df_{Within}=10-3=7\)
\(MS_{Between}=\frac{SS_{Between}}{df_{Between}}=\frac{2.4}{2}=1.2\)
\(MS_{Within}=\frac{SS_{Within}}{df_{Within}}=\frac{22.33}{7}=3.19\)
\(F_{obs}=\frac{MS_{Between}}{MS_{Within}}=\frac{1.2}{3.19}=0.38\)
|
Source of variation |
Sum of squares (\(SS\)) |
Degrees of freedom (\(df\)) |
Mean square (\(MS\)) |
\(F\) |
|---|---|---|---|---|
|
Factor
|
2.4 |
2 |
1.2 |
0.38 |
|
Error
|
22.33 |
7 |
3.19 |
|
|
Total |
2.4 + 22.33 = 24.73 |
2 + 7 = 9 |
For \(F_{\alpha=.05,2,7}\), \(F_{critical}\) = 4.74. \(F_{obs}\) does not exceed this \(F_{critical}\), so we cannot reject \(H_{0}\); we don't have sufficient evidence to say that the three different diet plans result in different average amounts of weight loss.
Exercise \(\PageIndex{1}\)
As part of an experiment to see how different types of soil cover would affect slicing tomato production, Marist College students grew tomato plants under different soil cover conditions. Groups of three plants each had one of the following treatments
- bare soil
- a commercial ground cover
- black plastic
- straw
- compost
All plants grew under the same conditions and were the same variety. Students recorded the weight (in grams) of tomatoes produced by each of the n = 15 plants:
|
Bare: \(n_1 = 3\) |
Ground Cover: \(n_2 = 3\) |
Plastic: \(n_3 = 3\) |
Straw: \(n_4 = 3\) |
Compost: \(n_5 = 3\) |
|---|---|---|---|---|
|
2,625 |
5,348 |
6,583 |
7,285 |
6,277 |
|
2,997 |
5,682 |
8,560 |
6,897 |
7,818 |
|
4,915 |
5,482 |
3,830 |
9,230 |
8,677 |
Create the one-way ANOVA table.
The one-way ANOVA hypothesis test is always right-tailed because larger \(F\)-values are way out in the right tail of the F -distribution curve.
Example \(\PageIndex{2}\)
Let’s return to the slicing tomato exercise we were working on in the previous exercise. The means of the tomato yields under the five mulching conditions are represented by \(\mu_{1}, \mu_{2}, \mu_{3}, \mu_{4}, \mu_{5}\). We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5%, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest.
Answer
-
The null and alternative hypotheses are:
\(H_{0} : \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}=\mu_{5}\)
\(H_{a} : \) at least two of the population means differ
The one-way ANOVA results are shown in the table below.
Table \(\PageIndex{5}\) Source of variation
Sum of squares (\(SS\))
Degrees of freedom (\(df\))
Mean square (\(MS\))
\(F\)
Factor (Between)
36,648,561
\(5 – 1 = 4\)
\(\frac{36,648,561}{4}=9,162,140\)
\(\frac{9,162,140}{2,044,672.6}=4.481\)
Error (Within)
20,446,726
\(15 – 5 = 10\)
\(\frac{20,446,726}{10}=2,044,672.6\)
Total
57,095,287
\(15 – 1 = 14\)
Distribution for the test: \(F_{\alpha=.05,4,10}\)
\(df_{num} = 5 – 1 = 4\)
\(df_{denom} = 15 – 5 = 10\)
\(F_{critical}=3.48\)
Test statistic: \(F_{obs} = 4.481\)
Make a decision: Since \(F_{obs} > F_{critical}\), we reject \(H_0\).
Conclusion: At the 5% significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least one of mulches led to a different mean yield than the others.
Exercise \(\PageIndex{2}\)
MRSA, or Staphylococcus aureus , can cause a serious bacterial infections in hospital patients. The table below shows various colony counts from different patients who may or may not have MRSA. The data from the table is plotted in the figure below.
|
Conc = 0.6 |
Conc = 0.8 |
Conc = 1.0 |
Conc = 1.2 |
Conc = 1.4 |
|---|---|---|---|---|
|
9 |
16 |
22 |
30 |
27 |
|
66 |
93 |
147 |
199 |
168 |
|
98 |
82 |
120 |
148 |
132 |
Plot of the data for the different concentrations:
Test whether the mean number of colonies are the same or are different. Construct the ANOVA table, find the p -value, and state your conclusion. Use a 5% significance level.
Example \(\PageIndex{3}\)
Four sororities took a random sample of their members regarding their grade means for the past term. The results are shown in the table below.
|
Sorority 1 |
Sorority 2 |
Sorority 3 |
Sorority 4 |
|---|---|---|---|
|
2.17 |
2.63 |
2.63 |
3.79 |
|
1.85 |
1.77 |
3.78 |
3.45 |
|
2.83 |
3.25 |
4.00 |
3.08 |
|
1.69 |
1.86 |
2.55 |
2.26 |
|
3.33 |
2.21 |
2.45 |
3.18 |
Using a significance level of 1%, is there a difference in mean grades among the sororities?
Answer
-
Note
This is an example of a balanced design , because each group (i.e., sorority) has the same number of observations.
\(H_{0} : \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}\)
\(H_a\): Not all of the means \(\mu_{1}, \mu_{2}, \mu_{3}, \mu_{4}\) are equal.
Distribution for the test: \(F_{\alpha=.01,3,16}\)
where \(g = 4\) groups and \(n = 20\) samples in total
\(df_{num}= g – 1 = 4 – 1 = 3\)
\(df_{denom} = n – g = 20 – 4 = 16\)
\(F_{critical}=5.29\)
Calculate the test statistic: \(F_{obs} = 2.23\)
Graph:
Make a decision: Since \(F_{obs} < F_{critical}\), we reject \(H_0\).
Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities.
Exercise \(\PageIndex{3}\)
Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown here in the table.
|
Basketball |
Baseball |
Hockey |
Lacrosse |
|---|---|---|---|
|
3.6 |
2.1 |
4.0 |
2.0 |
|
2.9 |
2.6 |
2.0 |
3.6 |
|
2.5 |
3.9 |
2.6 |
3.9 |
|
3.3 |
3.1 |
3.2 |
2.7 |
|
3.8 |
3.4 |
3.2 |
2.5 |
Use a significance level of 5%, and determine if there is a difference in GPA among the teams.
Example \(\PageIndex{4}\)
A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data (in inches) in the table shown here.
|
Tommy's plants |
Tara's plants |
Nick's plants |
|---|---|---|
|
24 |
25 |
23 |
|
21 |
31 |
27 |
|
23 |
23 |
22 |
|
30 |
20 |
30 |
|
23 |
28 |
20 |
Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 5% level of significance.
Answer
-
\(H_{0} : \mu_{Tommy}=\mu_{Tara}=\mu_{Nick}\)
\(H_a\): At least two of the means (\(\mu_{Tommy}, \mu_{Tara}, \mu_{Nick}\)) are not equal.
Means: \(\bar{x}_{Tommy}=\frac{121}{5}=24.2, \bar{x}_{Tara}=\frac{127}{5}=25.4, \bar{x}_{Nick}=\frac{122}{5}=24.4, \bar{x}=\frac{370}{15}=24.7\)
Variances: \(s_{Tommy}^2=11.7, s_{Tara}^2=18.3, s_{Nick}^2=16.3\)
\(SS_{Between}=5(24.2-24.7)^2+5(25.4-24.7)^2+5(24.4-24.7)^2=4.15\)
\(SS_{Within}=(5-1)11.7+(5-1)18.3+(5-1)16.3=185.2\)
\(df_{Between}=3-1=2\)
\(df_{Within}=15-3=12\)
\(MS_{Between}=\frac{SS_{Between}}{df_{Between}}=\frac{4.15}{2}=2.075\)
\(MS_{Within}=\frac{SS_{Within}}{df_{Within}}=\frac{185.2}{12}=15.43\)
\(F_{obs}=\frac{MS_{Between}}{MS_{Within}}=\frac{2.075}{15.43}=0.13\)
Table \(\PageIndex{10}\) Source of variation
Sum of squares (\(SS\))
Degrees of freedom (\(df\))
Mean square (\(MS\))
\(F\)
Factor
(Between)4.15
2
2.075
0.13
Error
(Within)185.2
12
15.43
Total
189.35
14
For \(F_{\alpha=.05,2,12}\), \(F_{critical}\) = 3.89. \(F_{obs}\) does not exceed this \(F_{critical}\), so we cannot reject \(H_{0}\). With a 5% level of significance, from this sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.


