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.

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 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

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:

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:

Distribution of the difference:

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

The confidence interval is:

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

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation: