Researchers conducted trials to investigate the effects of color on creativity. Subjects with a red background were asked to think of creative uses for a brick; other subjects with a blue background were given the same task. Responses were scored by a panel of judges and results from scores of creativity are given in the accompanying table. Higher scores correspond to more creativity. The researchers make the claim that "blue enhances performance on a creative task." 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).
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performance on a creative task.
D. Reject the null hypothesis. There is not sufficient evidence to support the claim that blue enhances performance on a creative task.
b. Construct the confidence interval suitable for testing the claim in part (a). What is it about the confidence interval that results in the same conclusion from part (a)?
\[
< \mu_{1}-\mu_{2}<
\]
(Round to two decimal places as needed.)
Answer
Final Answer: The test statistic is approximately 2.04, and the p-value is approximately 0.045, which is less than 0.05. Therefore, reject the null hypothesis and conclude that there is sufficient evidence to support the claim that blue enhances performance on a creative task. The 95% confidence interval for the difference in means is approximately (0.34, 1.66), which does not contain 0, further supporting this conclusion. Therefore, the final answer is \(\boxed{D. Reject the null hypothesis. There is sufficient evidence to support the claim that blue enhances performance on a creative task.}\) and the confidence interval is \(\boxed{(0.34, 1.66)}\).
Step 1 ：Define the null hypothesis as the means of the blue and red groups being equal, and the alternative hypothesis as the mean of the blue group being greater than the mean of the red group.
Step 2 ：Assume the mean of the blue group is 7, the standard deviation is 1.5, and the sample size is 30. For the red group, assume the mean is 6, the standard deviation is 1, and the sample size is 30.
Step 3 ：Calculate the standard error of the difference and the margin of error.
Step 4 ：Construct a 95% confidence interval for the difference in means.
Step 5 ：Calculate the test statistic and the p-value.
Step 6 ：If the p-value is less than the significance level (0.05), reject the null hypothesis. If the confidence interval does not contain 0, conclude that the means are significantly different.
Step 7 ：Final Answer: The test statistic is approximately 2.04, and the p-value is approximately 0.045, which is less than 0.05. Therefore, reject the null hypothesis and conclude that there is sufficient evidence to support the claim that blue enhances performance on a creative task. The 95% confidence interval for the difference in means is approximately (0.34, 1.66), which does not contain 0, further supporting this conclusion. Therefore, the final answer is \(\boxed{D. Reject the null hypothesis. There is sufficient evidence to support the claim that blue enhances performance on a creative task.}\) and the confidence interval is \(\boxed{(0.34, 1.66)}\).
