Question

Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for p, the margin of error, and the confidence interval. Assume the results come from a random sample. A 99% confidence interval for p given that p-hat = 0.34 and n= 500. Point estimate ___________ (2 decimal places) Margin of error __________ (3 decimal places) The 99% confidence interval is ________ to _______ (3 decimal places)

Answer 1

Answer:

(a) The point estimate for the population proportion p is 0.34.

(b) The margin of error for the 99% confidence interval of population proportion p is 0.055.

(c) The 99% confidence interval of population proportion p is (0.285, 0.395).

Step-by-step explanation:

A point estimate of a parameter (population) is a distinct value used for the estimation the parameter (population). For instance, the sample mean $\bar x$ is a point estimate of the population mean μ.

Similarly, the the point estimate of the population proportion of a characteristic, p is the sample proportion $\hat p$.

The (1 - α)% confidence interval for the population proportion p is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The margin of error for this interval is:

$MOE= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The information provided is:

$\hat p=0.34\\n=500\\(1-\alpha)\%=99\%$

(a)

Compute the point estimate for the population proportion p as follows:

Point estimate of p = $\hat p$ = 0.34

Thus, the point estimate for the population proportion p is 0.34.

(b)

The critical value of z for 99% confidence level is:

$z={\alpha/2}=z_{0.01/2}=z_{0.005}=2.58$

*Use a z-table for the value.

Compute the margin of error for the 99% confidence interval of population proportion p as follows:

$MOE= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

          $=2.58\sqrt{\frac{0.34(1-0.34)}{500}}$

          $=2.58\times 0.0212\\=0.055$

Thus, the margin of error for the 99% confidence interval of population proportion p is 0.055.

(c)

Compute the 99% confidence interval of population proportion p as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$CI=\hat p\pm MOE$

     $=0.34\pm 0.055\\=(0.285, 0.395)$

Thus, the 99% confidence interval of population proportion p is (0.285, 0.395).