Question

A certain chromosome defect occurs in only 1 in 200 adult Caucasian males. A random sample of 100 adult Caucasian males will be selected. The proportion of men in this sample who have the defect, \(\hat{p},\) will be calculated. a. What are the mean and standard deviation of the sampling distribution of \(\hat{p}\) ? b. Is the sampling distribution of \(\hat{p}\) approximately normal? Explain. c. What is the smallest value of \(n\) for which the sampling distribution of \(\hat{p}\) is approximately normal?

Short answer

a. Mean is 0.005 and the standard deviation is 0.022. b. No, the sample distribution of \(\hat{p}\) is not approximately normal as \(np = 0.5 < 5\) and \(n(1 - p) = 99.5 > 5\). c. Minimum sample size, \(n\), for \(\hat{p}\) to be approximately normal is 1000.

Step-by-step solution

Calculate Mean and Standard Deviation of \(\hat{p}\)

In this step, we utilize the formulas for mean and standard deviation of a sample proportion. The mean (\(\mu_{\hat{p}}\)) is equal to the success proportion in the population (\(p\)), which is \(\frac{1}{200}\) or 0.005 in this case. The standard deviation (\(\sigma_{\hat{p}}\)) is calculated by the formula, \(\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}\), where \(n\) is the sample size of 100 men.

Check the Normality of the Sampling Distribution

The normality of the sampling distribution can be justified by applying the rule of thumb which states that, for the sampling distribution of a proportion to be approximately normal, both \(np\) and \(n(1 - p)\) should be greater than or equal to 5. Given \(p = 0.005\) and \(n = 100\), we check whether both the quantities are greater than 5 or not.

Find the Minimum Sample Size for Normality

We must find the smallest value of \(n\) which makes the sampling distribution approximately normal. This value can be calculated by ensuring that both \(np\) and \(n(1 - p)\) are greater than or equal to 5.