The world series in baseball is won by the first team to win four games (ignoring the \(1903\) and \(1919-1921\) world series, when it was best of nine). Thus it takes at least four games and no more than seven games to establish a winner. If two teams are evenly matched the probabilities of the series lasting \(4, 5, 6, or 7\) games are as given in the second column of the following table. From the document World Series History on the Baseball Almanac website, as of November \(2013\), the actual numbers of times that the series lasted \(4, 5, 6, or 7\) games are as shown in the third column of the table.

a. At the \(1%\) significance level, do the data provide sufficient evidence to conclude that the distribution of genders in two-children families differs from the distribution predicted by the model described?

b. In view of your result from part (a) what conclusion can you draw?

Short Answer

Expert verified

Part a. There is no enough evidence to conclude that the two World Series teams are not matched evenly at a \(5%\) significance level.

Part b. There is enough evidence to conclude that the two World Series teams are not matched evenly at a \(10%\) significance level.

Step by step solution

01

Part a. Step 1. Given information

The below table gives the probabilities of World Series in baseball lasting \((4,5,6 or 7)\) games if two teams are matched evenly and the actual numbers of times the series lasted \((4,5,6 or 7)\) games.

02

Part a. Step 2. Calculation

Define the hypotheses:

\(H_{0}\): The two World Series teams are matched evenly.

\(H_{a}\) : The two World Series teams are not matched evenly.

\(\alpha=5%=0.05\)

Determine the observed frequencies and the chi-square subtotals:

\(\chi ^{2}=7.5943\)

Determine the critical value using Table V with \(df=4-1=3\)

\(\chi^{2}_{0.05}=7.815\)

\(7.5943<7.815 \Rightarrow \) do not reject \(H_{0}\):

Therefore, the two World Series teams are matched evenly at a \(5%\) significance level.

03

Part b. Step 1. Calculation

Define the hypotheses:

\(H_{0}\): The two World Series teams are matched evenly.

\(H_{a}\) : The two World Series teams are not matched evenly.

\(\alpha=10%=0.10\)

It was calculated in the above part that the test statistics

\(\chi ^{2}=7.5943\)

Determine the critical value using Table V with \(df=4-1=3\)

\(\chi^{2}_{0.10}=6.251\)

\(7.5943>6.251 \Rightarrow\) Reject \(H_{0}\)

Therefore the two World Series teams are not matched evenly at a \(10%\) significance level.

04

Part c. Step 1. Calculation

The assumptions to the Chi-Square goodness-of-fit test are:

  • The expected frequencies should be all greater than or equal to \(1\).
  • Not more than \(20%\) of the expected frequencies are less than \(5\).
  • The sample is drawn from the simple random sampling method.

Explanation:

The sample is not drawn from the simple random sampling method, thus it is not appropriate to use the chi-square goodness-of-fits test to this case.

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Suppose that you have bivariate data for an entire population.

a. How would you decide whether an association exists between the two variables under consideration?

b. Assuming that you make no calculation mistakes, could your conclusion be in error? Explain your answer.

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12.51 Farms. The U.S. Department of Agriculture publishes information about U.S. farms in Census of Agriculture. A joint frequency distribution for number of farms, by acreage and tenure of operator, is provided in the following contingency table. Frequencies are in


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Under 50
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