1. Eighth grade students in a certain state are required to take a standardized acheivement test in mathematics. State-wide, scores on the exam follow a normal distribution with a mean of 760 out of a possible 1000 and a standard deviation of 70. The mean score at a particular school with 150 eighth graders is 800. Is this significantly greater than state-wide mean?

Solution The null hypothesis in this case is that the eighth graders in this school had the same distribution of scores as students in the rest of the state and hence the same mean, 760. The alternative hypothesis was that the mean for the students in the school was higher than 760.

Since the standard deviation for the sample distribution is known and the exam scores follow a normal distribution, we will use the z-score as a measure of significance.

This is off the scale, and is clearly significant at any reasonable level you care to set.

2. The 1958 Detroit Area Study was an important sociological investigation of the influence of religion on everyday life. The sample was basically a simple random sample of the population of the metropolitan area. Of the 656 respondents, 267 were white Protestants and 230 were white Catholics. One question asked whether the government was doing enough in areas such as housing, unemployment, and education; 161 of the Protestants and 136 of the Catholics said No. Is there evidence that white Protestants and white Catholics differed on this issue?

Solution Since the question doesn't tell us to make any assumptions about the direction in which these two groups differed, we will use a two-sided significance test with a null hypothesis that says that the proportion of people answering No is the same for both groups and equals the pooled proportion for the entire survey:

The observed proportions for the two groups are

The standard error of the difference of the two proportions is

The statistic we will use to evaluate the two-sided alternative is

This is not significant at all.

3. In January 1975, the Committee on Drugs of the American Academy of Pediatrics recommended that tetracycline drugs not be given to children under the age of 8. A two-year study conducted in Tennessee investigated the extent to which physicians had prescribed these drugs between 1973 and 1975. The study categorized family practice physicians according to whether the county of their practice was urban, intermediate, or rural. The researchers examined how many doctors in each of these categories prescribed tetracycline to at least one patient under the age of 8. Here is the table of observed counts

	Urban	Intermediate	Rural
Tetracycline	65	90	172
No tetracycline	149	136	158

Carry out a significance test to determine whether county type and tetracycline use are related.

Solution To carry out a test we start by computing row and column totals.

Tetracycline	No tetracycline
327	443

Urban	Intermediate	Rural
214	226	330

Using the formula

with n = 770 we can compute the expected counts.

	Urban	Intermediate	Rural
Tetracycline	90.9	96.0	140.1
No tetracycline	123.1	130.0	189.9

Next we compute the statistic.

= 26.1

With (2 - 1)(3 - 1) = 2 degrees of freedom, this is significant below the 0.0005 level.

4. A bank compares two proposals to increase the amount that its credit card customers charge on their cards. Proposal A offers to eliminate the annual fee for customers who charge $2400 or more during the year. Proposal B offers a small percent of the total amount charged as a cash rebate at the end of the year. The bank offers each proposal to an SRS of 150 of its existing credit card customers. At the end of the year, the total amount charged by each customer is recorded. Here are the summary statistics:

	n		s
A	150	$1987	$392
B	150	$2056	$413

Is there a significant difference between the amounts charged by the two groups?

Solution Since we are dealing with the difference in the means of two numerical variables and we only have s to work with, we will do a significance test based on a t test. The standard error for the difference of the two groups is

The t-score is

To carry out a significance test we will use the two sided alternative. The probility that is at least 1.48 with 149 degrees of freedom is about 2(0.075) = 0.15. This is not significant.