1. Hypothesls testing about a population varlance five National Collegiate Athle
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Question
1. Hypothesls testing about a population varlance five National Collegiate Athletic Assoclation (NCAA) basketball tournaments between 2004 and 2008 are displayed as follows: Margin of Victory (Polnts)* Matchup 1 vs. 16 2 vs. 15 3 vs. 14 4 vs. 13 5 vs. 12 6 vs. 11 7 vs. 10 8 vs. 9 Number of Games 20 20 20 20 20 20 20 20 Mean 23.7 16.4 11.9 9.1 5.9 5.6 6.1 Variance 114.8 90.8 62.1 149.3 159.1 146.5 83.7 100.5 -0.7 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 4-seed and the 13-seed (4 vs 13) Historically, the vanance in the margins of victory for first-round 4 vs. 13 matchups has been o-1563. (1563 is the variance of the margins of victory for the 4 vs. 13 matchup in first-round tournament games played from 1985 to 1997) Isource: H. S. Stern and B. Mock, "College Basketball Upsets: Will a 16-Seed Ever Beat a 1-Seed?" ChanceExplanation / Answer
Given that,
population variance (^2) =156.3
sample size (n) = 20
sample variance (s^2)=149.3
null, Ho: ^2 <=156.3
alternate, H1 : ^2 >156.3
level of significance, = 0.05
from standard normal table,right tailed ^2 /2 =30.144
since our test is right-tailed
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(20 - 1 ) * 149.3 / 156.3 = 19*149.3/156.3 = 18.15
| ^2 cal | =18.15
critical value
the value of |^2 | at los 0.05 with d.f (n-1)=19 is 30.144
we got | ^2| =18.15 & | ^2 | =30.144
make decision
hence value of | ^2 cal | < | ^2 | and here we do not reject Ho
^2 p_value =0.5124
ANSWERS
---------------
null, Ho: ^2 <=156.3
alternate, H1 : ^2 >156.3
test statistic: 18.15
critical value: 30.144
p-value:0.5124
decision: do not reject Ho
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