Question

Researchers often use z tests to compare their samples to known population norms. The Graded Naming Test (GNT) asks respondents to name objects in a set of 30 black-and-white drawings. The test, often used to detect brain damage, starts with easy words like kangaroo and gets progressively more difficult, ending with words like sextant. The GNT population norm for adults in England is 20.4. Roberts (2003) wondered whether a sample of Canadian adults had different scores than adults in England. If they were different, the English norms would not be valid for use in Canada. The mean for 30 Canadian adults was 17.5. For the purposes of this exercise, assume that the standard deviation of the adults in England is 3.2.

A. Conduct all six steps of a z test. Be sure to label all six steps.

B. When we conduct a one-tailed test instead of a two-tailed test, there are small changes in steps 2 and 4 of hypothesis testing. (Note: For this example, assume that those from populations other than the one on which it was normed will score lower, on average. That is, hypothesize that the Canadians will have a lower mean.) Conduct steps 2, 4, and 6 of hypothesis testing for a one-tailed test.

C. When we change the p level that we use as a cutoff, there is a small change in step 4 of hypothesis testing. Although 0.05 is the most commonly used p level, other values, such as 0.01, are often used. For this example, conduct steps 4 and 6 of hypothesis testing for a two-tailed test and p level of 0.01, determining the cutoff and drawing the curve.

D. If it is easier to reject the null hypothesis with certain plevels, does this mean that there is a bigger difference between the samples with one p level versus the other plevel? Explain.

Answer 1

Given sample mean = 17.5, sample size n = 30

population std. dev= 3.2

population mean = 20.4

step -1)

null ypotheis H₀ : = 20.4 The mean score of the canadians is equal to mean score of The English.

step 2)

Alternative Hypothesis H₁ : 20.4 The mean score of the canadians is different from mean score of The English

step -3)

Critical value given level of significance = 0.05

Critical value = Z_0.025 = 1.96

step -4) Test statistic Z = = (17.5 - 20.4)/(3.2/30^0.5) = -4.96

|Z| = 4.96

step 5) |Z| >Z_0.025 hence we reject null hypothesis.

step 6)Since we rejected null hypothesis we conclude that there is a significant evidence that canadian score differ from England standards and Canada require different standards.

B) for one tailed test

Alternative Hypothesis H₁ : 20.4 The mean score of the canadians is different from mean score of The English

critical value Z_0.05 = 1.645

Since Z < -Z_0.05 we reject null hypothesis and conclude that that there is a significant evidence that canadian score are lesser than England standards and Canada require lesser standards.

C) Level of significance = 0.01

Critical value = Z_0.005 = 2.58

|Z| >Z_0.005 hence we reject null hypothesis.

Since we rejected null hypothesis we conclude that there is a significant evidence that canadian score differ from England standards and Canada require different standards.