The survey included a random sample of 640 western residents and 540 northeastern residents. 39% of the western residents and 51% of the northeastern residents reported that they were completely satisfied with their local telephone service. Find the 99% confidence interval for the difference in two proportions

The 99% confidence interval for the difference in two proportions is (0.0456, 0.1944).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Western residents:

39% out of 640, so:

Eastern residents:

51% out of 540, so:

Distribution of the difference:

Confidence interval:

In which

z is the z-score that has a p-value of .

99% confidence level

So , z is the value of Z that has a p-value of , so .

The lower bound of the interval is:

The upper bound of the interval is:

The 99% confidence interval for the difference in two proportions is (0.0456, 0.1944).