04 Jun Question Use the following information to answer question1-4.
Question
Use the following information to answer question1-4.
Reviewing her performance on her last ten (10) three foot putts, Lisa see the following pattern:
Make, Make, Miss, Make, Miss, Make, Make, Make, Miss
1. Enter a one?1?for every make and zero (0) for every miss. What is the mean of this sample of putts?
a. 3/5
b. b. 7/10
c. c. 3/2
d. d. 1/2
e. e. 3/1
2. What is the conditional probability of a make on a put following a miss?
a. 2/3
b. 1/10
c. 1/2
d. 1/5
e. 4/5
3. What is the conditional probability that Lisa made the putt given that she made the two previous putts?
a. 0
b. 1/3
c. 2/5
d. 1
e. 2/3
4. How many runs are there?
a. 3
b. 5
c. 6
d. 10
e. 4
Frustrated with her performance. Lisa practices 100 putts. She calculates 42 runs. The W-W runs test gives the following information.
Expected number of Runs:49; sd; 4.7737; z-score=-1.4664
5. At the 5% level of significance, which of the following is true?
a. The number of runs is not significantly different from the expected number of runs.
b. The number of runs is significantly less than the expected number of runs.
c. The number of runs is not significantly different from zero
d. The number of runs is significantly less than zero
e. The number of runs is significantly greater than the expected umber of runs.
6. Does the data provide evidence that Lisa has a hot hand?
a. Yes, because the number of runs is significantly less than the expected number of runs.
b. Yes, because the number of runs is significantly greater than the expected number of runs.
c. No, because the number of runs is significantly less than the expected number of runs.
d. No, because the number of runs is significantly greater than the expected number of runs.
e. No, because the number of runs is not significantly different than the expected number of runs.
7. The z-score means that the number of runs is roughly
a. 1.4664 standard deviations below the expected number of runs.
b. About 1.47% percent likely to be significantly different from the expected number of runs.
c. 1.4664 more than the expected number of runs.
d. 1.4664 standard deviations above the expected number of runs.
e. 1.4664 less than the expected number of runs.
Lisa is also worried about her consistency on the golf course. She thinks that she has “ post-birdie syndrome”. This condition shows up on the hole following a birdie (one under par on a golf hole). She thinks that she becomes overconfident and score worse ( remeber that in golf, a higher score is worse!) Like a good economist, she has gathered data to test her theory.
Score par or below Score above par
(success) ( failure)
Hole following a birdie 11 22
Hole not following a birdie 62 54
8. What test should she perform to determine whether she has post-birdie syndrome?
a. Multiple Regression
b. Standard Deviation test
c. Runs test
d. Single regression
e. Chi Square test
9. Which best describes the hypothesis that she is testing ( the alternative hypothesis, Not the null hypothesis)?
a. “ On average, I score lower ( I get a better score) after a birdie than I do otherwise”
b. “I have a hot hand”
c. “I do not have a hot hand”
d. “On average, I score higher( I get a worse score) after birdie than I do otherwise”
e. “On average, I score the same after a birdie as I do otherwise”
Lisa runs the appropriate test, which returns a p-value of 0.032
10. What does this p-value mean?
a. If the null hypothesis is true, then Lisa’s probability of making a type II error is 96.8%(0.968)
b. If the null hypothesis is correct, the probability of observing this much or more difference between scores is 3.2%(0.032)
c. The correlation coefficient for her scores is 0.032.
d. Her score after a birdie is 3.2% higher than her score before a birdie.
e. If the null hypothesis is true, then Lisa’s probability of making a type II error is 3.2% (0.032)
11. Suppose that Lisa had decided on a 5% significance level. Does she conclude that she has post-birdie syndrome?
a. Yes, because she scores lower after a birdie.
b. Impossible to tell, because we do not have the Z-statistic.
c. Yes, because her p-value is less than 5%
d. No, because her p-value is less than5%
e. No, because her p-value is less than the 5% critical value of 1.96.
12. Suppose that we divided her scores into groups of five, if we plotted the frequency distribution of number of scores in each group of five that were above par, we would plot a
a. Binomial distribution
b. Geometric distribution
c. Student’s t-distribution
d. Normal distribution
e. Chi square distribution
Height is measured in inches, BMI is body Mass index (weight in kilograms divided by height in meters squared). Wonderlic is the score on an intelligence test, 40 yard dash is measured in seconds and Division I-AA dummy answers the question, “ was your college in division I-AA? (1 for yes and 0 for a no)
13. What is the dependent variable?
a. Constant
b. Draft position
c. Division I-AA dummy
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