The work week for adults in the US that work full time is normally distributed with a mean of 47 hours. A newly hired engineer at a start-up company believes that employees at start-up companies work more on average then most working adults in the US. She asks 12 engineering friends at start-ups for the lengths in hours of their work week. Their responses are shown in the table below. Test the claim using a $1 \%$ level of significance. Give answer to at least 4 decimal places. \begin{tabular}{|c|} \hline Hours \\ \hline 49 \\ \hline 40 \\ \hline 50 \\ \hline 51 \\ \hline 49 \\ \hline 68 \\ \hline 45 \\ \hline 59 \\ \hline 50 \\ \hline 46 \\ \hline 51 \\ \hline 51 \\ \hline \end{tabular} What are the correct hypotheses? \[ \mathrm{H}_{0} \text { : } \] Select an answer $? \hat{\approx}$ hours $\mathrm{H}_{\mathrm{A}}:$ Select an answer $? \hat{\diamond}$ hours Based on the hypotheses, find the following: Test Statistic $=\square$ (round to 2 decimal) $\mathrm{p}$-value $=\square$ (round to 4 decimals) The correct decision is to Select an answer The correct summary would be: Select an answer that the mean number of hours of all employees at start-up companies work more than the US mean of 47 hours.

Step 1 ：The null hypothesis (H0) is that the mean number of hours worked by employees at start-up companies is equal to the US mean of 47 hours. The alternative hypothesis (HA) is that the mean number of hours worked by employees at start-up companies is greater than the US mean of 47 hours. So, we have: \(H0: \mu = 47\) hours, \(HA: \mu > 47\) hours

Step 2 ：First, we need to calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (s). The sample mean (\(\bar{x}\)) is the sum of the sample data divided by the sample size (n). \(\bar{x} = (49+40+50+51+49+68+45+59+50+46+51+51) / 12 = 51.5\) hours

Step 3 ：The sample standard deviation (s) is the square root of the sum of the squared differences between each data point and the sample mean, divided by the sample size minus 1. \(s = \sqrt{((49-51.5)^2 + (40-51.5)^2 + (50-51.5)^2 + (51-51.5)^2 + (49-51.5)^2 + (68-51.5)^2 + (45-51.5)^2 + (59-51.5)^2 + (50-51.5)^2 + (46-51.5)^2 + (51-51.5)^2 + (51-51.5)^2) / (12-1)} = 6.77\) hours

Step 4 ：The test statistic (t) is the difference between the sample mean and the population mean, divided by the standard error (the sample standard deviation divided by the square root of the sample size). \(t = (\bar{x} - \mu) / (s / \sqrt{n}) = (51.5 - 47) / (6.77 / \sqrt{12}) = 2.52\) (rounded to 2 decimal places)

Step 5 ：The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. For a one-tailed test with a test statistic of 2.52 and 11 degrees of freedom (n-1), the p-value is 0.0147 (rounded to 4 decimal places).

Step 6 ：Since the p-value (0.0147) is less than the significance level (0.01), we reject the null hypothesis. Therefore, the correct decision is to reject the null hypothesis. The correct summary would be: \(\boxed{\text{There is sufficient evidence at the 1% level of significance to support the claim that the mean number of hours employees at start-up companies work is more than the US mean of 47 hours.}}\)