18.5: Management Entrepreneurship Skills for Technology and Innovation
- How and why do entrepreneurs need to develop MTI skills?
Entrepreneurial activities in the marketplace often signal newness in a product or process. For an entrepreneurial firm, the value proposition is a key factor. The value proposition answers the following two questions:
- How will the firm make money on the product and/or services offered?
- How will the firm be positioned in the marketplace?
New business entities (a type of entrepreneurial activity) are usually more flexible and agile in the marketplace. The entrepreneur is very dedicated to the success of the firm because the new business is the “baby” of the entrepreneur. The starting of a new business is the riskiest approach for introducing new products and processes. The failure rate for entrepreneurs is high. Because the firm is new, there is usually very little slack in resources available—money is tight, labor is limited, and time is fleeting. Therefore, entrepreneurial activities are most successful when the lean start-up process can be used. It is applicable where development costs are low and revisions are not very costly. One reason there are many startups in the development of applications for mobile phones is that the costs are low and improvement of the product (once successful) is relatively easy. Entrepreneurs have the ability to adapt their plans incrementally, especially when using lean start-up methods. 5 For entrepreneurs, the capabilities to sense, seize, and transform can be an advantage if they stay agile and avoid over-committing to a course of action. 6 Entrepreneurs, by definition, are more agile than more-established organizations.
This agility is true within large firms that want to continue to be entrepreneurial in their activities. Firms such as Google and 3M allow their employees to work on innovative projects during their working hours. Google modeled its policy of allowing employees time to explore other potential products and processes after 3M’s longtime policy. Both of these firms are known as innovative firms because they encourage employees to look for and test innovative and valuable propositions. The flexibility allowed employees gives both Google and 3M the agility to find new ways of doing things.
- In the beginning, Acer was very entrepreneurial. However, the firm realized that if it was to continue to grow, it needed to develop some structures and processes. What adjustments did Acer make to become a global firm?


10.5: Performance Management
- How do performance management practices impact company performance?
Performance management practices and processes are among the most important that human resources manages, yet they are also among the most contentious processes in an organization. Many people view performance management as a human resources role and believe that it is in some parallel path with the business. On the contrary, for the process to be successful, it should not only be human resources that is responsible for driving performance. For the (typically) annual performance management process, human resources and line management should partner on the implementation and ongoing communication of the process. Although HR is responsible for creating and facilitating the performance management processes, it is the organizational managers that need to strongly support the process and communicate the linkage of performance management to overall organizational goals and performance.
In my experience, it was helpful when business leadership emphasized that performance management isn’t a human resources process—it is a mission-critical business process. If a business manager can’t track and drive performance at the individual level, then the overall organization won’t know how it’s tracking on overall organizational goals. Before discussing the state of performance management in the workplace today, it is important to understand the origin of performance management. Performance management began as a simple tool to drive accountability (as it still does) but has evolved more recently into a tool used for employee development. Performance management can be tracked back to the U.S. military’s “merit rating” system, which was created during World War I to identify poor performers for discharge or transfer (Capelli & Tavis, 2016).
After World War II, about 60% of all U.S. companies were using a performance appraisal process, and by the 1960s, nearly 90% of all U.S. companies were using one. Although the rules around job seniority determined pay increases and promotions for the unionized worker population, strong performance management scores meant good advancement prospects for managers. In the beginning, the notion of using this type of system to improve performance was more of an afterthought and not the main purpose. By the 1960s, when we started to see a shortage of managerial talent, companies began to use performance systems to develop employees into supervisors, and managers into executives.
In 1981, when Jack Welch became CEO of General Electric, he championed the forced-ranking system—another military creation. He did this to address the long-standing concern that supervisors failed to label real differences in performance (Goler et al., 2016). GE utilized this performance management system to shed the people at the bottom. They equated performance with people’s inherent capabilities and ignored their potential to grow. People were categorized as “A” players (to be rewarded), “B” players (to be accommodated), and “C” players (to be dismissed). In the GE system, development was reserved for the “A” players—and those with high potential were chosen to advance to senior positions. Since the days of GE’s forced ranking, many companies have implemented a similar forced-ranking system, but many have backed away from the practice. After Jack Welch retired, GE backed away from the practice as well. Companies, GE included, saw that it negatively fostered internal competition and undermined collaboration and teamwork and thus decided to drop forced ranking from their performance management processes.
Most people agree, in theory, that performance management is important. What people may not agree upon is how performance management should be implemented. As the dissatisfaction with performance management processes began to increase, some companies began to change the way they thought about performance. In 2001, an “Agile Manifesto” was developed by software developers and “emphasized principles of collaboration, self-organization, self-direction, and regular reflection on how to work more effectively, with the aim of prototyping more quickly and responding in real-time to customer feedback and changes in requirements.” (Goler et al., 2016). The impact on performance management was clear, and companies started to think about performance management processes that were less cumbersome, incorporated frequent feedback, and delivered performance impacts.
In a recent public survey by Deloitte Services, 58% of executives surveyed believed that their current performance management approach drives neither employee engagement nor high performance. They need something more nimble, real-time, and individualized—and focused on fueling performance in the future rather than assessing it in the past (Buckingham & Goodall, 2015). In light of this study, Deloitte became one of the companies that has recently sought to redesign their performance processes. As part of their “radical redesign,” they seek to see performance at the individual level, and thus they ask team leaders about their own future actions and decisions with respect to each individual. They ask leaders what they’d do with their team members, not what they think of them (Buckingham & Goodall, 2015). The four questions that Deloitte asks of its managers are as follows:
- Given what I know of this person’s performance, and if it were my money, I would award this person the highest possible compensation increase and bonus.
- Given what I know of this person’s performance, I would always want him or her on my team.
- This person is at risk for low performance.
- This person is ready for promotion today.
Although there has been some discussion over the last several years about some companies wanting to drop performance appraisals completely, most of the research seems to support that the total absence of performance management doesn’t help either. A recent global survey by CEB Global reports that more than 9,000 managers and employees think that not having performance evaluations is worse than having them (Goler et al., 2016). Their findings indicate that even though every organization has people who are unhappy with their bonuses or disappointed that they weren’t promoted, research shows that employees are more willing to accept an undesirable outcome when the process is fair. The key question really becomes,
How can HR help the business create a process to fairly evaluate performance and enhance employee development while not burdening the business with undue bureaucracy and non-value-added activities?
Global versus Local HR
Multinational companies are always challenged to determine the balance between global and local needs when creating a human resource management strategy. Some large companies weigh heavily on the side of centralization, with very few local deviations from the global strategy. Others may allow more localization of processes and decision-making if there are very specific local cultural needs that must be addressed. In either case, companies are well-served by maintaining global standards while also allowing for local market adaptation in the human resources areas where it makes the most sense.
