Stat 1040
Chapter 27 Solutions

  1. No. If they were correct, the number of positives should be like the sum of 500 draws from a box with one ``1'' and one ``0''. The expected number of positives is 250 and the SE is sqrt{500}*0.5 = 11. The observed number of positives is 2.4 SE's above what we would expect, so we conclude that more than 50% of the tickets in the box show positive numbers.

    1. There are two samples, so we need a 2-sample z-test.
    2. There are two boxes. In the 1985 box, there is a ticket for each person in the population, marked ``1'' for those who would rate clergymen ``very high or high'' and marked ``0'' otherwise. The 1985 data are like 1000 draws from this box. The 1992 box is set up the same way, for 1992. The null hypothesis says that the percentage of ``1''s in the 1985 box is the same as the percentage of ``1''s in the 1992 box. The alternative hypothesis says that the percentage of ``1''s in the 1992 box is smaller than that for the 1985 box.
    3. The SD of the 1985 box is estimated as sqrt{0.67*0.33} = 0.47 and the SE of the percent is 1.5%. Similarly, the SE for the 1992 percent is 1.6%. The SE for the difference between these two percentages is sqrt{1.5^2 + 1.6^2} = 2.2%. The observed percentages are 67% (1985) and 54% (1992), so the z-value is z=(67-54)/2.2 = 6 and the P-value is almost zero. We conclude that the difference is real (we can not say what the cause is).

    1. The SE for the treatment percent is 2.0% and for the control percent it's 4.0%. The SE for the difference between the two percents is 4.5%. The observed difference is 1.1%, so z=0.24 and the P-value is large, so the observed difference could easily be due to chance. There is no evidence that income support reduces recidivism.
    2. The SE for the treatment average is 0.7, for the control it's 1.4, and for the difference it's 1.6. The observed difference is -7.5 weeks, so z=-7.5/1.6 = -4.7 and the P-value is almost zero. We conclude that the difference is real - income support makes the released prisoners work less.

    1. The sample percentages are 47.8% and 4.4%. The standard errors for these percentages are 7.4% and 3.0%. The SE for the difference is 8%, so z=(47.8-4.4)/8 = 5.4. The P-value is almost zero, so we reject the null hypothesis and conclude that the difference is real - people are more likely to say they would do it than to actually do it.
    2. The sample percentages are 30.4% and 4.4%. The standard errors are 6.8% and 3.0%. The standard error of the difference is 7.4%, so z=(30.4-4.4)/7.4 = 3.5 and the P-value is very small. We reject the null hypothesis and conclude that people are more likely to volunteer if they have been first asked to predict whether they would.
    3. These are the same people, so a 2-sample z-test is not valid (we need independent simple random samples).

  5. These are children from the same 400 families and the sampling was not random, so a 2-sample z-test is not valid (we need independent simple random samples).