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3.2 For 239 golf - tournaments on the PGA tour between 2004 and 2009, the economists D. Pope and M. Schweitzer evaluated risk aversion by comparing percentages of p...

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3.2 For 239 golf - tournaments on the PGA tour between 2004 and 2009, the economists D. Pope and M. Schweitzer evaluated risk aversion by comparing percentages of putts made when putting for a par versus putting for a birdie (Am. Econ. Rev: 101: 129-157, 2011). For 2828 pairs of putts taken from within L inch of each other (from an average distance of about 50 inches) in the same tournament; the sample proportions made were 0.835 when putting for birdie and 0.880 when putting for par (thus avoid

3.2 For 239 golf - tournaments on the PGA tour between 2004 and 2009, the economists D. Pope and M. Schweitzer evaluated risk aversion by comparing percentages of putts made when putting for a par versus putting for a birdie (Am. Econ. Rev: 101: 129-157, 2011). For 2828 pairs of putts taken from within L inch of each other (from an average distance of about 50 inches) in the same tournament; the sample proportions made were 0.835 when putting for birdie and 0.880 when putting for par (thus avoiding the loss of a bogey). Construct a 95% confidence interval for the difference between the proportions in a corresponding conceptual population. State assumptions, and indicate a key way they do not apply for this study: (Chapter 1 presents more refined methods.)



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A Better Golf Tee? An independent golf equipment testing facility compared the difference in the performance of golf balls hit off a regular $2-3 / 4^{\prime \prime}$ wooden tee to those hit off a $3^{\prime \prime}$ Stinger Competition golf tee. A Callaway Great Big Bertha driver with 10 degrees of loft was used for the test, and a robot swung the club head at approximately 95 miles per hour. Data on total distance traveled (in yards) with each type of tee, based on the test results, are provided on the WeissStats CD. a. Obtain normal probability plots, boxplots, and the standard deviations for the two samples. b. At the $1 \%$ significance level, do the data provide sufficient evidence to conclude that, on average, the Stinger tee improves total distance traveled? c. Find a $99 \%$ confidence interval for the difference between the mean total distance traveled with the regular and Stinger tees. d. Are your procedures in parts (b) and (c) justificd? Why or why not?

The following is a solution for number six and this looks at I think seven or eight different golfers and they're supposed to record their lowest score with um they're old clubs and then they're supposed to play with these new clubs for um you know a couple months and then they record their new lowest scores with the new clubs. So the data values. So L. One this is with their old clubs, these are like their lowest scores and then L. Two, this is with the new clubs and we're seeing if there's any difference. So what I'm gonna do is I'm gonna use this third column and I'm gonna code this third column to find those differences. So L one minus L. Two Is what I'm gonna do there. So L 1 -62. So these are my differences Between the two scores And to find those statistics I can just go to one of our stats. My list is L. three and I could calculate this X. Bar is actually my D. Bar and then this s of X. Is actually my S. F. D. So let's go and write those down and then answers, you know, first to questions basically. So 1.125 is my average difference. And then it's got a standard deviation about 2.167. And Part B. The point estimate is actually this d. bar as well, so it's just 1.125. So now I'm supposed to find the 99% confidence in the rule of the two of the differences. So if I go back to stat and air over the tests, I'm actually gonna be using the T interval. This follows the T distribution. Just a normal one sample T distribution with these matched pairs. That's what's kind of nice about him. So the data should be highlighted. L three is my list frequencies one and then my sea level my confidence level. I'm gonna change that 2.99 to represent 99%. So whenever I calculate I get negative 1.556-3.8063. So zero is contained in that interval. So I'm thinking there's not gonna be much of a difference here but we'll see. So negative 1.556 all the way up to 3.80 63. So now I'm testing to see if there is actually a difference. So the null hypothesis is that there is no difference. And then the alternative is saying that their scores originally are worse than their new scores with the new clubs. That's why it's greater than because the lower scores better engulf. So the first one should be greater than the second one. And we're testing at the 1% level significant. So I need to test statistic and also a P value. So I'm going to explicitly compare the p value to the alpha and that's going to tell me whether to reject or not reject. So if I go back to stats and then air over to test and this is just a normal plain old T test. It's not a two sample T test because it's just one sample of differences. So T test The Mute, Not zero. And once again it's greater than that. So everything in there is actually okay, so we just press calculate and we get 1.47 and the p value of .09. So that's enough information to go forward. So the test statistic is 1.4 68 And then the P value is zero 093 which is greater than my alpha value significantly. So I'm going to fail to reject h not fail to reject the null hypothesis. So I'm not rejecting that, that there is in fact no difference. So double negative, basically saying that these new golf clubs at least for these golfers, does not seem to help.

