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. Conduct all six steps of a z test. Step one: Identify the populations, Distribution, and assumptions. Step two: State the null and research hypothesis, in both words and symbolic notation. Step 3:Determine the characteristics of the comparison distribution. Step four: Determine the critical values, or cutt-offs, indicate the points beyond which we will reject the null hypothesis. Step five: Calculate the test statistic. Step six: Decide whether to reject or fail to reject the null hypothesis.

Answer 1

Step one: Identify the populations, Distribution, and assumptions.

We are interested in checking if Canadian adults GNT scores different than adults in England. Hence the population of interest is the adults in Canada.

We do not know the distribution of scores. But we will assume that the scores for adults in Canada have the same standard deviation as the scores for adults in England and it is 3.2

Step two: State the null and research hypothesis, in both words and symbolic notation.

Let be the true mean GNT population norm for adults in Canada. We want to test if   is different from 20.4, which is the mean scores for adults in England. That is .

The hypotheses are

Step 3:Determine the characteristics of the comparison distribution.

We have a sample of size n=30 Canadian adults. We also know the population standard deviation of scores of Canadian adults. Hence using the central limit theorem we can say that the sampling distribution of mean has normal distribution.

We have the following sample information

n=30 is the sample size

is the sample mean score for Canadian adults

is the population standard deviation of scores for Canadian adults

is the hypothesized value of mean score of Canadian adults (from the null hypothesis)

We can say that the sampling distribution of mean has normal distribution with mean and standard deviation

Step four: Determine the critical values, or cutt-offs, indicate the points beyond which we will reject the null hypothesis.

We will assume that we want to test the hypothesis at 5% level of significance, (not provided in the question).

This is a 2 tailed test (The alternative hypothesis has "not equal to"). Hence the significance level corresponds to the total area under both the tails is  0.05. The area under each tail is 0.05/2=0.025.

the right tail critical value is

Using the standard normal table, we get that for z=1.96, P(Z<1.96) = 0.975. Hence is the right tail critical value. The left tail critical value is -1.96.

We will reject the null hypothesis, if the test statistics does not lies within -1.96 to +1.96.

Step five: Calculate the test statistic.

The test statistics is

Step six: Decide whether to reject or fail to reject the null hypothesis.

We can see that the test statistics -4.96 does not lie with in the interval -1.96 to +1.96. Hence we reject the null hypothesis.

We conclude that at 5% level of significance there is sufficient evidence to support the claim that Canadian adults had different scores than adults in England.