According to the MIT Sloan Management Review (2012) article “Six Principles of Effective Global Talent Management," most multinational companies introduce global performance standards, competency profiles, and performance management tools and processes. These are the human resources areas that are most closely linked to the overall strategies and goals, and thus remain at the global level. Those HR processes that are not perceived as being as closely linked to the strategy and that may need to have local market inputs include processes such as training and compensation. Hiring practices may also need to be locally adapted, due to country-specific labor laws and challenges. One caveat, however, is that a company may limit itself in terms of its global talent management if it has too many country-specific adaptations to hiring, assessment, and development processes for top talent. It is important that the company takes a global approach to talent management so that cross-learning opportunities and cross-cultural development opportunities can take place.
One of the most important aspects of global talent management is that a company can break down silos and pollinate the business with talented employees from around the globe. Some companies even have global leadership programs that bring together high-potential leaders from across the organization to build camaraderie, share knowledge, and engage in learning. Others have created rotational programs for leaders to be able to experience new roles in other cultures in order to build their personal resumes and cultural intelligence. Human resources can have an enormous impact on the company’s ability to harness the power of a global talent pool when they create a global network for talent while also balancing this with the requirements of the local market.
Discussion Questions
- What are the challenges of a company developing a different competency model or performance management process for each of its local offices?
- Why might compensation programs and hiring practices need to have local adaptation? What would be the risks if these were not adapted to local markets?
As organizations evaluate their options for a performance management system, human resources and business leadership need to consider several challenges that will need to be addressed—no matter what the system (Capelli & Tavis, 2016).
The first is the challenge of aligning individual and company goals. Traditionally, the model has been to “cascade” goals down through the organization, and employees are supposed to create goals that reflect and support the direction set at the top. The notion of SMART goals ( Specific, Measurable, Achievable, Relevant, Time-bound ) has made the rounds over the years, but goal setting can still be challenging if business goals are complex or if employee goals seem more relatable to specific project work than to the overall top-line goals. The business and the individual need to be able to respond to goal shifts, which occur very often in response to the rapid rate of change and changing customer needs. This is an ongoing issue that human resources and business leadership will need to reconcile.
The next key challenge to think about when designing a performance management process is rewarding performance. Reward structures are discussed later in this chapter, but reward systems must be rooted in performance management systems. Currently, the companies that are redesigning their performance processes are trying to figure out how their new practices will impact their pay-for-performance models. Companies don’t appear to be abandoning the concept of rewarding employees based on and driven by their performance, so the linkage between the two will need to be redefined as the systems are changed.
The identification of poor performers is a challenge that has existed since the earliest days of performance management, and even the most formal performance management process doesn’t seem to be particularly good at weeding out poor performers. A lot of this is due to the managers who evaluate employees and are reluctant to address the poor performers that they’re seeing. Also, the annual performance management process tends to make some managers feel that poor performance should be overlooked during the year and only addressed (often ineffectively) during a one-per-year review. Whatever new performance management models an organization adopts, they will have to ensure that poor performance is dealt with in real time and is communicated, documented, and managed closely.
Avoiding legal troubles is another ongoing challenge for organizations and is another reason for real-time communication and documentation of performance issues. Human resources supports managers as they deal with employee relations issues, and the thought of not having a formal, numerical ratings system is unfathomable for some people who worry about defending themselves against litigation. However, because even formal performance processes can be subjective and may reveal ratings bias, neither the traditional formal process nor some of the radical new approaches can guarantee that legal troubles will never develop. From my experience, the best strategy for effective and fair performance management is real-time communication and documentation of issues. The employee is told about his or her performance issues (in as close to real time as possible), and the manager has documented the performance issues and conversations objectively and has engaged human resources with any larger or more complex issues.
“Managing the feedback fire hose” and keeping conversations, documentation, and feedback in a place where it can be tracked and utilized is an ongoing challenge. The typical annual performance process is not conducive to capturing ongoing feedback and conversations. There have been some new technologies introduced (such as apps) that can be used to capture ongoing conversations between managers and employees. General Electric uses an app called PD@GE (PD = performance development) that allows managers to pull up notes and materials from prior conversations with employees. IBM has a similar app that allows peer-to-peer feedback. Although there are clearly some technology solutions that can be used to help communicate and collect feedback, human resources will need to continue to communicate and reinforce rules around objectivity and appropriate use of the tools.
Performance management processes—traditional and inventive new approaches alike—will face the same challenges over time. Human resource management professionals need to be aware of these challenges and design a performance management system that addresses them in the format and within the context of their culture.
- Where did the concept of performance management originate?
- What are some of the key challenges of any performance management process?


4.1.2: READ- The Differences Between Short-Term and Long-Term Disability
Short-Term and Long-Term Disability
Disability insurance is a crucial safety net that supports employees who get hurt and can’t perform their job requirements. Both short-term and long-term disability insurance serve different purposes depending on the employee’s situation.
As a business owner with employees, it’s important to familiarize yourself with your legal obligations and your employees' responsibilities regarding both types of disability insurance.
Short-term disability insurance
Short-term disability insurance covers an employee’s compensation — supplementing between 40-70% of their salary — for a short period of time in the event that they are unable to perform their job duties due to serious medical conditions, such as an injury or illness. An employee’s coverage can last anywhere from a few weeks to a full year in certain circumstances.
To receive coverage, the employee must prove they cannot perform their duties with evidence from medical providers like a therapist or doctor. Qualifying medical conditions include surgery rehabilitation, injury from an accident, or pregnancy. However, unlike workers' compensation coverage, the injury or illness does not have to take place on the job. Because many states are required to offer workers’ compensation insurance, on- the-job injuries and illnesses are often covered under that policy instead of disability.
Whether or not employers must provide short-term disability insurance depends on the state in which you are registered to conduct business. Only a few states — like New York, California, and Rhode Island — require employers to provide short-term disability coverage to all employees, whether it’s a plan employers purchase through a private carrier or a state-sponsored plan.
Long-term disability insurance
Long-term disability insurance also protects an employee's compensation, covering roughly 60% of their gross monthly income for an extended period of time. In contrast to short-term disability, long-term disability is awarded to those who cannot perform any job — not just their current one. The length of coverage can span from two years all the way to retirement, or when Social Security payments commence.
Qualifying injuries and illnesses include cancer, mental illness, arthritis, back problems, stroke, and other serious conditions that prevent employees from completing normal day-to-day tasks. To receive compensation, employees must provide medical records which show the qualifying injury lasted beyond the elimination period. Then, employees can receive benefits until they have been medically cleared or until their policy benefits have been exhausted.
Oftentimes, an employer will offer both short- and long-term disability benefits; short-term benefits are usually exhausted first, at which point long-term benefits kick in. Employers can, but are not required to, provide long-term disability benefits. Doing so usually yields cheaper rates for employees, but often at the expense of a less comprehensive plan. Furthermore, if an employee leaves the company, they are not able to take their coverage with them, making private policies more appealing to some.
Short-term disability insurance is more suited for situations in which an employee was injured but can ultimately return to work, whereas long-term disability helps those who will be out of work for a long time, or even permanently.
The differences between short-term and long-term disability insurance
While short- and long-term disability insurance have similarities — neither have a deductible or minimum spending threshold and they both cost 1% to 3% of your annual salary — they also have several differences.
The two primarily differ based on the length of the coverage period. Short-term disability insurance is more suited for situations in which an employee was injured but can ultimately return to work, whereas long-term disability helps those who will be out of work for a long time, or even permanently.
In addition to the coverage length, benefits begin at different times for each insurance. For short-term disability, benefits begin after a predetermined amount of time, called an elimination period, ranging from seven to 30 days, with 14 days being the average. The elimination period for long-term disability is longer, lasting anywhere from 30 days to two years, with 90 days being the most common. This range will depend on the policy you purchase — cheaper policies incur longer elimination periods.
Finally, short- and long-term disability insurance each cover a different portion of an employee’s income. While the exact amount is determined by their salary and plan coverage, short-term disability insurance typically covers about 80% of one’s income, while long-term is closer to 60%.
CO— aims to bring you inspiration from leading respected experts. However, before making any business decision, you should consult a professional who can advise you based on your individual situation.
