Question #228221

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


1
Expert's answer
2021-08-25T05:46:07-0400

n=200x=145n=200 \\ x=145

a) Point estimate

p^=200xn=200145200=0.275\hat{p}= \frac{200-x}{n} = \frac{200-145}{200}=0.275

b) 98 % confidence interval:

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

From Z table

Z=2.3261452000.73CI=0.73±2.326×0.73×0.27200=0.73±0.073(0.657,0.8030)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).


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