HOMEWORK #8 ANSWERS

HOMEWORK #8 ANSWERS

Chapter 12

1. In a run of 1 SD, the regression line rises r x SD. The slope is 0.60 x 20/10 = 1.2

final points per midterm point. The intercept is 55 - 1.2 x 70 = -29. So the equation is:

predicted final score = 1.2 x midterm score - 29.

6. The slope stays the same, and the intercept goes up by 10%.

8. (a) True. (b) False. (c) False. (d) True. (e) True.

Comment. See section 2.

10. $41,000 is likely to be too high. You need the other regression line, for income on IQ (section 10.5). A better estimate is $26,000.

12. No. There doesn't even seem to be any consistent direction to the association.

Chapter 26

2. The data are like 3800 draws made at random with replacement from a box

| ?? 0's ?? 1's |, with 1 = red.

(a) Null: The fraction of 1's in the box is 18/38.

Alt: The fraction of 1's in the box is more than 18/38.

(b) The expected number of reds (computed using the null) is 1800. The SD of

the box (also computed using the null) is nearly 0.5, so the SE for the number of reds is x 0.5 » 31. So z = (obs-exp)/SE = (1890-1800)/31 » 2.9, and P » 2/1000.

Comments. (i) This problem is about the number of reds. In the formula for the z-statistic, obs, exp, and SE all refer to the number of reds. The expected, as always, is computed from the null. In this problem, the null gives the composition of the box, so the SD is computed from the null; it is not estimated from the data (p.487).

(ii) This problem, and several others below, can be done using one-sided or two-sided tests. The distinction does not matter here; it is discussed in chapter 29.

5. The box has one ticket for each freshman at the university, showing how many hours per week that student spends at parties. So there are about 3000 tickets in the box. The data are like 100 draws from the box. The null hypothesis says that the average of the box is 7.5 hours. The alternative says that the average is less than 7.5 hours. The observed value for the sample average is 6.6 hours. The SD of the box is not known, but can be estimated from the data as 9 hours. On this basis, the SE for the sample average is estimated as 0.9 hours. Then z = (obs-exp)/SE » (6.6 - 7.5)/0.9 = -1. The difference looks like chance.

8. Model: There is one ticket in the box for each person in the county, age 18 and over. The ticket shows that person's educational level. The data are like 1000 draws from the box.

Null: The average of the box is 13 years.

Alt: The average of the box isn't 13 years.

The expected value for the average of the draws is 13 years, based on the null. The SD of the box is unknown (there is no reason the spread in the county should equal the spread in the nation), but can be estimated as 5 years--the SD of the data. On this basis, the SE for the sample average is estimated as 0.16 years. The observed value for the sample average is 14 years, so

z = (obs-exp)/SE = (14 - 13)/0.16 » 6,

and P » 0. This is probably a rich, suburban county, where the educational level would be higher than average.

10. Null: the 3 Sunday numbers are like 3 draws made at random (without replacement) from a box containing all 25 numbers in the table. The average of these numbers is nearly 436, and their SD is just about 40. The EV (expected value) for the average is 436, and the SE is 22; we are using the correction factor here. The 3 Sunday numbers average about 357, so

z = (obs - exp)/SE = (357 - 436)/22 » -3.6, and P » 2/10,000.

11. (a) The total weight of the 1000 guineas in the Pyx will be like the sum of 1000 draws made at random with replacement from a box. The average of the numbers in the box is 128 grains, and the SD is 1/200 x 128 = 128/200 grains. (The numbers in the box represent the possible weights of coins minted by the machine.) The expected value for the sum is 128,000 grains. The SE is x 128/200 » 20 grains. The total weight of the coins in the Pyx will be 128,000 grains, give or take 20 grains or so, and it is almost impossible for the Master of the Mint to fail the Trial of the Pyx: the remedy of 640 grains is 32 SEs.

(b) The method is as in (a). The total weight of the coins in the Pyx will be 127,500

grains give or take 20 grains or so. Again, the Master of the Mint can hardly fail. The total weight is extremely unlikely to fall below 128,000 - 640 = 127,360 grains-a cutoff which is still 7 SEs below the expected value.