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.

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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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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 [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

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:

[tex]p_S = \frac{82}{90} = 0.9111[/tex]

[tex]s_S = \sqrt{\frac{0.9111*0.0888}{90}} = 0.045[/tex]

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:

[tex]p_N = \frac{86}{110} = 0.7818[/tex]

[tex]s_N = \sqrt{\frac{0.7818*0.2182}{110}} = 0.0394[/tex]

Distribution of the difference:

[tex]p = p_S - p_N = 0.9111 - 0.7818 = 0.1293[/tex]

[tex]s = \sqrt{s_S^2+s_N^2} = \sqrt{0.045^2+0.0394^2} = 0.0598[/tex]

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

The confidence interval is:

[tex]p \pm zs[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

99% confidence level

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].

The lower bound of the interval is:

[tex]p - zs = 0.1293 - 2.575*0.0598 = -0.0247[/tex]

[tex]p + zs = 0.1293 + 2.575*0.0598 = 0.2833[/tex]

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