Question

A telephone service representative believes that the proportion of customers completely satisfied with their local telephone service is different between the South and the Midwest. The representative's belief is based on the results of a survey. The survey included a random sample of 1300 southern residents and 1380 midwestern residents. 39% of the southern residents and 50% of the midwestern residents reported that they were completely satisfied with their local telephone service. Find the 80% confidence interval for the difference in two proportions. Step 1 of 3 : Find the point estimate that should be used in constructing the confidence interval

Answer 1

Answer:

The point estimate that should be used in constructing the confidence interval is 0.11.

The 80% confidence interval for the difference in two proportions is (0.0856, 0.1344).

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.

Midwest:

50% of 1380, so:

$p_M = 0.5$

$s_M = \sqrt{\frac{0.5*0.5}{1380}} = 0.0135$

South:

39% of 1300, so:

$p_S = 0.39$

$s_S = \sqrt{\frac{0.39*0.61}{1300}} = 0.0135$

Distribution of the difference:

$p = p_M - p_S = 0.5 - 0.39 = 0.11$

So the point estimate that should be used in constructing the confidence interval is 0.11.

$s = \sqrt{s_M^2+s_S^2} = \sqrt{0.0135^2+0.0135^2} = 0.0191$

Confidence interval:

$p \pm zs$

In which

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

80% confidence level

So $\alpha = 0.2$, z is the value of Z that has a p-value of $1 - \frac{0.2}{2} = 0.9$, so $Z = 1.28$.  

The lower bound of the interval is:

$p - zs = 0.11 - 1.28*0.0191 = 0.0856$

The upper bound of the interval is:

$p + zs = 0.11 + 1.28*0.0191 = 0.1344$

The 80% confidence interval for the difference in two proportions is (0.0856, 0.1344).