In: Statistics and Probability
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Player | Round 1 | Round 2 |
Michael Letzig | 70 | 72 |
Scott Verplank | 71 | 72 |
D.A. Points | 70 | 75 |
Jerry Kelly | 72 | 71 |
Soren Hansen | 70 | 69 |
D.J. Trahan | 67 | 67 |
Bubba Watson | 71 | 67 |
Reteif Goosen | 68 | 75 |
Jeff Klauk | 67 | 73 |
Kenny Perry | 70 | 69 |
Aron Price | 72 | 72 |
Charles Howell | 72 | 70 |
Jason Dufner | 70 | 73 |
Mike Weir | 70 | 77 |
Carl Pettersson | 68 | 70 |
Bo Van Pelt | 68 | 65 |
Ernie Els | 71 | 70 |
Cameron Beckman | 70 | 68 |
Nick Watney | 69 | 68 |
Tommy Armour III | 67 | 71 |
Chapter 10 (Excel file included):
Scores in the first and fourth (final) rounds for a sample of 20 golfers who competed in the PGA tournaments are included in the attached Excel file. Suppose you would like to determine if the mean score for the first round of a PGA Tour event is significantly different than the mean score for the fourth and final round. You are curious: Does the pressure of playing in the final round cause scores to go up? Or does the increased player concentration cause scores to come down?
H0: or the mean score for the first round of a PGA Tour event is not significantly different than the mean score for the fourth and final round
H1: or the mean score for the first round of a PGA Tour event is significantly different than the mean score for the fourth and final round.
1.We enter the values is excel as follows:
Player | Round 1 | Round 2 |
Michael Letzig | 70 | 72 |
Scott Verplank | 71 | 72 |
D.A. Points | 70 | 75 |
Jerry Kelly | 72 | 71 |
Soren Hansen | 70 | 69 |
D.J. Trahan | 67 | 67 |
Bubba Watson | 71 | 67 |
Reteif Goosen | 68 | 75 |
Jeff Klauk | 67 | 73 |
Kenny Perry | 70 | 69 |
Aron Price | 72 | 72 |
Charles Howell | 72 | 70 |
Jason Dufner | 70 | 73 |
Mike Weir | 70 | 77 |
Carl Pettersson | 68 | 70 |
Bo Van Pelt | 68 | 65 |
Ernie Els | 71 | 70 |
Cameron Beckman | 70 | 68 |
Nick Watney | 69 | 68 |
Tommy Armour III | 67 | 71 |
2.Then click on data--->dataanalysis---->t-test:two samples assuming equal variances.
The output is as follows for two tailed test:
t-Test: Two-Sample Assuming Equal Variances | ||
Variable 1 | Variable 2 | |
Mean | 69.65 | 70.7 |
Variance | 2.765789474 | 9.168421053 |
Observations | 20 | 20 |
Pooled Variance | 5.967105263 | |
Hypothesized Mean Difference | 0 | |
df | 38 | |
t Stat | -1.359275376 | |
P(T<=t) two-tail | 0.182071918 | |
t Critical two-tail | 2.024394164 |
since, |t|=1.35927<tc=2.024394 we accept the null hypothesis hence,the mean score for the first round of a PGA Tour event is not significantly different than the mean score for the fourth and final round
2.p-value=0.1820719>0.05 hence we accept the null hypothesis.
3.A point estimate for the difference in two population means is simply the difference in the corresponding sample means.
hence,point estimate for difference of means=70.7-69.65=1.05
4.
90% confidence interval:
Player | Round 1 | Round 2 | Difference |
Michael Letzig | 70 | 72 | -2 |
Scott Verplank | 71 | 72 | -1 |
D.A. Points | 70 | 75 | -5 |
Jerry Kelly | 72 | 71 | 1 |
Soren Hansen | 70 | 69 | 1 |
D.J. Trahan | 67 | 67 | 0 |
Bubba Watson | 71 | 67 | 4 |
Reteif Goosen | 68 | 75 | -7 |
Jeff Klauk | 67 | 73 | -6 |
Kenny Perry | 70 | 69 | 1 |
Aron Price | 72 | 72 | 0 |
Charles Howell | 72 | 70 | 2 |
Jason Dufner | 70 | 73 | -3 |
Mike Weir | 70 | 77 | -7 |
Carl Pettersson | 68 | 70 | -2 |
Bo Van Pelt | 68 | 65 | 3 |
Ernie Els | 71 | 70 | 1 |
Cameron Beckman | 70 | 68 | 2 |
Nick Watney | 69 | 68 | 1 |
Tommy Armour III | 67 | 71 | -4 |
for mean---->AVERAGE() function is used.
Mean= | -1.05 |
for sample sd----->STDEV.S() function is used.
Sample sd= | 3.316228041 |
for Margin of error---->CONFIDENCE.T() function is used.
Margin of error(E)= | 1.282205813 |
lower limit=mean-E and upper limit=mean+E at 90% confidence interval:
Lower limit(mean-E) | -2.33220581 |
Upper limit(mean+E) | 0.232205813 |
please rate my answer and comment for doubts.