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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def welch_anova(dv=None, between=None, data=None, export_filename=None):
 """One-way Welch ANOVA. Parameters dv : string Name of column containing the dependant variable. between : string Name of column containing the between factor. data : pandas DataFrame DataFrame. Note that this function can also directly be used as a Pandas method, in which case this argument is no longer needed. export_filename : string Filename (without extension) for the output file. If None, do not export the table. By default, the file will be created in the current python console directory. To change that, specify the filename with full path. Returns ------- aov : DataFrame ANOVA summary :: 'Source' : Factor names 'SS' : Sums of squares 'DF' : Degrees of freedom 'MS' : Mean squares 'F' : F-values 'p-unc' : uncorrected p-values 'np2' : Partial eta-square effect sizes See Also -------- anova : One-way ANOVA rm_anova : One-way and two-way repeated measures ANOVA mixed_anova : Two way mixed ANOVA kruskal : Non-parametric one-way ANOVA Notes ----- The classic ANOVA is very powerful when the groups are normally distributed and have equal variances. However, when the groups have unequal variances, it is best to use the Welch ANOVA that better controls for type I error (Liu 2015). The homogeneity of variances can be measured with the `homoscedasticity` function. The two other assumptions of normality and independance remain. The main idea of Welch ANOVA is to use a weight :math:`w_i` to reduce the effect of unequal variances. This weight is calculated using the sample size :math:`n_i` and variance :math:`s_i^2` of each group .. math:: w_i = \\frac{n_i}{s_i^2} Using these weights, the adjusted grand mean of the data is: .. math:: \\overline{Y}_{welch} = \\frac{\\sum_{i=1}^r w_i\\overline{Y}_i} {\\sum w} where :math:`\\overline{Y}_i` is the mean of the :math:`i` group. The treatment sums of squares is defined as: .. math:: SS_{treatment} = \\sum_{i=1}^r w_i (\\overline{Y}_i - \\overline{Y}_{welch})^2 We then need to calculate a term lambda: .. math:: \\Lambda = \\frac{3\\sum_{i=1}^r(\\frac{1}{n_i-1}) (1 - \\frac{w_i}{\\sum w})^2}{r^2 - 1} from which the F-value can be calculated: .. math:: F_{welch} = \\frac{SS_{treatment} / (r-1)} {1 + \\frac{2\\Lambda(r-2)}{3}} and the p-value approximated using a F-distribution with :math:`(r-1, 1 / \\Lambda)` degrees of freedom. When the groups are balanced and have equal variances, the optimal post-hoc test is the Tukey-HSD test (`pairwise_tukey`). If the groups have unequal variances, the Games-Howell test is more adequate. Results have been tested against R. References .. [1] Liu, Hangcheng. "Comparing Welch's ANOVA, a Kruskal-Wallis test and traditional ANOVA in case of Heterogeneity of Variance." (2015). .. [2] Welch, Bernard Lewis. "On the comparison of several mean values: an alternative approach." Biometrika 38.3/4 (1951):
 330-336. Examples -------- 1. One-way Welch ANOVA on the pain threshold dataset. Source ddof1 ddof2 F p-unc 0 Hair color 3 8.33 5.89 0.018813 """

    # Check data
    _check_dataframe(dv=dv, between=between, data=data, effects='between')

    # Reset index (avoid duplicate axis error)
    data = data.reset_index(drop=True)

    # Number of groups
    r = data[between].nunique()
    ddof1 = r - 1

    # Compute weights and ajusted means
    grp = data.groupby(between)[dv]
    weights = grp.count() / grp.var()
    adj_grandmean = (weights * grp.mean()).sum() / weights.sum()

    # Treatment sum of squares
    ss_tr = np.sum(weights * np.square(grp.mean() - adj_grandmean))
    ms_tr = ss_tr / ddof1

    # Calculate lambda, F-value and p-value
    lamb = (3 * np.sum((1 / (grp.count() - 1)) *
                       (1 - (weights / weights.sum()))**2)) / (r**2 - 1)
    fval = ms_tr / (1 + (2 * lamb * (r - 2)) / 3)
    pval = f.sf(fval, ddof1, 1 / lamb)

    # Create output dataframe
    aov = pd.DataFrame({'Source': between,
                        'ddof1': ddof1,
                        'ddof2': 1 / lamb,
                        'F': fval,
                        'p-unc': pval,
                        }, index=[0])

    col_order = ['Source', 'ddof1', 'ddof2', 'F', 'p-unc']
    aov = aov.reindex(columns=col_order)
    aov[['F', 'ddof2']] = aov[['F', 'ddof2']].round(3)

    # Export to .csv
    if export_filename is not None:
        _export_table(aov, export_filename)
    return aov