Published August 25, 2022
A printable copy of this article is attached below.
Source: Short-Term vs. Long-Term Disability: What’s the Difference? U.S. Chamber of Commerce. 2024


8.7: Media Literacy
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Studying how we use and consume mass communication allows us to scrutinize the conflicts, contradictions, problems, or even positive outcomes in our use of mass communication. With so much to learn about mass communication, how informed are you? Our consciousness of our media consumption is vital to understanding its effects on us as members of society. Media literacy is our awareness regarding our mediated environment or consumption of mass communication. It is our ability to responsibly comprehend, access, and use mass communication in our personal and professional lives. Potter states that we should maintain cognitive, emotional, aesthetic, and moral awareness as we interact with media. Baran suggests a number of skills we can develop in order to be media literate.
- Understand and respect the power of mass communication messages . An important skill for media literacy is to acknowledge just how dominant mass communication is in our lives and around the globe. Through mass communication, media shape, entertain, inform, represent, reflect, create, move, educate, and affect our behaviors, attitudes, values, and habits in direct and indirect ways. Virtually everyone in the world has been touched in some way by mass communication, and has made personal and professional decisions largely based on representations of reality portrayed though mass communication. We must understand and respect the power media have in our lives and understand how we make sense of certain meanings.
- Understand content by paying attention and filtering out noise . As we learned in Chapter 1, anything that hinders communication is noise. Much of the noise in mass communication originates with our consumption behaviors. How often do you do something other than pay complete attention to the media that you’re accessing? Do you listen to the radio while you drive, watch television while you eat, or text message a friend while you’re in class? When it comes to mass communication we tend to multitask, an act that acts as noise and impacts the quality of the messages and our understanding of their meanings. We often turn ourselves into passive consumers, not really paying attention to the messages we receive as we perform other tasks while consuming media.
- Understand emotional versus reasoned reactions to mass communication content in order to act accordingly . A great deal of mass communication content is intended to touch us on an emotional level. Therefore, it’s important to understand our emotional reactions to mass communication. Advertising often appeals to our emotions in order to sell products (Jhally). “Sex sells” is an old advertising adage, but one that highlights how often we make decisions based on emotional reactions, versus reasoned actions. Glance through magazines like Maxim or Glamour and you’ll quickly realize how the emotions associated with sex are used to sell products of all kinds. Reasoned actions require us to think critically about the mass communication we consume before we come to conclusions simply based on our emotional responses.
- Develop heightened expectations of mass communication content . Would you consider yourself an informed consumer of mass communication? Do you expect a lot from mass communication? You may like a mystery novel because it’s “fun,” or a movie might take your mind off of reality for a few hours. However, Baran challenges us to require more from the media we consume. “When we expect little from the content before us, we tend to give meaning making little effort and attention” (57). It depends upon you what you’re willing to accept as quality. Some people may watch fewer and fewer mainstream movies because they think the current movies in theaters are low culture or are aimed at less educated audiences. They may begin to look for more foreign films, independent films, and documentaries rather than go to see the popular movies released by Hollywood. We’ve even seen a backlash against television programming in general. With the rise of services like Netflix, Hulu and Amazon On-Demand, many media consumers have chosen to become what’s known as “cord cutters” and cancel their cable subscriptions. These new services often offer popular tv shows and sometimes even the most current episodes available to watch at your own leisure.
- Understand genre conventions and recognize when they are being mixed . All media have their own unique characteristics or “certain distinctive, standardized style elements” that mark them as a category or genre (Baran 57). We expect certain things from different forms of mass communication. Most of us believe, for example, that we are able to tell the difference between news and entertainment. But, are we? Television news shows often recreate parts of a story to fill in missing video of an event. Do you always catch the “re-enactment” disclaimer? Shows such as The Daily Show or Last Week Tonight effectively blurred the lines between comedy and news, and both became recognized as credible sources for news information. Even eighty years ago, Walter Lippmann recognized that media are so invasive in our lives that we might have difficulty distinguishing between what is real and what is manipulated by the media. The “reality TV” genre is now blurring these lines even more. Another example is the election of Arnold Schwarzenegger as governor of California. He, and others, often refer to him as the “governator,” a blurring of his fictional role as the Terminator and his real role as California’s governor.
- Think critically about mass communication messages, no matter how credible their source . It is essential that we critically consider the source of all mass communication messages. No matter how credible a media source, we can’t always believe everything we see or hear because all mass communication is motivated by political, profit, or personal factors. Publicists, editors, and publishers present the information from their perspective–informed by their experiences and agendas. Even if the motive is pure or the spin is minimal, we tend to selectively interpret meanings based on our own lived experiences. Audiences do not always hold similar perceptions regarding mediated messages.
- Understand the internal language of mass communication to understand its effects, no matter how complex . This skill requires us to develop sensitivity to what is going on in the media. This doesn’t just refer to whether you can program a DVR or surf the internet. This means being familiar with the intent or motivation behind the action or message. “Each medium has its own specific internal language. This language is expressed in production values–the choice of lighting, editing, special effects, music, camera angle, location on the page, and size and placement of headline. To be able to read a media text, you must understand its language” (Baran 58). What effect do these have on your interpretive or sense making abilities? Most news coverage of the Iraq war included background symbols of American flags, eagles, as well as words like “Freedom,” and “Liberation.” What is the impact of using these symbols in “objective” coverage of something like war? Shows like Scandal makes editorial choices to glamorize and demoralize politics while making it appear provocatively thrilling. On the surface, we might not realize the amount of effort that goes into dealing with political scandals, but shows like Scandal shed a light on these unspoken issues.
Case In Point
The Tao of Media Literacy
How do media affect us? Are we media literate? Werner Heisenberg in The Physicist’s Conception of Nature relates a timeless, allegorical story about the role of technology in our lives and questions if our interactions are mindful or thoughtless in regards to change. In Heisenberg’s analogy, the wise old, Chinese sage warns us about the delicate balance between humans, nature, and technology.
In this connection it has often been said that the far-reaching changes in our environment and in our way of life wrought by this technical age have also changed dangerously our ways of thinking, and that here lie the roots of the crises, which have shaken our times and which, for instance, are also expressed in modern art. True, this objection’s much older than modern technology and science, the use of implements going back to our earliest beginnings. Thus, two and a half thousand years ago, the Chinese sage Chuang-Tzu spoke of the danger of the machine when he said: As Tzu-Gung was [traveling] through the regions north of the river Han, he saw an old man working in his vegetable garden. He had dug an irrigation ditch. The man would descend into the well, fetch up a vessel of water in his arms and pour it out into the ditch. While his efforts were tremendous the results appeared to be very [meager]. Tzu-Gung said, “There is a way whereby you can irrigate a hundred ditches in one day, and whereby you can do much with little effort. Would you not like to hear of it?” Then the gardener stood up, looked at him and said, “And what would that be?” Tzu-Gung replied, “You take a wooden lever, weighted at the back and light in front. In this way you can bring up water so quickly that it just gushes out. This is called a draw-well.” Then anger rose up on the old man’s face, and he said, “I have heard my teacher say that whoever uses machines does all his work like a machine. He who does his work like a machine grows a heart like a machine, and he who carries the heart of a machine in his breast loses his simplicity. He who has lost his simplicity becomes unsure in the strivings of his soul. Uncertainty in the strivings of the soul is something which does not agree with honest sense. It is not that I do not know of such things: I am ashamed to use them.”
