To test the efficacy of a certain diet cat food, a random sample of individuals is asked to weigh their cat feed the cat the diet food for a month and weigh their cat a second time The weight of the cat is recorded each time. The data (in pounds) were recorded as: Subject: 1 \( \qquad \) 2 3 4 \begin{tabular}{l} \( 5 \quad 6 \quad 7 \) \\ \hline \end{tabular} Pre-diet weight: 155 m> 418137 104 131896 Post-diet weight: \( \begin{array}{lllllll}53.3 & 115 & 14.4 & 13.3 & 101 & 12.9 & 9.7\end{array} \) A paired \( t \) test is conducted on the data, with the difference Pre-diet - Post-diet paired by subject and \( \mathrm{H}_{2} \)-uDIFF \( >0 \) Assume that the paired differences come from a normally distributed population. The p -value for this paired t test is \( = \) 0.036 0.157 0.190

When conducting a paired t-test, you essentially compare the means of two related groups to determine if there is a statistically significant difference between them. In this scenario, we assess the effectiveness of the diet cat food by comparing the weights of the cats before and after the diet. A lower p-value, generally below 0.05, suggests that there is significant evidence against the null hypothesis. Thus, a p-value of 0.036 indicates that the diet likely had a meaningful impact on the cats' weights. If you're calculating the p-value yourself, make sure to gather the differences in weights (Pre-diet - Post-diet) for each subject. Sum these differences, then compute the mean and standard deviation of these deviations. Finally, use a t-distribution to find the p-value based on your calculated t-statistic. A common mistake is forgetting to check the assumptions of normality for the differences or miscalculating the differences themselves!