6. 9. Confidence interval for the difference between the

population proportions: (large samples)

As it was discussed earlier, for a large sample the sample proportion is approximately normally distributed with mean p and standard deviation

.

Suppose that a random sample of size observations from a population with proportion of “success” has a sample proportion of success , and that an independent random sample of size observations from a population with proportion of “success” yields sample proportion .

Since and both are large, their sample proportions and are approximately normally distributed with means and , and standard deviations

and

respectively.

Then the random variable , and the variance

.

The standardized random variable

is approximately standard normal.

In order to find confidence interval for , we must either know or estimate the quantity of

.

We can estimate the population proportion by the sample proportion ; and we can estimate the population proportion by the sample proportion . Then

Then an approximate confidence interval is given by

where

.

Example:

Mike and Tom like to throw darts. Mike throws 100 times and hits the target 54 times; Tom throws 100 times and hits the target 49 times. Find a 95 % confidence interval for , where represents the true proportion of hits in Mike’s tosses, and represents the true proportion of hits in Tom’s tosses.

Solution:

and

Thus, with 95 % confidence we can state that the difference between the proportions of Mike’s and Tom’s tosses is between -0.088 and 0.188.

Exercises

1. Find a 90 % confidence interval for , if a sample of size 200 yielded and a sample of size 300 yielded .

2. Construct a 99 % confidence interval for if

;

;

3. A sample of 400 observations taken from the first population gave . Another sample of 700 observations taken from the second population gave . Make a 96 % confidence interval for .

4. A sample of 500 items produced by a supplier A possessed 270 defective items. A random sample of 360 items produced by supplier B possessed 162 defective items. Compute a 95 % confidence interval estimate for the difference in proportion defective from the two suppliers.

5. Assume that 66 % of single women and 81.9 % of single men own cars. Also assume that these estimates are based on random samples of 1640 single women and 1800 single men. Develop a 99 % confidence interval for the difference between the two population proportions.

6. The management of a market wanted to investigate if the percentage of men and women who prefer to buy national brand products over the store brand products are different. A sample of 500 men shoppers at supermarkets showed that 175 of them prefer to buy national brand products over the store brand products. Another sample of 800 women shoppers showed that 360 of them prefer to buy national brand products over the store brand products.

Construct a 95 % confidence interval for the difference between the proportions of all men and women shoppers at supermarket who prefer to buy national brand products over the store brand products.

7. A sample of 600 females was selected from ethnic group A and a sample of 700 from ethnic group B. Each female was asked “Did you get married before you were 22?”. 246 of females from group A and 266 of females from group B answered “yes”. Find a 95 % confidence interval for the two population proportions.

8. According to a survey, 1010 adults conducted and 74.2 % of male and 88.8 % of women said that they are concerned about living near a nuclear power plant. Assume that there were 520 men and 490 women in this sample.

Construct a 99 % confidence interval for the difference between the proportions of all men and all women who are concerned about living near power plant.

Answers

1. (-0.02; 0.12); 2. (-0.18; 0.06); 3. (-0.006; 0.116); 4. (0.023; 0.157);

5. (-0.20; -0.12); 6. (-0.154; 0.046); 7. (-0.02; 0.08); 8.(-0.21; -0.08).