Contributions and Affiliations
- Survey of Communication Study. Authored by : Scott T Paynton and Linda K Hahn. Provided by : Humboldt State University. Located at : en.wikibooks.org/wiki/Survey_of_Communication_Study. License : CC BY-SA: Attribution-ShareAlike
- Image of graffiti. Authored by : ElAlispruz. Located at : commons.wikimedia.org/wiki/File:Talkingheadtv.jpg. License : CC BY: Attribution


10.7: The Fisher Exact Test
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What should you do if your cell counts are too small, but you’d still like to test the null hypothesis that the two variables are independent? One answer would be “collect more data”, but that’s far too glib: there are a lot of situations in which it would be either infeasible or unethical do that. If so, statisticians have a kind of moral obligation to provide scientists with better tests. In this instance, Fisher (1922) kindly provided the right answer to the question. To illustrate the basic idea, let’s suppose that we’re analysing data from a field experiment, looking at the emotional status of people who have been accused of witchcraft; some of whom are currently being burned at the stake.
181
Unfortunately for the scientist (but rather fortunately for the general populace), it’s actually quite hard to find people in the process of being set on fire, so the cell counts are awfully small in some cases. The
salem.Rdata
file illustrates the point:
load("./rbook-master/data/salem.Rdata")
salem.tabs <- table( trial )
print( salem.tabs )
## on.fire
## happy FALSE TRUE
## FALSE 3 3
## TRUE 10 0
Looking at this data, you’d be hard pressed not to suspect that people not on fire are more likely to be happy than people on fire. However, the chi-square test makes this very hard to test because of the small sample size. If I try to do so, R gives me a warning message:
chisq.test( salem.tabs )
## Warning in chisq.test(salem.tabs): Chi-squared approximation may be
## incorrect
##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: salem.tabs
## X-squared = 3.3094, df = 1, p-value = 0.06888
Speaking as someone who doesn’t want to be set on fire, I’d really like to be able to get a better answer than this. This is where Fisher’s exact test comes in very handy.
The Fisher exact test works somewhat differently to the chi-square test (or in fact any of the other hypothesis tests that I talk about in this book) insofar as it doesn’t have a test statistic; it calculates the p-value “directly”. I’ll explain the basics of how the test works for a 2×2 contingency table, though the test works fine for larger tables. As before, let’s have some notation:
| Happy | Sad | Total | |
|---|---|---|---|
| Set on fire | O 11 | O 12 | R 1 |
| Not set on fire | O 21 | O 22 | R 2 |
| Total | C 1 | C 2 | N |
In order to construct the test Fisher treats both the row and column totals (R 1 , R 2 , C 1 and C 2 ) are known, fixed quantities; and then calculates the probability that we would have obtained the observed frequencies that we did (O 11 , O 12 , O 21 and O 22 ) given those totals. In the notation that we developed in Chapter 9 this is written:
P(O 11 ,O 12 ,O 21 ,O 22 | R 1 ,R 2 ,C 1 ,C 2 )
and as you might imagine, it’s a slightly tricky exercise to figure out what this probability is, but it turns out that this probability is described by a distribution known as the hypergeometric distribution . 182 Now that we know this, what we have to do to calculate our p-value is calculate the probability of observing this particular table or a table that is “more extreme” . 183 Back in the 1920s, computing this sum was daunting even in the simplest of situations, but these days it’s pretty easy as long as the tables aren’t too big and the sample size isn’t too large. The conceptually tricky issue is to figure out what it means to say that one contingency table is more “extreme” than another. The easiest solution is to say that the table with the lowest probability is the most extreme. This then gives us the p-value.
The implementation of the test in R is via the
fisher.test()
function. Here’s how it is used:
fisher.test( salem.tabs )
##
## Fisher's Exact Test for Count Data
##
## data: salem.tabs
## p-value = 0.03571
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.000000 1.202913
## sample estimates:
## odds ratio
## 0
This is a bit more output than we got from some of our earlier tests. The main thing we’re interested in here is the p-value, which in this case is small enough (p=.036) to justify rejecting the null hypothesis that people on fire are just as happy as people not on fire.


18.5: Calcium Channel Blockers
By the end of this section, you should be able to:
- 18.5.1 Identify the characteristics of the calcium channel blocker drugs used to treat hypertension and angina.
- 18.5.2 Explain the indications, actions, adverse reactions, and interactions of the calcium channel blocker drugs used to treat hypertension and angina.
- 18.5.3 Describe nursing implications of calcium channel blocker drugs used to treat hypertension and angina.
- 18.5.4 Explain the client education related to calcium channel blocker drugs used to treat hypertension and angina.
Introduction and Use
Calcium channel blocker s are a classification of drug that blocks calcium from entering cells by binding to long-acting voltage-gated calcium channels in the heart, smooth muscle, and pancreas. Calcium causes vasoconstriction. Calcium channel blockers inhibit calcium and allow for vasodilation, thereby causing the heart rate to slow and blood pressure to lower. Calcium channel blockers are commonly used to treat hypertension, angina, and arrhythmias.
Calcium channel blockers are classified as dihydropyridines or nondihydropyridines. Dihydropyridines are peripheral vasodilators that lower blood pressure and heart rate; they are used to treat post-intracranial hemorrhage, vasospasm, and migraines. In addition to their action on the peripheral vasculature, nondihydropyridines inhibit the sinoatrial and AV nodes, slowing cardiac conduction and heart rate and decreasing oxygen demand.
Calcium channel blockers may be used alone or in combination with other drugs. They are highly protein bound, which increases the volume of distribution within the body. Calcium channel blockers are available in long- and short-acting formulas.
Special Considerations
Calcium Channel Blockers
Calcium channel blockers are among the most effective antihypertensive classifications of drugs for use in non-Hispanic Black adults .
(Source: Abrahamowicz et al., 2023)
Safety Alert
Calcium Channel Blockers
Abrupt discontinuation of calcium channel blockers may cause chest pain.
Table 18.9 lists common calcium channel blockers and typical routes and dosing for adult clients.
| Drug | Routes and Dosage Ranges |
|---|---|
|
Amlodipine
( Norvasc ) |
5–10 mg orally daily. |
|
Diltiazem
( Cardizem ) |
Initial dosage:
30 mg orally 4 times daily, before meals and at bedtime. Increase dosage gradually (given in divided doses 3–4 times daily) at 1- to 2-day intervals until optimum response is obtained.
Average optimum dosage: 180–360 mg daily. |
|
Nicardipine
( Cardene ) |
Immediate release:
20–40 mg orally 3 times daily.
Sustained release: 30–60 mg daily; maximum dose 120 mg daily. Intravenous (IV) infusion: 5–15 mg hourly. |
|
Nifedipine
( Procardia ) |
Immediate release:
Starting dose 10 mg orally 3 times daily. The usual effective dose range is 10–20 mg 3 times daily. Doses above 120 mg daily are rarely necessary. More than 180 mg daily is not recommended.
Sustained release: 30–60 mg orally daily; maximum dose 120 mg daily. |
|
Verapamil
( Calan SR , Verelan ) |
Immediate release:
40–120 mg 3 times daily.
Sustained release: 180–240 mg daily. IV: Follow manufacturer’s instructions and health care provider’s orders. |
Adverse Effects and Contraindications
Adverse effects of calcium channel blockers include dizziness, constipation, palpitations, fatigue, flushing, nausea, rash, headache, and swelling in the legs and feet. Serious side effects include thrombocytopenia, hypotension, and hyperglycemia. Clients with a hypersensitivity to calcium channel blockers or their components should not use this classification of drug. Other contraindications include sick sinus syndrome (except in clients with an artificial pacemaker), severe hypotension, a history of myocardial infarction, or a history of pulmonary congestion.
