6.7.2. Confidence intervals for the difference between

means of two normal populations with known variances

Suppose that the random variable is based on a random sample of

size from a normal population with mean and known variance .

Also suppose that the random variable is based on a random sample of

size from a normal population with mean and known variance .

The difference between population means has a mean and variance .

Therefore, the random variable

has a standard normal distribution.

We can use this fact to obtain confidence interval for the difference between the population means.

Definition:

When the variances and of two normal are known, then

confidence interval for is given by

.

Example:

A sample of size 13 from a normal population with variance 100

yielded . A sample of size 7 from a second normal population with variance 80 yielded .Find a 95 % confidence interval for .

Solution:

and

The 95 % confidence interval for is

Remark:

When and are both large, the normal approximation remains valid if and are replaced by their estimators and . When and are greater than 30, an approximate confidence interval for

is given by

where is the number for which

and Z follows standard normal distribution.

Example:

A sample of 50 yogurt cups produced by the company showed that they contain an average of 146 calories per cup with a standard deviation of

6.4 calories. A sample of 60 such yogurt cups produced by its competitor showed that they contained an average of 143 calories per cup with a standard deviation of 7.2 calories. Make a 97 % confidence interval for the difference between the mean number of calories in yogurt cups produced by the two companies.

Solution:

We can refer to the respective samples as sample 1 and sample 2.

Let and be the means of populations 1 and 2 respectively, and

let and be the means of the respective samples.

From the given information:

; ;

; ;

Since both sample sizes are large ( , ) we can replace and by and respectively.

Then confidence interval for is given by

and

Finally, substituting all the values in the confidence interval formula, we obtain 97 % confidence interval for as

.

Thus, with 97 % confidence we can state that the difference in the mean calories of the two population of yogurt cups produced by two different companies is between 0.18 and 5.82.

Exercises

1. A random sample of size 10 from a normal population with variance 50 gave a mean 43.2. A second random sample of size 18 from a normal population with variance 72 gave a mean 48.7. Find a 99 per cent confidence interval for the difference between two population means.

2. A random sample of size 100 yielded the sample values

. A random sample size 100 from another population yielded

. Find a 95 % confidence interval for .

3. An urban planning group is interested in estimating the difference between mean household incomes for two cities. Independent samples of households in two cities provide the following results:

City 1 City 2

Develop an interval estimate of the difference between mean incomes in the two cities. Show the results for confidence coefficients of 0.90 and 0.95.

4. The management at the National Bank investigates mean waiting time for all customers at its two branches. They took a sample of 200 customers from the branch A and found that they waited an average of 4.60 minutes with a standard deviation of 1.2 minutes before being served. Another sample of 300 customers taken from the branch B showed that these customers waited an average of 4.85 minutes with a standard deviation of 1.5 minutes before being served. Make a 97 % confidence interval for the difference between the two population means.

5. Rural and urban students are to be compared on the basis of their scores on a nationwide university entrance test. Two random samples of sizes 80 and 95 are selected from rural and urban students. The summary statistics from the test scores are

	Rural	Urban
Sample size Mean Standard deviation	80 78.6 9.1	95 85.7 8.3

Establish a 96 % confidence interval for the difference in population mean scores between urban and rural students.

6. A business consultant wanted to investigate if providing day care facilities on premises by companies reduces the absentee rate of working mothers from companies that provide day care facilities on premises. Sample of 50 mothers selected from the companies that provide day care facilities was taken. These mothers missed an average of 6.4 days from work last year with a standard deviation of 1.20 days. Another sample of 50 such mothers taken from companies that do not provide day care facilities on premises showed that these mothers missed an average of 9.3 days last year with a standard deviation of 1.83 days. Construct a 98 % confidence interval for the difference between the two population means.

Answers

1. (-13.24; 2.24); 2. (53.63; 70.37); 3. (68,68; 181.32); (57.89; 192.11);

4. (-0.51; 0.01); 5. (4.37; 9.83); 6. (-3.62 to -2.18 days).