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

Question

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.

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---