Answer to Question #282990 in Statistics and Probability for Job

Question

1. In a study of consumer confidence among middle-class families, 450 were interviewed in the metropolitan area of the country. When asked about having to cut back on discretionary purchases of big-ticket items, 180 of 450 responded that their families had done so over the previous 6 months.

A. What is the 90% confidence interval estimate of the population proportion p of this type of household cutting back on discretionary spending?

B. What sample size would you recommend to achieve a slandered error of 0.03?

Answer 1

Solution:

Given, n=450, X=180

$\hat p=X/n=180/450=0.4 \\ \hat q=1-\hat p=1-0.4=0.6$

(A):

At 90% CI, the z is:

$\alpha=1-90\%=1-0.9=0.1 \\ \alpha/2=0.1/2=0.05 \\ z_{\alpha/2}=z_{0.05}=1.645$

Margin of error = E =

$z_{\alpha/2}\times \sqrt{\hat p \hat q/n} \\=1.645\times \sqrt{0.4\times0.6/450} \\=0.0379$

The 90% CI for population proportion is

$=\hat p-E<p<\hat p+E \\=0.4-0.379<p<0.4+0.379 \\=0.021<p<0.779$

(B):

Standard error of proportion $=\sqrt{\hat p \hat q/n}$

$0.03=\sqrt{0.4\times 0.6/n} \\ \Rightarrow 0.0009=0.24/n \\\Rightarrow n=266.666... \\ \Rightarrow n\approx267$