Answer to Question #234625 in Statistics and Probability for Tuesaa

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) Suppose that a 98% confidence interval estimate of the proportion of local vendors who make an average income more than N$ 7 600 per month is required. How large a sample should be taken if the desired margin of error is 8%?

b) Suppose it has been established from past studies that not more than 65% of local vendors make an average income in excess of N$ 7 600 per month. Use the critical value approach and conduct a hypothesis test at a 5% level of significance to determine whether the proportion of local vendors who make an average income in excess of N$ 7 600 per month differs from the historical level. 

Answer 1

Solution:

(a):

Let $\alpha=0.02$

Then $z_{0.02/2}=2.33$

Margin error, ME = 0.08

Required sample size = n = $(\dfrac{z_{0.02/2}}{ME})^2\times p \times (1-p)$

$\Rightarrow n=(2.33/0.08)^2(0.73)(0.27) \ [\because p=145/200=0.73] \\\Rightarrow n=167.193\approx167$

(b):

$H_0:p\le0.65 \\H_1:p > 0.65$

$\alpha=0.05$

Then $z_{0.05}=1.64$ [Using one-tailed value]

$\hat p=145/200=0.73$

Test statistic$=\dfrac{\hat p-p}{\sqrt{\dfrac{p(1-p)}{n}}}=\dfrac{0.73-0.65}{\sqrt{\dfrac{0.65(1-0.65)}{200}}}=2.37$

Now, 2.37 > 1.64, we have sufficient evidence to reject H₀ .

Thus, $p>0.65$ or more than 65% of local vendors make an average income in excess of N$ 7 600 per month.