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

Answer:

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 $\mu$ 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 $s = \frac{\sigma}{\sqrt{n}}$.

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 $s = \sqrt{\frac{p(1-p)}{n}}$

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:

$p_1 = 0.39$

$s_1 = \sqrt{\frac{0.39*0.61}{640}} = 0.0193$

Eastern residents:

51% out of 540, so:

$p_2 = 0.51$

$s_2 = \sqrt{\frac{0.51*0.49}{540}} = 0.0215$

Distribution of the difference:

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

$s = \sqrt{s_2^2+s_1^2} = \sqrt{0.0215^2+0.0193^2} = 0.0289$

Confidence interval:

$p \pm zs$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$, 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).