Question

Test the claim that for the population of statistics final exams, the mean score is 72 using alternative hypothesis that the mean score is different from 72. Sample statistics include n=17, x bar =73, and s=15. Use a significance level of

. (Assume normally distributed population.)

1. The test statistic is ........?

2. The positive critical value is .......?

3. The negative critical value is ........?

4. The conclusion is

A. There is sufficient evidence to reject the claim that the mean score is equal to 72.

B. There is not sufficient evidence to reject the claim that the mean score is equal to 72.

Answer 1

To test the claim that the mean score for the population of statistics final exams is 72, we can perform a hypothesis test using the sample statistics. Based on the test statistic and critical values, we fail to reject the null hypothesis.

Step-by-step explanation:

To test the claim that the mean score for the population of statistics final exams is 72, with an alternative hypothesis stating that the mean score is different from 72, we can perform a hypothesis test using the sample statistics. The sample size (n) is 17, the sample mean (x bar) is 73, and the sample standard deviation (s) is 15.

1. The test statistic can be calculated using the formula: test statistic = (x bar - population mean) / (s / sqrt(n)). Substituting the values, we get: test statistic = (73 - 72) / (15 / sqrt(17)) = 0.3829.
2. The positive critical value can be found using a alpha of 0.05 and the t-distribution. Since we have a two-tailed test, the critical value is t(0.025, 16) = 2.120.
3. The negative critical value is the negative of the positive critical value, so it is -2.120.
4. Based on the test statistic, we compare it with the critical values. Since the test statistic (-0.3829) falls within the range of the critical values (-2.120 to 2.120), we fail to reject the null hypothesis.

Therefore, the conclusion is B. There is not sufficient evidence to reject the claim that the mean score is equal to 72.