Watching TV: The 2012 general Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours was 3.09 with a standard deviation of 2.85 . Assume that in a sample of 36 teenagers, the sample standard deviation of daily TV time is 4.5 hours, and that the population of TV watching times is normally distributed. Can you conclude that the population standard deviation of TV watching times for teenagers is greater than 2.85 ? Use the $\alpha=0.01$ level of significance. Part: 0 / 5 Part 1 of 5 State the appropriate null and alternate hypotheses. \[ \begin{array}{l} H_{0}: \sigma=\quad \text { \ } 2.85 \\ H_{1}: \sigma>2.85 \\ \end{array} \] This hypothesis test is a left-tailed $\quad$ test. Part: $1 / 5$ Part 2 of 5 Find the critical value. Round the answer to three decimal places. For $\alpha=0.01$, the critical value is $\square$.
Solution
Step 1 ：State the appropriate null and alternate hypotheses. The null hypothesis \(H_{0}\) is that the population standard deviation \(\sigma\) is equal to 2.85. The alternate hypothesis \(H_{1}\) is that the population standard deviation \(\sigma\) is greater than 2.85. This hypothesis test is a left-tailed test.
Step 2 ：Find the critical value. For a significance level of \(\alpha=0.01\), we need to find the critical value from the chi-square distribution table. The degrees of freedom is the sample size minus 1, which is 36-1=35. We are conducting a one-tailed test, so we need to find the chi-square value corresponding to the 0.01 level in the right tail of the chi-square distribution with 35 degrees of freedom.
Step 3 ：The critical value for a chi-square distribution with 35 degrees of freedom at the 0.01 significance level is approximately 57.342. Therefore, the critical value is \(\boxed{57.342}\).