Table 18.10 is a drug prototype table for calcium channel blockers featuring amlodipine. It lists drug class, mechanism of action, adult dosage, indications, therapeutic effects, drug and food interactions, adverse effects, and contraindications.
|
Drug Class
Calcium channel blocker Mechanism of Action Inhibits the influx of calcium ions into vascular smooth muscle and cardiac muscle, thereby decreasing peripheral vascular resistance and reducing blood pressure |
Drug Dosage
5–10 mg orally daily. |
|
Indications
To control hypertension In the treatment of angina Therapeutic Effects Lowers blood pressure Increases blood supply and oxygen to the heart |
Drug Interactions
Simvastatin Clarithromycin Ritonavir Food Interactions No significant interactions |
|
Adverse Effects
Peripheral edema Headache Fatigue Palpitations Hypotension Dizziness Nausea Flushing Pruritus Skin rash Muscle cramps Erectile dysfunction |
Contraindications
Hypersensitivity Caution: Aortic stenosis Symptomatic hypotension Angina Myocardial infarction Hepatic impairment Pregnancy Breastfeeding |
Nursing Implications
The nurse should do the following for clients who are taking calcium channel blockers:
- Monitor the client for interactions because many medications and herbal supplements, such as St. John’s wort, interact with calcium channel blockers. Grapefruit juice also can affect the action of certain calcium channel blockers.
- Assess and monitor the client for adverse effects and contraindications.
- Provide client teaching regarding the drug and when to call the health care provider. See below for client teaching guidelines.
Client Teaching Guidelines
The client taking a calcium channel blocker should:
- Report side effects such as severe headaches, dizziness, lightheadedness, flushing, nausea, and leg swelling to their health care provider.
- Take this medication with food to avoid GI upset, nausea, and vomiting.
FDA Black Box Warning
Calcium Channel Blockers
Immediate release nifedipine, a potent calcium blocker, may increase the risk of myocardial infarction, stroke, and arrhythmias.
Unfolding Case Study
Part B
Read the following clinical scenario to answer the questions that follow. This case study is a follow-up to Case Study Part A.
Hahn Tran follows up with her health care provider one week later to get the results of her diagnostic studies. The health care provider diagnose her with hypertension stage 2 and starts her on enalapril, an ACE inhibitor. The nurse then teaches Hahn about this new drug.
Avoid salt substitutes that contain potassium while taking this medication.
- Answer
-
Stop smoking.


14.10: Chapter Review - Practice Exercises
-
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Practice Exercises
18 .
Go to this
sample database from Durham University
. Open the database and create an aggregate query that sums the Cost field of the Asset Items table. What is the total cost of the Asset Items?
19 .
Imagine that you are a human resources (HR) professional responsible for reducing tardiness in the workplace. You have a database table listing all of the company’s employees, with corresponding details about each employee. How would you create a form that will keep track of clocking-in after each change in the
HR Employee Handbook
? The form should include the employees’ new clock-in times, date of test, and any notes taken about the HR changes.
20 .
Build a simple Main Menu form and create a text box for the current date and time, along with a label that reads “Main Menu.” Format the text box to Broadway font, centered, and in bright red. Try to center the label and also to center the text box directly below the label.
21 .
Create a custom sales report, following these steps:
- Open Access and create a new report.
- Use Design View to add the following sections to your report: Report Header, Page Header, Group Header, Detail, Group Footer, Page Footer, and Report Footer.
- In the Page Header section, add the column headings for Product Name, Quantity Sold, and Total Sales.
- Group the report by Product Category in the Group Header section.
- In the Detail section, display the product name, quantity sold, and total sales for each product.
- Calculate the sum of total sales for each product category and display it in the Group Footer section.
- Add page numbers to the Page Footer section.
- In the Report Footer section, calculate and display the overall total sales for the entire report.
22 .
You’re responsible for creating an invoice report for WorldCorp. Customize the report, following these steps:
- Create a new report in Access.
- Add a logo of WorldCorp to the Report Header section.
- Format the logo to ensure it fits well and looks professional.
- In the Page Header section, add the company name, address, and contact information.
- Design the Detail section to show the invoice details, including item name, quantity, price, and subtotal.
- Add a calculated field to calculate the total amount for each item.
- Display the due date of the invoice in the Page Footer section.
- Format the report to use a consistent font style and size throughout.
23 .
You are working on a database that tracks inventory for a retail store chain. Whenever a new transaction is entered, you need to generate an invoice for the customer that includes their name, the date of the sale, and a list of the items bought. Create a new macro that has multiple steps and save it with a descriptive name. Then, run the macro to verify that it generates an invoice.
24 .
You are working on a database that tracks employee information for WorldCorp. You want to enhance the user experience by using macros to display custom messages and validate data entry. Create a macro that displays a message box when the user enters certain data. Then, run the macro to make sure it displays everything correctly.
25 .
Go to the
Contextures website
and select some sample data to work with. First, create an Excel table, using the data you have selected. Then, try to link the Excel table into Access. Make changes in the Excel file to see if the changes are captured in Access.
26 .
Go to the
Contextures website
and select one of the sample sets of sales data to work with. Create a tactical sales dashboard using Access. This dashboard should provide a quick overview of key sales metrics and allow you to navigate to detailed reports.


9.1: Two Proportions
There are times you want to test a claim about two population proportions or construct a confidence interval estimate of the difference between two population proportions. As with all other hypothesis tests and confidence intervals, the process is the same though the formulas and assumptions are different.
Hypothesis Test for Two Populations Proportion (2-Prop Test)
-
State the random variables and the parameters in words.
\(x_{1}\)= number of successes from group 1
\(x_{2}\) = number of successes from group 2
\(p_{1}\) = proportion of successes in group 1
\(p_{2}\) = proportion of successes in group 2 -
State the null and alternative hypotheses and the level of significance
\(\begin{array}{ll}{H_{o} : p_{1}=p_{2}} & {\text { or } \quad H_{o} : p_{1}-p_{2}=0} \\ {H_{A} : p_{1}<p_{2}} &\quad\quad\: {H_{A} : p_{1}-p_{2}<0} \\ {H_{A} : p_{1}>p_{2}} &\quad\quad\: {H_{A} : p_{1}-p_{2}>0} \\ {H_{A} : p_{1} \neq p_{2}} & \quad\quad\:{H_{A} : p_{1}-p_{2} \neq 0}\end{array}\)
Also, state your \(\alpha\) level here. -
State and check the assumptions for a hypothesis test
- A simple random sample of size \(n_{1}\) is taken from population 1, and a simple random sample of size \(n_{2}\) is taken from population 2.
- The samples are independent.
- The assumptions for the binomial distribution are satisfied for both populations.
- To determine the sampling distribution of \(\hat{p}_{1}\), you need to show that \(n_{1} p_{1} \geq 5\) and \(n_{1} q_{1} \geq 5\), where \(q_{1}=1-p_{1}\). If this requirement is true, then the sampling distribution of \(\hat{p}_{1}\) is well approximated by a normal curve. To determine the sampling distribution of \(\hat{p}_{2}\), you need to show that \(n_{2} p_{2} \geq 5\) and \(n_{2} q_{2} \geq 5\), where \(q_{2}=1-p_{2}\). If this requirement is true, then the sampling distribution of \(\hat{p}_{2}\) is well approximated by a normal curve. However, you do not know \(p_{1}\) and \(p_{2}\), so you need to use \(\hat{p}_{1}\) and instead \(\hat{p}_{2}\). This is not perfect, but it is the best you can do. Since \(n_{1} \hat{p}_{1}=n_{1} \dfrac{x_{1}}{n_{1}}=x_{1}\) (and similar for the other calculations) you just need to make sure that \(x_{1}\), \(n_{1}-x_{1}\), \(n_{2}-x_{2}\),and are all more than 5.
