Stat 1040
Chapter 26 Solutions

    1. True.
    2. False. It's the null that says the difference is due to chance.

  2. The data are like 3800 draws made at random from a box containing 0's and 1's (1 stands for red), but with the number of each ticket unknown.
    1. Null: the fractions of 1's in the box is 18/38. Alt: The fraction of 1's in the box is more than 18/38.
    2. The expected number of reds (using the box from the null hypothesis) is 1800. The SD of the box (again using the null hypothesis) is nearly 0.5, so the SE is sqrt{3800}*0.5 = (about) 31. So z = (obs - exp)/SE = (1890 - 1800)/31 = (about) 2.9, and P = (about) 2/1000.
    3. Yes.

  3. Null hypothesis: the data are like 200 draws from the box containing three 1's and a single 0 (1 = blue and 0 = white). The expected number of blues is 150, with a SE of about 6. So z = (obs - exp)/SE = (142 - 150)/6 = (about) -1.3 and P = (about) 10%. This is marginal, could be chance.

  4. 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, which puts the estimate for the SE for the average at 0.9 hours. Then z = (obs - exp)/SE = (6.6 - 7.5)/0.9 = -1. The difference looks like chance.

    1. This is like tossing a coin 350 times and asking for the chance of getting 102 heads or fewer. The expected number of heads is 175, and the SE is 9. The chance is about 0.
    2. 100 draws are made at random without replacement from a box with 102 1's (for the women) and 248 0's (for the men). The expected number of 1's is 29. If the draws are made with replacement, the SE for the number of 1's is sqrt{100}*sqrt{0.29*0.71} = (about) 4.54. The correction factor is sqrt{(350-100)/(350-1)} = (about) 0.846. The SE for the number, when drawing without replacement, is 0.846*4.54 = (about) 3.8. The chance is about 0.
    3. Judge Ford was not choosing at random, he was excluding women.

  6. 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 (based on the null) is 13 years, give or take 0.16 years or so. (The 0.16 is the SE calculated from the SD of the data, 5 years. There is no reason to assume the SD of the nation equals the SD of the county, so the data provides a better estimate.) The observed value for the sample average is 14 years, so z = (obs - exp)/SE = (14 - 13)/0.16 = (about) 6, and P is about 0. This may be a rich, suburban county, where the educational level is above average.

  7. Null: the 3 Sunday numbers are like 3 draws made at random (without replacement) froma 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 for the average is 436, and the SE is 22 (using the correction factor). The 3 Sunday numbers average about 357, so z = (obs - exp)/SE = (357 - 436)/22 = (about) -3.6, and P = (about) 2/10,000. Many deliveries are induced, and some are surgical, so obstetricians really can influence the timing and avoid deliveries on Sunday (when they would rather be golfing).

    1. There are 59 pairs. The treatment animal has a heavier cortex in 52 of the 59 pairs. If the treatment has no effect, the number where the treatment animal has the heavier cortex is like the sum of 59 draws from a box containing one 0 and one 1. The expected value of the sum is 59*0.5 = 29.5, and the SE is sqrt{59}*0.5 = 3.84. The observed number is 52, which is several SE's above the expected value, so the P-value is almost zero - there's virtually no chance of obtaining as many pairs as Rosenzweig did, or more, with the treatment animal having the heavier cortex. We conclude that the treatment did have an effect.