2. Hypothesis testing about a population variance Aa Aa E Summary statistics for

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2. Hypothesis testing about a population variance Aa Aa E Summary statistics for the first-round games in the five National Collegiate Athletic Association (NCAA) basketball tournaments between 2004 and 2008 are displayed as follows: Margin of victory (Points Matchup Number of Games Mean Variance 1 vs. 16 20 23,7 114.8 2 vs. 15 20 16.4 90,8 3 vs. 14 20 11.9 62.1 4 vs. 13 9.1 20 149.3 5 vs. 12 20 5.9 159.1 6 vs. 11 20 5.6 146,5 7 vs. 10 20 6.1 83.7 8 vs. 9 20 0.7 100.5 The margin of victory is negative for an upset a win by the lower-seeded team (Data source: These calculations were obtained from data compiled by The News & Observer.) The NCAA tournament is divided into four regions; 16 teams, seeded 1 to 16, are assigned to each region. In the first round of tournament play, in each of the four regions, the 1-seed plays the 16-seed, the 2-seed plays the 15-seed, and so on. As a result, in each tournament, there are four opening-round games for each matchup A college basketball fan (who is also a statistics student) hypothesizes that for a given matchup the margins of victory in the first-round games are more consistent (as measured by their variance) in recent tournaments than in past tournaments. She decides to conduct a hypothesis test for the matchup between the 2-seed and the 15-seed (2 vs. 15) Historically, the variance in the margins of victory for first-round 2 vs. 15 matchups has been o2 130.0. (130.0 is the variance of the margins of victory for the 2 vs. 15 matchup in first-round tournament games played from 1985 to 1997.) (Source: H. S. Stern and B. Mock, "College Basketball Upsets: Will a 16-seed Ever Beat a 1-seed?" Chance 11, no. 1, (1998).] Assume that the population of first-round victory margins is normally distributed and that the 20 games summarized in the table constitute a random sample of recent first-round games The statistics student should formulate the hypothesis test as O Ho: a2 2 130.00, Ha: 02 130.0

Explanation / Answer

Here part(a)

Hypothesis Testing

H0: 2 >= 130 ; Ha : 2< 130

as we are not comparing two poplation variance, here we are comparing population variance to stndard population variance.

Test Statistic

X2= (n-1) s2/ 2= ( 20-1) * 90.8/130 = 13.2708

The rejection rule

for alpha = 0.05, the critical value of X2 = 10.117

so reject H0if X2 < 10.177

p - value here = 0.8244

Here the null hypothesis can't be rejected, the evidence provided by the sample data doesn't corroborate the conclusion thaat variance of margin of victory has been declined.

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