Question

A research institute poll asked respondents if they felt vulnerable to identity theft. In the​ poll, n = 1074 and x = 543 who said​ "yes." Use a 95% confidence level.
(a) Find the best point of estimate of the population of portion p.
(b) Identify the value of the margin of error E. Round to four decimal places as needed.
(c) Construct the confidence interval.

Answer 1

Answer:

(a) The best point estimate of the population proportion p is 0.51.

(b) The margin of error is 0.0299.

(c) The 95% confidence interval for population proportion of all respondent who said "yes" is (0.4801, 0.5399).

Step-by-step explanation:

The random variable X is defined as the number of respondents who felt vulnerable to identity theft.

The information provided is:

n = 1074

x = 543

(a)

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

The best point estimate of the population proportion p is the sample proportion $\hat p$.

Compute the sample proportion value as follows:

$\hat p=\frac{x}{n}=\frac{543}{1074}=0.505586\approx0.51$

Thus, the best point estimate of the population proportion p is 0.51.

(b)

The margin of error for the confidence interval of p is:

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

For 95% confidence level the critical value of z is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the margin of error as follows:

$E=z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.51(1-0.51)}{1074}}=0.0299$

Thus, the margin of error is 0.0299.

(c)

Construct the 95% confidence interval for population proportion as follows:

$CI=\hat p\pm E\\=0.51\pm0.0299\\=(0.51-0.0299, 0.51+0.0299)\\=(0.4801, 0.5399)$

Thus, the 95% confidence interval for population proportion of all respondent who said "yes" is (0.4801, 0.5399).