The following is a solution for number six and this looks at I think seven or eight different golfers and they're supposed to record their lowest score with um they're old clubs and then they're supposed to play with these new clubs for um you know a couple months and then they record their new lowest scores with the new clubs. So the data values. So L. One this is with their old clubs, these are like their lowest scores and then L. Two, this is with the new clubs and we're seeing if there's any difference. So what I'm gonna do is I'm gonna use this third column and I'm gonna code this third column to find those differences. So L one minus L. Two Is what I'm gonna do there. So L 1 -62. So these are my differences Between the two scores And to find those statistics I can just go to one of our stats. My list is L. three and I could calculate this X. Bar is actually my D. Bar and then this s of X. Is actually my S. F. D. So let's go and write those down and then answers, you know, first to questions basically. So 1.125 is my average difference. And then it's got a standard deviation about 2.167. And Part B. The point estimate is actually this d. bar as well, so it's just 1.125. So now I'm supposed to find the 99% confidence in the rule of the two of the differences. So if I go back to stat and air over the tests, I'm actually gonna be using the T interval. This follows the T distribution. Just a normal one sample T distribution with these matched pairs. That's what's kind of nice about him. So the data should be highlighted. L three is my list frequencies one and then my sea level my confidence level. I'm gonna change that 2.99 to represent 99%. So whenever I calculate I get negative 1.556-3.8063. So zero is contained in that interval. So I'm thinking there's not gonna be much of a difference here but we'll see. So negative 1.556 all the way up to 3.80 63. So now I'm testing to see if there is actually a difference. So the null hypothesis is that there is no difference. And then the alternative is saying that their scores originally are worse than their new scores with the new clubs. That's why it's greater than because the lower scores better engulf. So the first one should be greater than the second one. And we're testing at the 1% level significant. So I need to test statistic and also a P value. So I'm going to explicitly compare the p value to the alpha and that's going to tell me whether to reject or not reject. So if I go back to stats and then air over to test and this is just a normal plain old T test. It's not a two sample T test because it's just one sample of differences. So T test The Mute, Not zero. And once again it's greater than that. So everything in there is actually okay, so we just press calculate and we get 1.47 and the p value of .09. So that's enough information to go forward. So the test statistic is 1.4 68 And then the P value is zero 093 which is greater than my alpha value significantly. So I'm going to fail to reject h not fail to reject the null hypothesis. So I'm not rejecting that, that there is in fact no difference. So double negative, basically saying that these new golf clubs at least for these golfers, does not seem to help.

For this problem were given the total number of putts as one million, 613,234 were given that the number of pets made or 983,764 and the number of putts missed or 629,470. We're also told that the probability of the shot being part given that the cut was made his 0.64 the probability of the putt being a birdie, given that it was, made his 0.188 The probability of the putt being part, given that it was missed, his 0.203 and the probability of the pup being a birdie. Given that it was made this 0.734 hurt a asks what's the probability that the player makes a putt to calculate that we take the number of favorable outcomes divided by the total number of outcomes. So the number of favorable outcomes are the number of players that made the putt. So 983,764 divided by the total number of putts, which is one million, 613,234 to give a probability of making the putt as 0.6098 Part B says suppose that a player has a putt for par. What's the probability of the player will make the putt do this? We're going to use Berry space there, so to calculate the probability of making the putt. Given that it's par, we say that that's equal to the probability of a being part, given that the putt was made times the probability of being made divided by the probability of a being part, given that it was made times the probability of being made, plus the probability of it being part given that it was missed times the probability of it being missed. To calculate this, we need to know the probability of the shot being missed. So we take the number of favorable outcomes divided by the total number of outcomes. So the number of what's that were missed was there. 629,470 divided by the total number of putts, which is 1,613,234 to give a probability of a pup being missed as 0.3902 We can then take the probabilities above, plus the calculated probabilities of a pup being made or missed and plug it in to the equation. So the probability of making the shot given the shot was par. We 0.64 times. 0.6098 divided by 0.64 times. 0.6098 plus 0.203 times 0.3902 To give a probability of 18313 Part C says suppose that a player has a putt for birdie. What is the probability that the player will make the putt we again use based here? Um so the probability of making the putt, given that it's birdy is equal to the probability the pup being birdie. Given that it's made times the probability of the pup being made divided by the probability of the pup being birdie, given that it was made times, the probability of it being made, plus the probability of the putt being birdie. Given that it was missed times the probability of being missed putting in the numbers given by the problem and calculated above, we get the probability to be 0.188 times, 0.609 e divided by 0.188 times 0.6 year 98 plus 980.734 times 0.3902 To give a probability, a 0.285 nights party asks us to comment on the differences in probabilities from Parts B and C. So in part B, the probability of making a putt given that the shot is par is 0.8313 Well, for part C. The probability of making a putt given that it's Birdy is 0.2859 Therefore, you have a higher probability of making a shot when taking a par putt, then when making a birdie putt.

Okay, so once again, you're gonna be choosing between a non pulled and pulled t test. First, your results, your data that you get from, your calculator of your choice, It's gonna look somewhat like this, you're going to get two of these population sizes that look approximately the same and they should be normal if one of them looks like this, uh you might want to opt for a non pulled teeth tests because these things are deviations are going to be a little different. So there you go, that's going to be pulled versus non pooled. The box plots are gonna look approximately the same with these things lining up right here and here. Same thing with the Green one. Their mediums should be lining up right there and then the extreme values are gonna be out about the same place. Okay, that those standard deviations and better yet, when you get the standard deviation for this plot and or this plot, see if these two are approximately equal if they are opt for a non or a pooled test, If they aren't, then the warriors just go for a non pooled test. Now we were asked to find at the 5% significance level. So that means how far right here, at .05 to find if there is a significant difference between the two, we want to find out if it is better or not. So that is going to be either a left tail or right tail test, depending on which way you want to look at it as a reminder. Still left tail test is a negative right tail test. So let's just say we chop it off right here. Whether or not you get a negative value, just make sure you get your positive values. So you're just looking at this side of the graph. This right here is going to be .05. If your test test statistic lens in this region right here, we will reject the null hypothesis in favor of the alternative, which is going to be that new one is greater than youtube Alternatively could be written as me 1 -9 too Does not equal zero. Are actually is greater than zero like that. Oh, that's messy. Okay. But anyway, under the next part, which is probably making that confidence interval Uh at the 90% confidence interval. So you will be using that same critical value that you found using the significance level .5 since the you take half of the significance level, asked for for a confidence interval as a reciprocal of the sample sizes. And then you take the difference of the means right there. And then that will be your 90% confidence interval. Okay.


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