A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below.
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a. Use a 0.05 significance level to test the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests.

What are the null and alternative hypotheses?
A.
$\begin{array}{l} H_{0} : μ_{1} = μ_{2} \\ H_{1} : μ_{1} \neq μ_{2} \end{array}$
C.
$\begin{array}{l} H_{0} : μ_{1} = μ_{2} \\ H_{1} : μ_{1} > μ_{2} \end{array}$
B.
$\begin{array}{l} H_{0} : μ_{1} \neq μ_{2} \\ H_{1} : μ_{1} < μ_{2} \end{array}$
D.
$\begin{array}{l} H_{0} : μ_{1} = μ_{2} \\ H_{1} : μ_{1} < μ_{2} \end{array}$

The test statistic, $t$ , is -1.79 . (Round to two decimal places as needed.)

The P-value is 0.042 . (Round to three decimal places as needed.)
State the conclusion for the test.
A. Fail to reject $H_{0}$ . There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests.
B. Reject $H_{0}$ . There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests.
C. Reject $H_{0}$ . There is not sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests.
D. Fail to reject $H_{0}$ . There is not sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests.
b. Construct a confidence interval suitable for testing the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests.
$0.02 < μ_{1} - μ_{2} < 13.86$
(Round to two decimal places as needed.)

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The final answer is: The null and alternative hypotheses are: $H_{0} : μ_{1} = μ_{2}$ and $H_{1} : μ_{1} < μ_{2}$ . The conclusion of the test is: Reject $H_{0}$ . There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. The confidence interval for the difference in means is approximately $(- 0.98, 14.82)$ .

Steps

Step 1 ：Define the null and alternative hypotheses. The null hypothesis $H_{0}$ is that the means of the proctored and nonproctored tests are equal, i.e., $μ_{1} = μ_{2}$ . The alternative hypothesis $H_{1}$ is that the mean of the nonproctored tests is greater than the mean of the proctored tests, i.e., $μ_{1} < μ_{2}$ .

Step 2 ：Calculate the test statistic and the P-value. The test statistic is -1.79 and the P-value is 0.042.

Step 3 ：Compare the P-value with the significance level. The P-value (0.042) is less than the significance level (0.05), so we reject the null hypothesis $H_{0}$ . There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests.

Step 4 ：Construct a confidence interval for the difference in means. The confidence interval is approximately (-0.98, 14.82). This interval does not contain zero, which suggests that there is a significant difference in the means.

Step 5 ：The final answer is: The null and alternative hypotheses are: $H_{0} : μ_{1} = μ_{2}$ and $H_{1} : μ_{1} < μ_{2}$ . The conclusion of the test is: Reject $H_{0}$ . There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. The confidence interval for the difference in means is approximately $(- 0.98, 14.82)$ .

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