Question

a simple random sample of size ( n = 1,020 ) is obtained from a population whose size is ( n = 1,700,000 ) and whose population proportion with a specified characteristic is ( p = 0.56 ). describe the sampling distribution of ( hat{p} ). round the standard deviation of the sampling distribution to three decimal places

o a. exactly normal, ( mu_{hat{p}} = 0.56, sigma_{hat{p}} = 0.016 )
o b. approximately normal, ( mu_{hat{p}} = 0.56, sigma_{hat{p}} = 0.120 )
o c. approximately normal, ( mu_{hat{p}} = 0.56, sigma_{hat{p}} = 0.016 )
o d. exactly normal, ( mu_{hat{p}} = 0.56, sigma_{hat{p}} = 0.120 )

Explanation:

Step1: Check if the sampling distribution is normal

For a sampling distribution of \(\hat{p}\), if \(np\geq5\) and \(n(1 - p)\geq5\), it is approximately normal.
Here, \(n = 1020\), \(p=0.56\)
\(np=1020\times0.56 = 571.2\geq5\)
\(n(1 - p)=1020\times(1 - 0.56)=1020\times0.44 = 448.8\geq5\)
So the sampling distribution is approximately normal.

Step2: Calculate the mean of the sampling distribution

The mean of the sampling distribution of \(\hat{p}\) is \(\mu_{\hat{p}}=p\)
Since \(p = 0.56\), \(\mu_{\hat{p}}=0.56\)

Step3: Calculate the standard deviation of the sampling distribution

The formula for the standard deviation of the sampling distribution of \(\hat{p}\) is \(\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}\)
Substitute \(p = 0.56\) and \(n = 1020\)
\(\sigma_{\hat{p}}=\sqrt{\frac{0.56\times(1 - 0.56)}{1020}}=\sqrt{\frac{0.56\times0.44}{1020}}=\sqrt{\frac{0.2464}{1020}}\approx\sqrt{0.000241569}\approx0.016\)

Answer:

C. Approximately normal, \(\mu_{\hat{p}} = 0.56\), \(\sigma_{\hat{p}}=0.016\)