Question

During the 2010 MLB seasons, the home team won 1,269 games and the away team won 1,371 games. What is the test statistic to compare the observed data versus a 50/50 outcome? Round to two decimal places.

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Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The proportion of observed won for both the teams are same.

Alternative hypothesis: The proportion of observed won for both the teams are different.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 2 - 1
D.F = 1
(E_i) = n * p_i


X² = 3.94

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 1 degrees of freedom is more extreme than 3.94.

We use the Chi-Square Distribution Calculator to find P(X² > 3.94) = 0.047.

Interpret results. Since the P-value (0.047) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the clam that both team have won different proportions of games.