Question

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

Answer 1

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

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
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

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
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

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:

`p_1 = 0.39`

`s_1 = \sqrt{(0.39*0.61)/(640)} = 0.0193`

Eastern residents:

51% out of 540, so:

`p_2 = 0.51`

`s_2 = \sqrt{(0.51*0.49)/(540)} = 0.0215`

Distribution of the difference:

`p = p_2 - p_1 = 0.51 - 0.39 = 0.12`

`s = √(s_2^2+s_1^2) = √(0.0215^2+0.0193^2) = 0.0289`

Confidence interval:

`p \pm zs`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a p-value of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

The lower bound of the interval is:

`p - zs = 0.12 - 2.575*0.0289 = 0.0456`

The upper bound of the interval is:

`p + zs = 0.12 + 2.575*0.0289 = 0.1944`

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