The statistci student conducts the hypothesis test using a level of significance

Question

The statistci student conducts the hypothesis test using a level of significance of ? = .10

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)* Mean 23.7 16.4 11.9 Matchup 1 vs. 16 2 vs. 15 3 vs. 1 4 vs. 13 5 vs. 12 6 vs. 11 7 vs. 10 8 vs. 9 Variance Number of Games 20 20 20 20 20 20 20 20 90.8 62.1 149.3 159.1 146.5 83.7 100.5 5.9 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 3-seed and the 14-seed (3 vs. 14) Historically, the variance in the margins of victory for first-round 3 vs. 14 matchups has been 2-144.0, (144.0 is the variance of the margins of victory for the 3 vs. 14 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 Ho: 2-62.1, Hai 2 > 62.1 Ho: 2-144.0, Hai 2

Explanation / Answer

Here we have to test the hypothesis that,

H0 : 2 144.0 Vs H1 : 2 < 144.0

That games are more consistent which has less variance.

Therefore alternative hypothesis has less than sign.

Therefore option b) is correct.

Given that alpha = level of significance =0.10

Here the rejection rule is,

reject H0 if X2 > critical value.

where critical value syntax in EXCEL is,

=CHIINV(probability, deg_freedom)

where, probability = alpha = 0.10

deg_freedom = n-1 = 20-1 = 19

critical value = 27.204

reject H0 if X2 27.204

Option e) is correct.

P-value also we can calculate by using EXCEL.

For calculating P-value we need test statistic.

test statistic (X2) = (n-1)*s2 /  2

n = number of games = 20

s2 = sample variance = 62.1

X2 = (20-1)*62.1 / 144.0 = 8.1938

P-value syntax in EXCEL is,

=CHIDIST(x, deg_freedom)

where x is test statistic value.

deg_freedom = n-1 = 20-1 =19

P-value = 0.9846

P-value > alpha

Accept H0 at 10% level of significance.

Conclusion : Population variance is greator than 144.0.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.