Answer to Question #107338 in Statistics and Probability for NKULULEKO HAROLD LOUW

Question

In a random sample of 500 observations, 290 successes and 210 failures were detected.a)i) Calculate the point estimate of the population proportion of successes.ii) Estimate with 95% confidence the population proportion of successes.b) Use a hypothesis test with significance level 0.05 to test whether the population proportion of successes is less than 0.6.i) State the null and alternative hypothesis.ii) State and calculate the appropriate test statistic.iii) Determine the critical value of the test and state the rejection region.iv) State whether or not you reject the null hypothesis.v) Draw an appropriate conclusion.c) Does the conclusion of part (b) also follow from (a)?

Answer 1

$a)i) p\prime=\frac{290}{500}=0.58\text{ is a point estimate of population proportion}\\ ii)p\prime n=(0.58)500>5, (1-p\prime)n=(0.42)500>5$

So the binomial distribution can be estimated by the normal

distribution.

$\alpha=0.05\\ (p\prime-z_{\alpha/2}\sqrt{\frac{p\prime (1-p\prime)}{n}},p\prime+z_{\alpha/2}\sqrt{\frac{p\prime (1-p\prime)}{n}})\\ z_{0.025}=1.96\\ (0.58-1.96\sqrt{\frac{(0.58)(0.42)}{500}},0.58+1.96\sqrt{\frac{(0.58)(0.42)}{500}})\\ (0.537;0.623)\text{ is the confidence interval}.\\ b)i) H_0:p=0.6, H_1:p<0.6\text{ (left-sided)}\\ ii) z_{stat}=\frac{p\prime-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.58-0.6}{\sqrt{\frac{(0.6)(0.4)}{n}}}=-0.913\\ iii) z_{cr}=1.64\\ (-\infty,-1.64)\text{ is the critical region}\\ iv) z_{stat}>-z_{cr}. \text{ So we accept } H_0: p=0.6.$

$v)$There is no evidence that the population proportion of successes is less than 0.6.

c) The conclusion of part (b) does not follow from (a) because we have one-tailed hypothesis test ($\alpha=0.05$) and $95$% confidence interval.