Question

In a random sample of 500 observations, 290 successes and 210 failures were detected.

1. Calculate the point estimate of the population proportion of success.

2. Estimate with 95% confidence the population proportion of success.

Use a hypothesis test with significance level 0.05 to test whether the population proportion of success is less than 0.6

● state the null and alternative hypothesis
●state and calculate the appropriate test statistic
●determine the critical value of the test and state the test and state rejection region
●draw an appropriate conclusion
●state whether or not you reject the null hypothesis

Answer 1

1) Point estimate of population proportion of successes $\hat{p}=\frac{290}{500}=0.58.$

2) $n\hat{p}=500\cdot 0.58\geq10,$$n(1-\hat{p})=500(1-0.58)\geq10$. So we can approximate the binomial distribution with a normal distribution.

$z-value=1.96\\ \hat{q}=1-\hat{p}=1-0.58=0.42.\\ EBP=z-value\sqrt{\frac{\hat{p}\hat{q}}{n}}=1.96\sqrt{\frac{(0.58)(0.42)}{500}}\approx 0.043.\\ (\hat{p}-EBP;\hat{p}+EBP)\\ (0.58-0.043;0.58+0.043)\\ (0.537;0.623)\text{ --- our confidence interval}.$

3) $H_0: p=0.6, H_1:p<0.6$ (left-tailed test)

$pn=(0.6)500>5\\ (1-p)n=(0.4)500>5$

So we will use z-test.

$z-stat=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.58-0.6}{\sqrt{\frac{(0.6)(0.4)}{500}}}\approx -0.9129.\\ \Phi(z_{cr})=\frac{1-2\alpha}{2}=0.45.\\ z_{cr}=1.64.\\ (-\infty,-1.64)\text{is the critical region}.\\ z-stat\text{ is not in the critical region. So we accept, do not reject } H_0.$