# Chapter 8 - Hypothesis Testing - Review - Review Exercises - Page 433: 2

There is not sufficient evidence to reject that 2/3 of adults are satisfied with the leisure time the have.

#### Work Step by Step

$H_{0}:p=2/3$. $H_{a}:p\ne2/3.$ $\hat{p}$ is the number of objects with a specified value divided by the sample size. Hence $\hat{p}=\frac{x}{n}=\frac{657}{1010}=0.6505.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.6505-2/3}{\sqrt{2/3(1-2/3)/1010}}=-1.09.$ The P is the probability of the z-score being more than 1.09 or less than -1.09 is the sum of the probability of the z-score being less than -1.09 plus 1 minus the probability of the z-score being less than 1.09, hence:P=0.1379+1-0.8631=0.2758. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P=0.2758 is more than $\alpha=0.01$, hence we fail to reject the null hypothesis. Hence we can say that there is not sufficient evidence to reject that 2/3 of adults are satisfied with the leisure time the have.

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