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Test the claim that for the population of statistics final exams, the mean score is 71 using alternative hypothesis that the mean score is different from 71. Sample statistics include n = 26, ã = 72, and s = 18. Use a significance level of a = 0.05. (Assume x : normally distributed population.) = a a The test statistic is 0.283 The positive critical value is 1.645 The negative critical value is 1.645

Sagot :

Since the test statistic is not greater than the critical value, the null hypothesis cannot be rejected. Therefore, the claim that the mean score is 71 is not rejected.

1. We are given the population of statistics final exams and asked to test the claim that the mean score is 71 using an alternative hypothesis that the mean score is different from 71. We are also given the sample statistics including the sample size (n = 26), the sample mean (ã = 72), and the sample standard deviation (s = 18). We are asked to use a significance level of α = 0.05.

2. Since we are assuming that the population is normally distributed, we will use a two-tailed z-test to test the claim.

3. The null hypothesis is that the mean score is 71 and the alternative hypothesis is that the mean score is different from 71.

4. The test statistic is calculated as:

Test statistic = (sample mean - population mean) / (standard deviation/√n)

= (72 - 71) / (18/√26)

= 0.283

5. The critical values for a two-tailed z-test with a significance level of α = 0.05 are 1.645 for both the positive and negative critical values.

6. Since the test statistic (0.283) is not greater than the critical value

Learn more about null hypothesis here

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