Answer to Question #342560 in Statistics and Probability for nana

Question

(a) Identify the claim and state H0 and Ha.

(b) Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed, and whether to use a z-test, a t-test, or a chi-square test. Explain your reasoning.

(c) Choose one of the options. If convenient, use technology.

Option 1: Find the critical value(s), identify the rejection region(s), and find the appropriate standardized test statistic.

Option 2: Find the appropriate standardized test statistic and the P-value.

(d) Decide whether to reject or fail to reject the null hypothesis.

(e) Interpret the decision in the context of the original claim.

4. A research center claims that more than 55% of U.S. adults think that it is essential that the United States continue to be a world leader in space exploration. In a random sample of 25 U.S. adults, 64% think that it is essential that the United States continue to be a world leader in space exploration. At a = 0.05, is there enough evidence to support the center’s claim?

Answer 1

(a) The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p\le0.55$

$H_a:p>0.55$

(b) This corresponds to a right-tailed test, for which a z-test for one population proportion will be used.

(c) Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a right-tailed test is $z_c = 1.96.$

The rejection region for this right-tailed test is $R = \{z: z > 1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}=\dfrac{0.64-0.55}{\sqrt{\dfrac{0.55(1-0.55)}{25}}}=0.9045$$

Since it is observed that $z =0.9045< 1.96=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(Z>0.9045)=0.182865,$ and since $p =0.182865> 0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is greater than 0.55, at the $\alpha = 0.05$ significance level.