The table below lists the number of games played in a yearly best-of-seven baseball championship series, along with the expected proportions for the number of games played with teams of equal abilities. Use a 0.05 significance level to test the claim that the actual numbers of games fit the distribution indicated by the expected proportions. \begin{tabular}{|l|c|c|c|c|} \hline Games Played & 4 & 5 & 6 & 7 \\ \hline Actual contests & 20 & 22 & 22 & 37 \\ \hline Expected proportion & $\frac{2}{16}$ & $\frac{4}{16}$ & $\frac{5}{16}$ & $\frac{5}{16}$ \\ \hline \end{tabular} Determine the null and alternative hypotheses. Calculate the test statistic, $\chi^{2}$. $\chi^{2}=\square$ (Round to three decimal places as needed.) Calculate the P-value. P-value $=\square$ (Round to four decimal places as needed.) What is the conclusion for this hypothesis test? A. Reject $\mathrm{H}_{0}$. There is insufficient evidence to warrant rejection of the claim that the actual numbers of games fit the distribution indicated by the expected proportions. B. Fail to reject $\mathrm{H}_{0}$. There is sufficient evidence to warrant rejection of the claim that the actual numbers of games fit the distribution indicated by the expected proportions. C. Fail to reject $\mathrm{H}_{0}$. There is insufficient evidence to warrant rejection of the claim that the actual numbers of games fit the distribution indicated by the expected proportions. D. Reject $\mathrm{H}_{0}$. There is sufficient evidence to warrant rejection of the claim that the actual numbers of games fit the distribution indicated by the expected proportions.
Solution
Step 1 ：Define the null hypothesis (H0): The actual numbers of games fit the distribution indicated by the expected proportions.
Step 2 ：Define the alternative hypothesis (H1): The actual numbers of games do not fit the distribution indicated by the expected proportions.
Step 3 ：Calculate the total number of games: \(20 + 22 + 22 + 37 = 101\).
Step 4 ：Calculate the expected numbers of games for each category: \(4\) games: \(101 * \frac{2}{16} = 12.625\), \(5\) games: \(101 * \frac{4}{16} = 25.25\), \(6\) games: \(101 * \frac{5}{16} = 31.5625\), \(7\) games: \(101 * \frac{5}{16} = 31.5625\).
Step 5 ：Calculate the test statistic, \(\chi^2\), using the formula: \(\chi^2 = \sum \frac{(O - E)^2}{E}\), where O is the observed frequency and E is the expected frequency.
Step 6 ：Compute the \(\chi^2\) value: \(\chi^2 = \frac{(20 - 12.625)^2}{12.625} + \frac{(22 - 25.25)^2}{25.25} + \frac{(22 - 31.5625)^2}{31.5625} + \frac{(37 - 31.5625)^2}{31.5625} = 8.610\) (rounded to three decimal places).
Step 7 ：Calculate the degrees of freedom (df) for this test: \(df = 4 - 1 = 3\).
Step 8 ：Use a chi-square distribution table to find the P-value. For \(\chi^2 = 8.610\) and \(df = 3\), the P-value is less than 0.05.
Step 9 ：Since the P-value is less than 0.05, reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the actual numbers of games fit the distribution indicated by the expected proportions.
Step 10 ：\(\boxed{\text{The answer is D.}}\)