-
Find the sample statistics, test statistic, and p-value
Sample Proportion:
\(\begin{array}{ll}{n_{1}=\text { size of sample } 1} & {n_{2}=\text { size of sample } 2} \\ {\hat{p}_{1}=\dfrac{x_{1}}{n_{1}}(\text { sample } 1 \text { proportion) }} & {\hat{p}_{2}=\dfrac{x_{2}}{n_{2}} \text { (sample } 2 \text { proportion) }} \\ {\hat{q}_{1}=1-\hat{p}_{1} \text { (complement of } \hat{p}_{1} )} & {\hat{q}_{2}=1-\hat{p}_{2} \text { (complement of } \hat{p}_{2} )}\end{array}\)
Pooled Sample Proportion, \(\overline{p}\):
\(\begin{aligned} \overline{p} &=\dfrac{x_{1}+x_{2}}{n_{1}+n_{2}} \\ \overline{q} &=1-\overline{p} \end{aligned}\)
Test Statistic:
\(z=\dfrac{\left(\hat{p}_{1}-\hat{p}_{2}\right)-\left(p_{1}-p_{2}\right)}{\sqrt{\dfrac{\overline{p} \overline{q}}{n_{1}}+\dfrac{\overline{p} \overline{q}}{n_{2}}}}\)
Usually \(p_{1} - p_{2} = 0\), since \(H_{o} : p_{1}=p_{2}\)
p-value: On TI-83/84: use normalcdf(lower limit, upper limit, 0, 1)On R: use pnorm(z, 0, 1)Note
If \(H_{A} : p_{1}<p_{2}\) then lower limit is \(-1 E 99\) and upper limit is your test statistic. If \(H_{A} : p_{1}>p_{2}\), then lower limit is your test statistic and the upper limit is \(1 E 99\). If \(H_{A} : p_{1} \neq p_{2}\), then find the p-value for \(H_{A} : p_{1}<p_{2}\), and multiply by 2.
Note
If \(H_{A} : p_{1}<p_{2}\), then use pnorm(z, 0, 1). If \(H_{A} : p_{1}>p_{2}\), then use 1 - pnorm(z, 0, 1). If \(H_{A} : p_{1} \neq p_{2}\), then find the p-value for \(H_{A} : p_{1}<p_{2}\), and multiply by 2.
- 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 Between Two Population Proportion (2-Prop Interval)
The confidence interval for the difference in proportions has the same random variables and proportions and the same assumptions as the hypothesis test for two proportions. 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 proportions and state and check the assumptions before completing the confidence interval step
-
Find the sample statistics and the confidence interval
Sample Proportion:
\(\begin{array}{ll}{n_{1}=\text { size of sample } 1} & {n_{2}=\text { size of sample } 2} \\ {\hat{p}_{1}=\dfrac{x_{1}}{n_{1}}(\text { sample } 1 \text { proportion) }} & {\hat{p}_{2}=\dfrac{x_{2}}{n_{2}} \text { (sample } 2 \text { proportion) }} \\ {\hat{q}_{1}=1-\hat{p}_{1}\left(\text { complement of } \hat{p}_{1}\right)} & {\hat{q}_{2}=1-\hat{p}_{2} \text { (complement of } \hat{p}_{2} )}\end{array}\)
Confidence Interval:
The confidence interval estimate of the difference \(p_{1}-p_{2}\) is
\(\left(\hat{p}_{1}-\hat{p}_{2}\right)-E<p_{1}-p_{2}<\left(\hat{p}_{1}-\hat{p}_{2}\right)+E\)
where the margin of error E is given by \( E=z_{c} \sqrt{\dfrac{\hat{p}_{1} \hat{q}_{1}}{n_{1}}+\dfrac{\hat{p}_{2} \hat{q}_{2}}{n_{2}}}\)
\(z_{c}\) = critical value - Statistical Interpretation: In general this looks like, “there is a C% chance that \(\left(\hat{p}_{1}-\hat{p}_{2}\right)-E<p_{1}-p_{2}<\left(\hat{p}_{1}-\hat{p}_{2}\right)+E\) contains the true difference in proportions.”
- Real World Interpretation: This is where you state how much more (or less) the first proportion is from the second proportion.
The critical value is a value from the normal distribution. Since a confidence interval is found by adding and subtracting a margin of error amount from the sample proportion, and the interval has a probability of being true, then you can think of this as the statement \(P\left(\left(\hat{p}_{1}-\hat{p}_{2}\right)-E<p_{1}-p_{2}<\left(\hat{p}_{1}-\hat{p}_{2}\right)+E\right)=C\). So you can use the invNorm command on the TI-83/84 calculator or qnorm on R to find the critical value. These are always the same value, so it is easier to just look at the table A.1 in the Appendix.
Example \(\PageIndex{1}\) hypothesis test for two population proportions
Do husbands cheat on their wives more than wives cheat on their husbands ("Statistics brain," 2013)? Suppose you take a group of 1000 randomly selected husbands and find that 231 had cheated on their wives. Suppose in a group of 1200 randomly selected wives, 176 cheated on their husbands. Do the data show that the proportion of husbands who cheat on their wives are more than the proportion of wives who cheat on their husbands. Test at the 5% level.
- 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 a hypothesis test.
- Find the sample statistics, test statistic, and p-value.
- Conclusion
- Interpretation
Solution
1. \(x_{1}\) = number of husbands who cheat on his wife
\(x_{2}\) = number of wives who cheat on her husband
\(p_{1}\) = proportion of husbands who cheat on his wife
\(p_{2}\) = proportion of wives who cheat on her husband
2. \(\begin{array}{ll}{H_{o} : p_{1}=p_{2}} & {\text { or } \quad H_{o} : p_{1}-p_{2}=0} \\ {H_{A} : p_{1}>p_{2}} &\quad\quad\: {H_{A} : p_{1}-p_{2}>0} \\ {a=0.05}\end{array}\)
3.
- A simple random sample of 1000 responses about cheating from husbands is taken. This was stated in the problem. A simple random sample of 1200 responses about cheating from wives is taken. This was stated in the problem.
- The samples are independent. This is true since the samples involved different genders.
- The properties of the binomial distribution are satisfied in both populations. This is true since there are only two responses, there are a fixed number of trials, the probability of a success is the same, and the trials are independent.
-
The sampling distributions of \(\hat{p}_{1}\) and \(\hat{p}_{2}\) can be approximated with a normal distribution.
\(x_{1}=231, n_{1}-x_{1}=1000-231=769, x_{2}=176\), and
\(n_{2}-x_{2}=1200-176=1024\) are all greater than or equal to 5. So both sampling distributions of \(\hat{p}_{1}\) and \(\hat{p}_{2}\) can be approximated with a normal distribution.
4. Sample Proportion:
\(\begin{array}{ll}{n_{1}=1000} & {n_{2}=1200} \\ {\hat{p}_{1}=\dfrac{231}{1000}=0.231} & {\hat{p}_{2}=\dfrac{176}{1200} \approx 0.1467} \\ {\hat{q}_{1}=1-\dfrac{231}{1000}=\dfrac{769}{1000}=0.769} & {\hat{q}_{2}=1-\dfrac{176}{1200}=\dfrac{1024}{1200} \approx 0.8533}\end{array}\)
Pooled Sample Proportion, \(\overline{p}\):
\(\begin{array}{l}{\overline{p}=\dfrac{231+176}{1000+1200}=\dfrac{407}{2200}=0.185} \\ {\overline{q}=1-\dfrac{407}{2200}=\dfrac{1793}{2200}=0.815}\end{array}\)
Test Statistic:
\(z=\dfrac{(0.231-0.1467)-0}{\sqrt{\dfrac{0.185 * 0.815}{1000}+\dfrac{0.185 * 0.815}{1200}}}\)
\(=5.0704\)
p-value:
On TI-83/84: normalcdf \((5.0704,1 E 99,0,1)=1.988 \times 10^{-7}\)
On R: \(1-\text { pnorm }(5.0704,0,1)=1.988 \times 10^{-7}\)
Figure \(\PageIndex{1}\): Setup for 2-PropZTest on TI-83/84 Calculator Figure \(\PageIndex{2}\): Results for 2-PropZTest on TI-83/84 Calculator Figure \(\PageIndex{3}\): Results for 2-PropZTest on TI-83/84: CalculatorOn R: prop.test\(\left(c\left(x_{1}, x_{2}\right), c\left(n_{1}, n_{2}\right), \text { alternative }=\right.\) "less" or "greater". For this example, prop.test(c(231,176), c(1000, 1200), alternative="greater")
2-sample test for equality of proportions with continuity correction
data: c(231, 176) out of c(1000, 1200)
X-squared = 25.173, df = 1, p-value = 2.621e-07
alternative hypothesis: greater
95 percent confidence interval:
0.05579805 1.00000000
sample estimates:
prop 1 prop 2
0.2310000 0.1466667
Note
The answer from R is the p-value. It is different from the formula or the TI-83/84 calculator due to a continuity correction that R does.
