An experiment was conducted to determine whether giving candy to dining parties resulted in greater tips. The mean tip percentages and standard deviations are given in the accompanying table along with the sample sizes. Assume that the two samples are independent simple random
\begin{tabular}{|c|c|c|c|c|}
\hline & $\boldsymbol{\mu}$ & $\mathbf{n}$ & $\overline{\mathbf{x}}$ & $\mathbf{s}$ \\
\hline No candy & $\mu_{1}$ & 22 & 19.41 & 1.44 \\
\hline Two candies & $\mu_{2}$ & 22 & 21.91 & 2.57 \\
\hline
\end{tabular}
samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complęte parts (a) and (b).
Ine $r$-value is $u$. (Kound to three decimal places as needed.)
State the conclusion for the test.
A. Reject the null hypothesis. There is not sufficient evidence to support the claim that giving candy does result in greater tips.

B. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that giving candy does result in greater tips.
C. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that giving candy does result in greater tips.

D. Reject the null hypothesis. There is sufficient evidence to support the claim that giving candy does result in greater tips.
b. Construct the confidence interval suitable for testing the claim in part (a).
\[
< \mu_{1}-\mu_{2}<
\]
(Round to two decimal places as needed.)

Step 1 ：Given the following data for two groups, 'No candy' and 'Two candies':

Step 2 ：\(n_1 = 22, \overline{x}_1 = 19.41, s_1 = 1.44\) for 'No candy' group

Step 3 ：\(n_2 = 22, \overline{x}_2 = 21.91, s_2 = 2.57\) for 'Two candies' group

Step 4 ：We are asked to perform a hypothesis test to determine if there is a significant difference in the mean tip percentages between the two groups. The null hypothesis \(H_0\) is that there is no difference in the means (i.e., the mean difference is zero), and the alternative hypothesis \(H_1\) is that there is a difference in the means (i.e., the mean difference is not zero).

Step 5 ：We can use a two-sample t-test to test this hypothesis. The t-statistic and p-value will tell us whether we should reject or fail to reject the null hypothesis.

Step 6 ：If the p-value is less than 0.05, we reject the null hypothesis and conclude that there is a significant difference in the mean tip percentages between the two groups.

Step 7 ：If the p-value is greater than 0.05, we fail to reject the null hypothesis and conclude that there is not a significant difference in the mean tip percentages between the two groups.

Step 8 ：The final answer depends on the p-value. If the p-value is less than 0.05, we reject the null hypothesis and conclude that there is a significant difference in the mean tip percentages between the two groups (option D). If the p-value is greater than 0.05, we fail to reject the null hypothesis and conclude that there is not a significant difference in the mean tip percentages between the two groups (option B).

Step 9 ：Finally, we need to construct the confidence interval suitable for testing the claim. The confidence interval is given by \(\mu_{1}-\mu_{2}\).