Question 129432: Problem 1:
Harold Hacker recently read in a national golf magazine that the average weekend golfer carries a handicap of 15 strokes and the standard deviation is 4 strokes. Harold manages a local men’s church league and just tallied the end-of-season totals. The 64 players in Harold’s league finished the year with an average handicap of 14 strokes.
a. Set up the null and alternative hypotheses to test if the average handicap in Harold’s league is not the same as the national average reported in the magazine.
b. Test your hypothesis using = 0.02.
c. Find the p value.
d. Based on Harold’s end-of-season data, what can you conclude?
Problem 2:
The players on last year’s football team at State College were able to bench press a mean of 312 lb. Coach Juarez made it clear to the players during spring training that the team’s average best lift had to improve. A special weight-training program was launched, and all the players participated. In an effort to measure the team’s progress, the coach recorded the heaviest lifts of the starting offensive and defensive lineups at the start of this season. Results are as follows:
346 412 332 285 396 461 321 275
246 315 298 347 430 419 406 311
319 385 377 365 385 400
a. State the appropriate null and alternative hypotheses.
b. Calculate the test statistic.
HINT sample mean=356.43 lb
Sample standard deviation = 56.05 lb
c. At = 0.01, should Coach Juarez reject the null hypothesis?
d. Assuming the starting lineup is a representative sample, what conclusion can the coach draw?
Problem 3:
a) Johnson’s Service Center has devised three potential options available to preferred customers who redeem coupons and buy at least 10 gallons of fuel when they stop in. Option A is a flat 3 cents off each gallon. Option B is a combination of 2 cents off plus another $1 discount on the regular price of a $5 deluxe car wash. Option C is a $2 discount on the same $5 deluxe car wash but no reduction in the fuel purchase. The owner, Harold Johnson, ran each option on three different two-week trial periods and tracked daily sales receipts from those customers who redeemed their coupons. Results are shown in the table below:
Option A Option B Option C
$453
507
513
521
511
615
601
552
551
505
515
512
476
427 $492
514
536
511
528
678
611
653
596
516
534
543
498
437 $467
525
516
500
435
462
411
674
512
559
624
711
512
416
a) State the null hypothesis to test for equal population means.
b) Below are the results from EXCEL of ANOVA of the data at the 0.05 level of significance. Do the sample data indicate the at least one of the three population mean total returns is different from the others? Why? (Explain using “F” and “Fcrit” – a drawing is fine.)
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Column 1 14 7259 518.5 2555.962
Column 2 14 7647 546.2143 4340.335
Column 3 14 7324 523.1429 8389.209
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 6169 2 3084.5 0.605377 0.550917 3.2381
Within Groups 198711.6 39 5095.168
Total 204880.6 41
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Problem 1:
Harold Hacker recently read in a national golf magazine that the average weekend golfer carries a handicap of 15 strokes and the standard deviation is 4 strokes. Harold manages a local men’s church league and just tallied the end-of-season totals. The 64 players in Harold’s league finished the year with an average handicap of 14 strokes.
a. Set up the null and alternative hypotheses to test if the average handicap in Harold’s league is not the same as the national average reported in the magazine.
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Ho: mu = 15
Ha: mu is not 15
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b. Test your hypothesis using = 0.02.
t(14) = (14-15)/[4/sqrt(64)] = -1*8/14 = -4/7 = -0.5714..
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c. Find the p value.
p-value = P(-10
------------------
d. Based on Harold’s end-of-season data, what can you conclude?
The p-value is greater than 1% (half of 2% of this two-tailed test)
so we Fail to reject Ho.
Harold's results are statistically in keping with the published
national average.
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Cheers,
StanH.
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