Question

A researcher wishes to see if there is a difference in the manual dexterity of athletes and that of band members. Two random samples of 23 are selected from each group and are given a manual dexterity test. The mean of the athletes' test was 87, and the mean of the band members' test was 89. The population standard deviation for the test is 7.5. Use ₁ for the mean of the athletes' test. At =α0.01, is there a difference in the mean scores?

Answer 1

To determine if there is a difference in the mean scores of athletes and band members, perform a hypothesis test using a z-test with a known population standard deviation.

Step-by-step explanation:

To determine if there is a difference in the mean scores of athletes and band members, we can perform a hypothesis test. Since the population standard deviation is known, we can use a z-test.

The null hypothesis (H0) is that there is no difference in the mean scores, while the alternative hypothesis (Ha) is that there is a difference.

Using a significance level of α = 0.01, we calculate the test statistic using the formula:

z = (mean1 - mean2) / (σ / sqrt(n))

where mean1 is the mean of the athletes' test, mean2 is the mean of the band members' test, σ is the population standard deviation, and n is the sample size. If the test statistic falls in the rejection region (critical region) of the z-distribution, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

In this case, the test statistic is z = (87 - 89) / (7.5 / sqrt(23)) =~ -0.526.

Comparing the test statistic to the critical value of z at α = 0.01, we find that the critical value is -2.33. Since the test statistic -0.526 does not fall in the rejection region (|-0.526| < |-2.33|), we fail to reject the null hypothesis. Therefore, there is no evidence to suggest a difference in the mean scores of athletes and band members.