Answer to Question #232576 in Statistics and Probability for Patty

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%? (2)
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. (4)

Answer 1

Proportion p has approximately normal distribution with parameters $\mu_p=\hat p. \space \sigma_p=\sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}$. Thus for p analisis.we use normal distribution$N(\hat p,\sqrt{\frac{\hat p\cdot (1-\hat p)}{n}})$. So C- confidence interval for mean M(p) will be $\hat p \pm z^*_\frac{1-C}{2}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}$

where $z^*_\frac{1+C}{2}$ is such level that P(N(0,1)$> z^*_\frac{1-C}{2}=\frac{1-C}{2}$ Therfore margin error will be E=$z^*_\frac{1-C}{2}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}$ . Fron this equation we define size of sample $n=\frac{\hat p\cdot (1-\hat p)\cdot(z^*_\frac{1-C}{2})^2 }{E^2}$ .We are given with E=8%=0.08, $\hat p=145/200=0.725$ . C=98%=0.98 and calculate: $z^*_\frac{1-C}{2}=z^*_{0.01}=$

=qnorm(0.99,0,1)=2.326 where qnorm is statistical function from Mathcad therefore we have $n=\frac{0.725\cdot(1-0.725)\cdot 2.326^2}{0.08^2}=168.543\approx 169$

So size of sample should be 169.

b)

H0: p$\le$ 0.65 - zero hypothesis;

H1: p>0.65- alternative hypothesis;

$\alpha=0.05$ significance level;

z=qnorm(1-0.05,0,1)=1.645- z-value corresponding to significance level(by using Mathcad);

Let $Y=\frac{\frac{X}{200}-0.65}{\sqrt{0.65\cdot (1-0.85)/200}}$ 

If historical value 0.65 of the proportion is 0.65 is true then we will have Y~N(0,1) or Y will have standard normal distribution and for rejection of the hypothesis X/200=0.65 is enough to meet an event witch follows from 0.65 hypothesis but has small probability. Such event is Y$\ge z_{0.05}$ with probebility 0.05

We use equation Y=z=1.645 for finding minimal bound for Y and therefore for X.

From it eqution we have $X=(0.65+1.645\cdot\sqrt{0.65\cdot (1-0.65)/200})\cdot {200}$ =142.689$\approx 143$ .

So for rejecting hypothesis p<=65% with confidence it is enough to form sample with size 143 or more.