5. Conclusion
Reject \(H_{o}\), since the p-value is less than 5%.
6. Interpretation This is enough evidence to show that the proportion of husbands having affairs is more than the proportion of wives having affairs.
Example \(\PageIndex{2}\) confidence interval for two population properties
Do more husbands cheat on their wives more than wives cheat on the husbands ("Statistics brain," 2013)? Suppose you take a group of 1000 randomly selected husbands and find that 231 had cheated on their wives. Suppose in a group of 1200 randomly selected wives, 176 cheated on their husbands. Estimate the difference in the proportion of husbands and wives who cheat on their spouses using a 95% confidence level.
- State the random variables and the parameters in words.
- State and check the assumptions for the confidence interval.
- Find the sample statistics and the confidence interval.
- Statistical Interpretation
- Real World Interpretation
Solution
1. These were stated in Example \(\PageIndex{1}\), but are reproduced here for reference.
\(x_{1}\) = number of husbands who cheat on his wife
\(x_{2}\) = number of wives who cheat on her husband
\(p_{1}\) = proportion of husbands who cheat on his wife
\(p_{2}\) = proportion of wives who cheat on her husband
2. The assumptions were stated and checked in Example \(\PageIndex{1}\).
3. Sample Proportion:
\(\begin{array}{ll}{n_{1}=1000} & {n_{2}=1200} \\ {\hat{p}_{1}=\dfrac{231}{1000}=0.231} & {\hat{p}_{2}=\dfrac{176}{1200} \approx 0.1467} \\ {\hat{q}_{1}=1-\dfrac{231}{1000}=\dfrac{769}{1000}=0.769} & {\hat{q}_{2}=1-\dfrac{176}{1200}=\dfrac{1024}{1200} \approx 0.8533}\end{array}\)
Confidence Interval:
\(\begin{array}{l}{z_{c}=1.96} \\ {E=1.96 \sqrt{\dfrac{0.231 * 0.769}{1000}+\dfrac{0.1467 * 0.8533}{1200}}=0.033}\end{array}\)
The confidence interval estimate of the difference \(p_{1}-p_{2}\) is
\(\begin{array}{l}{\left(\hat{p}_{1}-\hat{p}_{2}\right)-E<p_{1}-p_{2}<\left(\hat{p}_{1}-\hat{p}_{2}\right)+E} \\ {(0.231-0.1467)-0.033<p_{1}-p_{2}<(0.231-0.1467)+0.033} \\ {0.0513<p_{1}-p_{2}<0.1173}\end{array}\)
Figure 9.1.4: Setup for 2-PropZInt on TI-83/84 Calculator Figure \(\PageIndex{5}\): Results for 2-PropZInt on TI-83/84 CalculatorOn R: prop.test\(\left(c\left(x_{1}, x_{2}\right), c\left(n_{1}, n_{2}\right), \text { conf.level }=\mathrm{C}\right)\), where C is in decimal form. For this example, prop.test(c(231,176), c(1000, 1200), conf.level=0.95)
2-sample test for equality of proportions with continuity correction
data: c(231, 176) out of c(1000, 1200)
X-squared = 25.173, df = 1, p-value = 5.241e-07
alternative hypothesis: two.sided
95 percent confidence interval:
0.05050705 0.11815962
sample estimates:
prop 1 prop 2
0.2310000 0.1466667
Note
The answer from R is the confidence interval. It is different from the formula or the TI-83/84 calculator due to a continuity correction that R does.
4. Statistical Interpretation: There is a 95% chance that \(0.0505<p_{1}-p_{2}<0.1182\) contains the true difference in proportions.
5. Real World Interpretation: The proportion of husbands who cheat is anywhere from 5.05% to 11.82% higher than the proportion of wives who cheat.
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.
- Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology exam 84,199 of them were female. In that same year, of the 211,693 students who took the calculus AB exam 102,598 of them were female ("AP exam scores," 2013). Is there enough evidence to show that the proportion of female students taking the biology exam is higher than the proportion of female students taking the calculus AB exam? Test at the 5% level.
- Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology exam 84,199 of them were female. In that same year, of the 211,693 students who took the calculus AB exam 102,598 of them were female ("AP exam scores," 2013). Estimate the difference in the proportion of female students taking the biology exam and female students taking the calculus AB exam using a 90% confidence level.
- Many high school students take the AP tests in different subject areas. In 2007, of the 211,693 students who took the calculus AB exam 102,598 of them were female and 109,095 of them were male ("AP exam scores," 2013). Is there enough evidence to show that the proportion of female students taking the calculus AB exam is different from the proportion of male students taking the calculus AB exam? Test at the 5% level.
- Many high school students take the AP tests in different subject areas. In 2007, of the 211,693 students who took the calculus AB exam 102,598 of them were female and 109,095 of them were male ("AP exam scores," 2013). Estimate using a 90% level the difference in proportion of female students taking the calculus AB exam versus male students taking the calculus AB exam.
- Are there more children diagnosed with Autism Spectrum Disorder (ASD) in states that have larger urban areas over states that are mostly rural? In the state of Pennsylvania, a fairly urban state, there are 245 eight year olds diagnosed with ASD out of 18,440 eight year olds evaluated. In the state of Utah, a fairly rural state, there are 45 eight year olds diagnosed with ASD out of 2,123 eight year olds evaluated ("Autism and developmental," 2008). Is there enough evidence to show that the proportion of children diagnosed with ASD in Pennsylvania is more than the proportion in Utah? Test at the 1% level.
- Are there more children diagnosed with Autism Spectrum Disorder (ASD) in states that have larger urban areas over states that are mostly rural? In the state of Pennsylvania, a fairly urban state, there are 245 eight year olds diagnosed with ASD out of 18,440 eight year olds evaluated. In the state of Utah, a fairly rural state, there are 45 eight year olds diagnosed with ASD out of 2,123 eight year olds evaluated ("Autism and developmental," 2008). Estimate the difference in proportion of children diagnosed with ASD between Pennsylvania and Utah. Use a 98% confidence level.
- A child dying from an accidental poisoning is a terrible incident. Is it more likely that a male child will get into poison than a female child? To find this out, data was collected that showed that out of 1830 children between the ages one and four who pass away from poisoning, 1031 were males and 799 were females (Flanagan, Rooney & Griffiths, 2005). Do the data show that there are more male children dying of poisoning than female children? Test at the 1% level.
- A child dying from an accidental poisoning is a terrible incident. Is it more likely that a male child will get into poison than a female child? To find this out, data was collected that showed that out of 1830 children between the ages one and four who pass away from poisoning, 1031 were males and 799 were females (Flanagan, Rooney & Griffiths, 2005). Compute a 99% confidence interval for the difference in proportions of poisoning deaths of male and female children ages one to four.
- Answer
-
For all hypothesis tests, just the conclusion is given. For all confidence intervals, just the interval using technology (Software R) is given. See solution for the entire answer.
- Reject Ho
- \(0.0941<p_{1}-p_{2}<0.0996\)
- Reject Ho
- \(-0.0332<p_{1}-p_{2}<-0.0282\)
- Fail to reject Ho
- \(-0.01547<p_{1}-p_{2}<-0.0001\)
- Reject Ho
- \(0.0840<p_{1}-p_{2}<0.1696\)