Answer to Question #106500 in Statistics and Probability for Oratioe

Question

in a random sample of 500 observations, 290 succeses and 210 failures were detected. a) calculate the point estimate if the population proportion of successes. b) estimate with 95% confidence the population proportion of successes.
c) use a hypothesis test with significance level 0.05 to test whether the population of proportion 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 test state the rejection region.
iv) state whether or not you reject the null hypothesis.
v) draw an appropriate conclusion

Answer 1

a) Point Estimate is $\hat{p}=\dfrac{x}{n},\ x$ is the number of successes, $n$ is the sample size

$$\hat{p}=\dfrac{x}{n}={290\over 500}=0.58$$

b) The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$ The corresponding confidence interval is computed as shown below:

$$CI(proportion)=$$

$$=(\hat{p}-z_c\sqrt{{\hat{p}(1-\hat{p})\over n}}, \hat{p}+z_c\sqrt{{\hat{p}(1-\hat{p})\over n}})=$$

$$=(0.58-1.96\sqrt{{0.58(1-0.58)\over 500}}, 0.58+1.96\sqrt{{0.58(1-0.58)\over 500}})=$$

$$=(0.5367, 0.6233)$$

Therefore, based on the data provided, the $95\%$  confidence interval for the population proportion is $0.5367<p<0.6233,$ which indicates that we are 95% confident that the true population proportion $p$  is contained by the interval  $(0.5367, 0.6233).$

c) The following null and alternative hypotheses need to be tested:

$H_0: p\geq0.6$

$H_1:p<0.6$

This corresponds to a left-tailed test, for which a z-test for one population proportion needs to be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a left-tailed test is $z_c=-1.645.$

The rejection region for this left-tailed test is $R=\{z:z<-1.645\}$

The z-statistic is computed as follows:

$$z={\bar{p}-p_0\over \sqrt{p_0(1-p_0)/n}}={0.58-0.6\over \sqrt{0.6(1-0.6)/500}}\approx-0.9129$$

Since it is observed that $z=-0.9129>-1.645=z_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population proportion $p$ is less than $0.6$ , at the $\alpha=0.05$  significance level.

Using the P-value approach: The p-value is $p=0.180621,$ and since  $p=0.180621>0.05,$ it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population proportion $p$ is less than  $0.6,$ at the $\alpha=0.05$ significance level.