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Because earnings cannot take on negative values and the standard deviation is almost the same as the mean, the distribution of earnings in each group is strongly skewed to the right. Because negative values are within two standard deviations of the mean and these values fall within the range of regular values for a symmetric distribution, the distribution must be strongly skewed. Because the sample sizes for $675$and $621$are both at least 30, we can continue to assume that the distribution is roughly normal.

$$\begin{gathered} {\stackrel{¯}{x}}_1=1884.52 \\ \stackrel{¯}{x_2}=1360.39 \\ {n}_1=376 \\ {n}_2=621 \\ {s}_1=1368.37 \\ {s}_2=1037.46 \\ c=0.90 \end{gathered}$$

Confidence interval calculation:

Now we'll calculate the $t$-value, and to do so, we'll need to know how many degrees of freedom there are. As a result, the degree of liberty will be:

$$\begin{gathered} df=min(n_1-1,n_2-1) \\ =min(675-1,621-1) \\ =620>100 \\ df=100 \end{gathered}$$

Then the $t$-value will be as:

$t^{*}=1.660$

As a result, the confidence interval will be:

localid="1654742561555" $$\begin{gathered} ({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2)-t_{\frac{\mathrm{Î\pm }}{2}}×\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}} \\ =(1884.52-1360.39)-1.660×\sqrt{\frac{1368.{37}^2}{675}+\frac{1037.{46}^2}{621}} \\ =412.69 \end{gathered}$$

$$\begin{gathered} ({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2)+t_{\frac{\mathrm{Î\pm }}{2}}×\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}} \\ =(1884.52-1360.39)+1.660×\sqrt{\frac{1368.{37}^2}{675}+\frac{1037.{46}^2}{621}} \\ =635.58 \end{gathered}$$

Therefore, we conclude that there is $90\%$the confidence that the mean difference is between $(412.69,635.58)$

The confidence interval of $90\%$means that $90\%$of all possible samples will have a confidence interval that contains the true population mean difference in earnings. This can be deduced based on the study's context.