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 \( \alpha \)
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Calculate Welch's t-test statistic: \(t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}} = \frac{71.063 - 77.855}{\sqrt{\frac{9.460^2}{11} + \frac{8.977^2}{9}}} = -1.888 \)
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: \(t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}} = \frac{71.063 - 77.855}{\sqrt{\frac{9.460^2}{11} + \frac{8.977^2}{9}}} = -1.888 \)