Answer to Question #228221 in Statistics and Probability for elizabeth

Question

Suppose a sample of 200 local vendors is selected as part of a survey conducted by the Ministry of Trade and it is found that 145 of the local vendors make an average income in excess of N$ 7 600 per month.

a) What is the point estimate of the proportion of local vendors who make an average income less than N$ 7 600 per month? (2)

b) What is the 98% confidence interval estimate of the population proportion of local vendors who make an average income in excess of N$ 7 600 per month? (10)

c) Interpret the answer in b). (3

Answer 1

$n=200 \\ x=145$

a) Point estimate

$\hat{p}= \frac{200-x}{n} = \frac{200-145}{200}=0.275$

b) 98 % confidence interval:

$\hat{p}± Z \times \sqrt{ \frac{\hat{p}(1- \hat{p})}{n} }$

From Z table

$Z = 2.326 \\ \frac{145}{200} \approx 0.73\\ CI = 0.73 ± 2.326 \times \sqrt{\frac{0.73 \times 0.27}{200} } \\ = 0.73 ± 0.073 \\ (0.657, 0.8030)$

c) The population proportion of local vendors who make an average in excess N$7600 per month lies in the interval (0.657, 0.8030).