Problem
You believe both populations are normally distributed, but you do not know the standard either. Let's assume that the variances of the two populations are not equal. You obtain \( t \) samples of data. Sample \#1 \begin{tabular}{|r|r|r|} \hline 87.1 & 65.4 & 72.3 \\ \hline 71.9 & 78.1 & 83 \\ \hline 65.1 & 77 & 57.1 \\ \hline 66.7 & 68 & \\ \hline \end{tabular} Sample \#2 \begin{tabular}{|l|l|l|} \hline 76.5 & 83.3 & 61.8 \\ \hline 86.5 & 75.4 & 81.7 \\ \hline 88.2 & 77.4 & 79.9 \\ \hline \end{tabular} What is the test statistic for this sample? (Keep sample statistics rounded to 3 decimal plac answer accurate to three decimal places.) test statistic \( =-1.8878 \quad \checkmark \quad 0^{\delta} \) What is the \( p \)-value for this sample? For this calculation, use the degrees of freed technology you are using. (Report answer accurate to four decimal places.) \( p \)-value \( = \) The \( p \)-value is... less than (or equal to) \( \alpha \) greater than \( \alpha \) Type here to search
Solution
Step 1 ：Calculate sample means: \(\bar{x}_{1} = (87.1 + 65.4 + 72.3 + 71.9 + 78.1 + 83 + 65.1 + 77 + 57.1 + 66.7 + 68)/11 = 71.063 \); \(\bar{x}_{2} = (76.5 + 83.3 + 61.8 + 86.5 + 75.4 + 81.7 + 88.2 + 77.4 + 79.9)/9 = 77.855 \)
Step 2 ：Calculate sample standard deviations: \(s_{1} = \sqrt{\frac{(87.1 - 71.063)^2 + \cdots + (68 - 71.063)^2}{10}} = 9.460 \); \(s_{2} = \sqrt{\frac{(76.5 - 77.855)^2 + \cdots + (79.9 - 77.855)^2}{8}} = 8.977 \)
Step 3 ：Calculate Welch's t-test statistic: \(t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}} = \frac{71.063 - 77.855}{\sqrt{\frac{9.460^2}{11} + \frac{8.977^2}{9}}} = -1.888 \)
