Question

Question 1 (1 point)

177 employees of your firm were asked about their job satisfaction. Out of the 177, 16 said they were unsatisfied. What is the estimate of the population proportion? What is the standard error of this estimate?

Question 1 options:

	1) The true population proportion is needed to calculate this.
	2) Estimate of proportion: 0.0904, Standard error: 0.0016.
	3) Estimate of proportion: 0.0904, Standard error: 0.0216.
	4) Estimate of proportion: 0.9096, Standard error: 0.0016.
	5) Estimate of proportion: 0.9096, Standard error: 0.0216.

Answer 1

Solution:

Given:

n = sample size = 177

x = Number of employees  said they were unsatisfied = 16

Part a) What is the estimate of the population proportion?

$\hat{p}=\frac{x}{n}$

$\hat{p}=\frac{16}{177}$

$\hat{p}=0.0904$

Part b) What is the standard error of this estimate?

$\sigma_{\hat{p}}= \sqrt{\frac{\hat{p}\times (1-\hat{p})}{n}}$

$\sigma_{\hat{p}}= \sqrt{\frac{0.0904 \times (1-0.0904 )}{177}}$

$\sigma_{\hat{p}}= \sqrt{\frac{0.0904 \times 0.9096 }{177}}$

$\sigma_{\hat{p}}= \sqrt{0.00046454 }$

$\sigma_{\hat{p}}=0.0215533$

$\sigma_{\hat{p}}=0.0216$

Thus correct option is:

3) Estimate of proportion: 0.0904, Standard error: 0.0216