Previously, $8.6 \%$ of workers had a travel time to work of more than 60 minutes. An urban economist believes that the percentage has increased since then. She randomly selects 60 workers and finds that 8 of them have a travel time to work that is more than 60 minutes. Test the economist's belief at the $\alpha=0.05$ level of significance. Because $n p_{0}\left(1-p_{0}\right)=\square \square 10$, the normal model be used to approximate the P-value. (Round to one decimal place as needed.)

Step 1 ：Given that the null hypothesis proportion (p0) is 0.086, the sample size (n) is 60, and the number of successes in the sample (x) is 8.

Step 2 ：Calculate the sample proportion (p_hat) by dividing the number of successes in the sample (x) by the sample size (n). This gives us \(p_{hat} = \frac{x}{n} = \frac{8}{60} = 0.13333333333333333\).

Step 3 ：Calculate the test statistic (z) using the formula \(z = \frac{p_{hat} - p0}{\sqrt{\frac{p0 * (1 - p0)}{n}}}\). Substituting the given values, we get \(z = \frac{0.13333333333333333 - 0.086}{\sqrt{\frac{0.086 * (1 - 0.086)}{60}}} = 1.3077369394636882\).

Step 4 ：Calculate the p-value by subtracting the cumulative distribution function (CDF) of the test statistic (z) from 1. This gives us \(p_{value} = 1 - CDF(z) = 1 - CDF(1.3077369394636882) = 0.09548127479879509\).

Step 5 ：Since the p-value (0.1) is greater than the level of significance (0.05), we do not reject the null hypothesis. This means that we do not have enough evidence to support the economist's belief that the proportion of workers with a travel time of more than 60 minutes has increased.

Step 6 ：\(\boxed{\text{Final Answer: The test statistic is approximately 1.31 and the P-value is approximately 0.1. Therefore, we do not reject the null hypothesis at the 0.05 level of significance. The data does not provide strong evidence to support the economist's belief that the proportion of workers with a travel time of more than 60 minutes has increased.}}\)