Answers to Homework 4

Transcription

1 Answers to Homework 4 1. (a) Let p be the true probability of an auto having automatic transmission. The observed proportion of autos is.847, and a 95% confidence interval for p has the form p ± z.025 p(1 p)/n. Substituting in the observed values gives.847 ± (1.96) (.847)(.153)/1203 =.847 ±.02. The exact interval, which we can get from Minitab, is ( , ), which is unsurprisingly very similar to the standard interval. (b) We can actually just get this from Minitab: One-Sample T: City mileage Variable N Mean StDev SE Mean 90% CI City mileage (20.307, ) Of course, this is based on the usual formula X ± t.05 n 1 s/ n. This uses the exact t-critical value of (based on 1202 degrees of freedom), but if you used (based on df = ) you will get virtually the same answer. (c) We now want a prediction interval, X ± t s A prediction interval for n mileage is thus ±( )(5.882) miles per gallon = ±9.687, or (10.9, 30.3) (d) We are assuming that these data can be viewed as a random sample from a normal distribution. We already discussed in Homework 1 why these are not a random sample, but have also said that we will ignore that here. What about normality? This is apparently problematic here, as a histogram of city mileages shows a noticeable right tail, with some very high mileages coming from hybrid cars: c 2017, Jeffrey S. Simonoff 1

2 This probably has not had a strong effect on the confidence interval, as with a sample this large from a not-greatly-nonnormal distribution the Central Limit Theorem has likely taken over, but this has likely affected the prediction interval. The obvious thing to do to try to fix the prediction interval is to take logs, construct the interval in the logged scale, and then exponentiate the interval endpoints to get back to the original scale. This will probably work better, since logged mileages are certainly closer to normally distributed (although still somewhat right-tailed): This output gives us the sample mean and standard deviation of the logged variable: c 2017, Jeffrey S. Simonoff 2

3 Descriptive Statistics: Logged city mileage Variable N Mean SE Mean StDev Minimum Q1 Logged city mileage Variable Median Q3 Maximum Logged city mileage The prediction interval in the logged scale is ±( )(.1132) 1 + 1/1203 = ± = (1.1118, ). Antilogging the two endpoints gives the prediction interval (12.9, 30.5) miles per gallon, being shifted up at the low end in particular (with an estimate of the typical mileage being the geometric mean = 19.9 miles per gallon). 2. (a) Let p be the probability that a yogurt has excellent taste. The number of such yogurts is 7 (of 23), so the observed proportion is.304, and a 99% confidence interval for p has the form p ± z.005 p(1 p)/n. Substituting in the observed values gives.304 ± (2.576) (.304)(.696)/23 =.304 ±.247 = (0.057, 0.551). The exact interval, which we can get from Minitab, is ( , ), which is noticeably different from the approximate interval, since the sample is small and the observed proportion is not very close to.5. (b) After using the calculator to create the price per serving variable, the standard confidence interval is available from Minitab: One-Sample T: Price per serving Variable N Mean StDev SE Mean 95% CI Price per serving (1.227, 1.649) (c) The prediction interval is X ± t s 1 + 1, or n 1.44 ± (2.074)(0.487)(1.022) = 1.44 ± 1.03 = ($0.31, $2.47). (d) This seems to be a suspect prediction interval, as the lower end is $0.50 less than the minimum value in the sample. The reason for this is clear from a normal plot of the variable (a histogram would work too): c 2017, Jeffrey S. Simonoff 3

4 The variable is reasonably normally distributed, except for one clear outlier, the extremely expensive Maple Hill Creamery Strawberry at $3.00 per serving. This could have inflated both the mean and standard deviation, so it should be omitted to see what effect it might have had. We should also look more carefully into this observation to see if we can understand why it is so unusual. If you go to the Maple Hill Creamery web site, you can see that their major selling point is as follows: Since 2009, we ve been crafting uniquely delicious 100% grass-fed organic dairy products, all while building a 100% grass-fed organic Milkshed in New York State. It makes sense that it would be more expensive to make diary products when cows are not fed corn or grain, and presumably this increased cost is passed along to consumers. It is also possible that the grass-fed organic system is a selling point that allows them to charge a premium price, although we can note that according to Consumer Reports its taste is relatively substandard, being one of only three of the yogurts that has a taste rating of only Good. Here is output after removing this point: One-Sample T: Price per serving Variable N Mean StDev SE Mean 95% CI Price per serving (1.2088, ) The confidence interval is about 25% narrower, reflecting the large drop in the standard deviation. The prediction interval is 1.37 ± (2.08)(0.357)(1.024) = 1.37 ± 0.76 = ($0.61, $2.13). It still appears to be too long at the low end, which could reflect a slightly short tail on the left side. c 2017, Jeffrey S. Simonoff 4

5 3. (a) We are looking for a prediction interval, which takes the form X ± t s 1 + 1, n or 15.1 ± (1.982)(15.4)(1.004) = 15.1 ± 30.6 = ( 15.5, 45.7). The interval goes into impossible negative values. The variable is, in fact, distinctly long right-tailed, with more than half of the floods having duration of 10 days or less and more than three-quarters having duration less than three weeks, but some lasting as long as months. You didn t have the data, of course, so you can t do this, but I can. Here is a histogram of the duration variable: The obvious solution is to take logs, construct a prediction interval in the logged scale, and then antilog. The logged variable is quite reasonably normally-distributed, so this should result in a reasonable interval: c 2017, Jeffrey S. Simonoff 5

6 Descriptive Statistics: Logged days Variable N Mean StDev Minimum Q1 Logged days Variable Median Q3 Maximum Logged days The interval is ± (1.982)(.3816)(1.004) = ±.759 = (.245, 1.763), which when antilogged gives the interval (1.8, 57.9) days. with a geometric mean of 10 days, clearly representing the shorter left tail and much longer right tail. (b) Let p be the true probability that there is at least one fatality from a flood. The observed proportion of movies is 78 =.696, and a 95% confidence interval for p 112 has the form p ± z.025 p(1 p)/n. Substituting in the observed values gives.696 ± (1.96) (.696)(.304)/112 =.696 ±.085 = (.611,.781). The exact interval, which we can get from Minitab, is ( , ). We are assuming for the approximate interval that the sample size is large enough to appeal to the normal approximation to the binomial, which seems to be okay (given the similarity in the exact and approximate intervals). More seriously, we are assuming that the probability of at least one fatality is the same for all floods, and whether or not a fatality occurs in any flood is independent of that in any other flood. The first assumption is clearly not true; as we discussed in class, there are big differences in character between floods from hurricanes and floods from snow melt (for example), and the chances of a fatality in one type is very different from that another. Further, as the 2017 hurricane season showed, there are local (in time) changes in weather patterns, and knowledge of the severity of one flood could tell us something about the severity of another flood, which would violate independence. c 2017, Jeffrey S. Simonoff 6