watching tv: the 2012 general social survey a...

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 45 teenagers, the sample standard deviation of daily tv time is 2.6 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 less than 2.85? use the $alpha = 0.05$ level of significance.\npart 1 of 5\nstate the appropriate null and alternate hypotheses.\n$h_0:sigma = 2.85$\n$h_1:sigma < 2.85$\nthis hypothesis test is a left - tailed test.\npart: 1 / 5\npart 2 of 5\nfind the critical value. round the answer to three decimal places.\nfor $alpha = 0.05$

Answer

# Explanation: ## Step1: Determine the degrees of freedom The degrees of freedom for a test of a single - variance is $df=n - 1$, where $n = 45$. So, $df=45-1 = 44$. ## Step2: Identify the distribution and the critical - value We are conducting a left - tailed test for a single variance with $\alpha=0.05$. We use the chi - square distribution $\chi^{2}$. Looking up the critical value in the chi - square distribution table or using a calculator function (inverse chi - square), for $\alpha = 0.05$ and $df = 44$, the critical value $\chi^{2}_{1-\alpha,df}=\chi^{2}_{0.95,44}$. Using a chi - square table or a statistical software (e.g., in R: qchisq(0.95,44)), we get $\chi^{2}_{0.95,44}\approx29.788$. # Answer: $29.788$