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.

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.

please help me by actually writing step by step what you are doing so I can understand clearer. do not ONLY use equations to reply please

Answer 1

Consider

X : GNT score for Canadians

n = number of Canadians selected = 30

Xbar: Sample mean GNT score for Canadians = 17.5

From the information

mu : Population norm for adults in England= 20.4

sigma : Standard deviation of adults in England = 3.2

We have to test the hypothesis that

Whether or not Canadians adults had lower GNT score than the adults in England ?

i.e. Null Hypothesis-

against

Alternative Hypothesis- ( Left-tailed test).

Since samples are coming from normal population and population standard deviation is known. We used one sample Z-test for testing population mean.

The value of test statistic is

Z = -4.9640

Alpha : Level of significance = 0.05

Critical region : Since the test is left-tailed we reject the null Hypothesis

if

Critical value: From normal probability table, at 5% level of significance critical value is

Decision: Since calculated value of test statistic is less than critical value, we reject the null hypothesis at 5% level of significance.

p-value approach

since test is left-tailed and value of test statistic is -4.9640, p-value is obtained by

By using R - software

> pnorm(-4.9640,0,1)
[1] 3.452797e-07

p-value = 3.4527 * 10^-7.

Since p-value is very small as compared to level of significance, we reject the null hypothesis at 5% level of significance.

Conclusion: There is sufficient evidence support to claim that Canadians adults had lower GNT score than the adults in England