Question

A BYU-Idaho professor took a survey of his classes and found that 82 out of 90 people who had served a mission had personally met a member of the quorum of the twelve apostles. Of the non-returned missionaries that were surveyed 86 of 110 had personally met a member of the quorum of the twelve apostles. Calculate a 99% confidence interval for the difference in the two proportions.

Answer 1

Answer:

The 99% confidence interval for the difference in the two proportions is (-0.0247, 0.2833).

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

A BYU-Idaho professor took a survey of his classes and found that 82 out of 90 people who had served a mission had personally met a member of the quorum of the twelve apostles.

This means that:

$p_S = \frac{82}{90} = 0.9111$

$s_S = \sqrt{\frac{0.9111*0.0888}{90}} = 0.045$

Of the non-returned missionaries that were surveyed 86 of 110 had personally met a member of the quorum of the twelve apostles.

This means that:

$p_N = \frac{86}{110} = 0.7818$

$s_N = \sqrt{\frac{0.7818*0.2182}{110}} = 0.0394$

Distribution of the difference:

$p = p_S - p_N = 0.9111 - 0.7818 = 0.1293$

$s = \sqrt{s_S^2+s_N^2} = \sqrt{0.045^2+0.0394^2} = 0.0598$

Calculate a 99% confidence interval for the difference in the two proportions.

The confidence interval is:

$p \pm zs$

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:

$p - zs = 0.1293 - 2.575*0.0598 = -0.0247$

$p + zs = 0.1293 + 2.575*0.0598 = 0.2833$

The 99% confidence interval for the difference in the two proportions is (-0.0247, 0.2833).