1. AP Stat 2007 Free Response

2. 1. A. Roughly speaking, the standard deviation (s = 2.141) measures a “typical” distance between the individual discoloration ratings and the mean discoloration rating for the strawberries in the control group.

3. 1. B The preservative does appear to have been effective in lowering the amount of discoloration in strawberries. The discoloration ratings for the strawberries that received the preservative, shown in the top dotplot, are clearly centered at a value that is lower than the center of the discoloration rating distribution for the control group, shown in the bottom dotplot. In addition, the dotplots can be used to find all 5 statistics in the 5- number summary for both groups. In fact, 4 of the 5 statistics are lower for the strawberries that received the preservative. (the maximum is the only exception)

4. 1.C Since 0 is not contained in the 95 percent confidence interval for the difference,, We can conclude that there is a significant difference between the mean ratings for the two groups at the level. The population mean discoloration rating for untreated strawberries is estimated to be between 0.16 and 2.72 units higher than the population mean discoloration rating for treated strawberries. Thus, we think there would be a difference in the population mean discoloration ratings fort treated and untreated strawberries.

5. 2. A. A control group gives the researchers a comparison group to be used to evaluate the effectiveness of the treatments. The control group allows the impact of the normal aging process on joint and hip health to be measured with appropriate response variables. The effects of glucosamine and chondroitin can be assessed by comparing the responses for these two treatment groups with those for the control group.

6. 2.B. Each dog will be assigned a unique random number, 001 – 300, using a random number generator on a calculator, statistical software, or a random number table. The numbers will be sorted from smallest to largest. The dogs assigned the first 100 numbers in the ordered list will receive glucosamine. The dogs with the next 100 numbers in the ordered list will be assigned to the control group. Finally, the dogs with the numbers 201 – 300 will receive chondroitin.

7. 2. C. The key question is which variable has the strongest association with joint and hip health. The goal of blocking is to create groups of homogeneous experimental units. It is reasonable to assume that most clinics will see all kinds and breeds of dogs so there is no reason to suspect that joint and hip health will be strongly associated with a clinic. On the other hand, different breeds of dogs tend to come in different sizes. The size of a dog is associated with joint and hip health, so it would be better to form homogeneous groups of dogs by blocking on breed.

8. 3. A. The random sample of n = 15 fish is more likely to have a sample mean length greater than 10 inches. The sampling distribution of the sample mean,, is normal with mean and Thus, both sampling distributions will be centered at 8 inches, but the sampling distribution of the sample mean when n = 50. The tail area ( ) will be larger for the distribution that is less concentrated about the mean of 8 inches when the sample size is 15.

9. 3 B.

10. 3. C Yes. The CLT says that the sampling distribution of the sample mean will become approximately normal as the sample size n increases. Since the sample size is reasonably large (n = 50), the calculation in part B will provide a good approximation to the probability of interest even though the population is non-normal.

11. 4. A hypothesis test for the mean difference in the level of E.coli bacteria contamination in beef detected by the two methods will be conducted.

12. 4. Part 1 Where is the mean difference (method A – method B) in the level of E. coli bacteria contamination in beef detected by the 2 methods.

13. 4 Part 2 Paired t – test = Assumptions: 1.Since the observations are obtained in 10 randomly selected specimens, it is reasonable to assume that the 10 data pairs are independent of one another. 2.The population distribution of differences is normal.

14. 4. Part 3

15. 4. Part 4 Since the p – value is greater than 0.5, we cannot reject the null hypothesis. We do NOT have statistically significant evidence to conclude that there is a difference in the mean amount of E. coli bacteria detected by the 2 methods for this type of beef. In other words, there does not appear to be a significant difference in these two methods for measuring the level of E. coli contamination in beef.

16. 5. A. This is an experiment because the researchers imposed treatments by randomly assigning drivers to the two different conditions (simulated driving while talking on a cell phone versus simulated driving while talking to a passenger).

17. 5. B. The proportion of drivers who miss an exit while using a cell phone The proportion of drivers who miss an exit while talking to a passenger

18. 5.C. Assumptions: ~ Independent random samples ~ Does not meet requirement Assumptions for sample size are not met for this problem. Will continue with work – sample size needs to increase.

19. 5.D. Assuming that talking on a cell phone and talking to a passenger are equally distracting (there is no difference in the two population proportions of drivers who will miss the exit), the p-value measures the chance of observing a difference in the two sample proportions as large as or larger than the one observed. Since the p-value 0.0683 is larger than 0.05, we cannot reject the null hypothesis. That is, we do not have statistically significant evidence to conclude that using a cell phone is more distracting to drivers than talking to another passenger in the car.

20. 6.A. The value 1.080 estimates the average increase (in feet) in the perceived distance for each additional foot in actual distance between the two objects.

21. 6. B. The model with 0 intercept makes more intuitive sense in this particular situation. If the 2 objects are placed side by side (so the actual distance is 0), then we would expect the subjects to say that the distance between the objects is 0.

22. 6.C. The true slope between the perceived distances and the actual distances.

23. 6.C. continued Since the p-value 0.398 is greater than 0.05, we cannot reject that the true slope = 1. That is, we do not have statistically significant evidence to conclude that the subjects overestimate the distance with the magnitude of the overestimation increasing as the actual distance increases.

24. 6.D. Contact wearers (contact = 1): Perceived distance = 1.05(actual distance) + 0.12(actual distance) Noncontact wearers (contact = 0): Perceived distance = 1.05 (actual distance)

26. 6. E. Model 3 allow prediction of perceived distance separately for contact wearers and for noncontact wearers. The value of 1.05 estimates the average increase (in feet) in the perceived distance for each one-foot increase in actual distance for the population of noncontact wearers. The value of 0.12 estimates the additional increase (in feet) in the average perceived distance for each one-foot increase in actual distance for the contact wearers.