Homework 8 Answers
Chapter 261. (a) True. (b) False. The null hypothesis says the difference is due to chance; the alternative says it's "real". (See pp. 478-479) 2. (a) null: avg of box equals 18/38 alt: avg of box more than 18/38 (b) z = [1890-1800]/{[(18/38)(20/38)]^0.5(3800)^0.5} is about 2.92, so the p-value is about 0.002. (c) It is very likely that it's biased in favor of reds. 4. (55-63)/[20/(30^0.5)] is about -2.19. So, we would expect that a TA section would do this badly (or worse) due to chance only about (100-97.22)/2 = 1.39%. 6. (a) [102-175]/[(1/2)(350^0.5)] is about -7.80, so the p-value is almost 0. (b) 102/(102+248) is about 0.29. This is the fraction of females in the group. (0.29)(100) = 29 So we expect 29 of 100 to be female. SE = (100^0.5) [(0.29)(0.71)]^0.5 is about 4.54 correction factor: [(350-100)/(350-1)]^0.5 is about 0.846 SE for number (drawing without repl): 0.846(4.54) is about 3.8 (9-29)/(3.8) is about -5.2, so p-value is almost 0. This is the chance of drawing 9 women or fewer in a random sample of the group. (c) Since the likelihood of these events is so small, it seems hard to believe the judge could have chosen like this by chance. Chapter 27 1. null hypothesis: tickets 50% positive alternative hypothesis: more than 50% positive exp. number of positives: 500(1/2) = 250 SE number of positives: (1/2)(500)^0.5 = 5(5)^0.5 z-score: (276-250)/[5(5)^(0.5)] is about 2.33. (The closest number on our normal table chart is 2.35.) (100-98.12)/2 = 0.94% --- p-value = 0.0094 This leads us to think that the null hypothesis is probably not correct; it is very likely that more than 50% of the tickets are positive. 3. percentage who rated clergyman "very high or high" 1985: 67% 1992: 54% Results are based on independent simple random sample of 1000 persons in each year. (a) Should make a 2-sample z-test because these results used a different sample each year (independent of the other year). We need to compare the means of the sample. (b) Need 2 box models, one for each year. Each box represents a different year's state of public opinion. Tickets show "1" or "0", with "1" meaning "very high" or "high" opinion of clergymen. It doesn't matter how many tickets - just need the right proportion of "1"'s and "0"'s (67% in 1985 box and 54% in 1992). null hypothesis: same proportion rating clergymen "high" or "very high" in both years - same percentage of "1"'s in both boxes alternative hypothesis: percentage of "1"'s is lower in the 1992 box (c) z-score: (0-13)/2.17 = -5.99 -- p-value almost 0 SE%(1985) = [(0.67)(0.33)]^0.5/(1000^0.5) x 100% = 1.49% SE%(1992) = [(0.54)(0.46)]^0.5/(1000^0.5) x 100% = 1.58% SE for difference of %: [(1.49)(1.49) + (1.58)(1.58)]^0.5 = 2.17% Since p-value is so low, chance is not a good explanation fo rthis result. The test does not prove that scandals were the cause of the change in opinion, however. 5. 383 students at UBC 200 chosen to got question A, 92 answered "yes" remaining 183 got question B, 161 answered "yes" null hypothesis: difference due to chance alternative hypothesis: difference due to something else (wording of questions) SE%(A): [(0.46)(0.54)]^0.5/(200^0.5) x 100% = 3.5% SE%(B): [(0.88)(0.12)]^0.5/(200^0.5) x 100% = 2.4% z-score: [(46-88) - 0]/[(3.5)(3.5) + (2.4)(2.4)]^0.5 = -9.90 -- p-value almost 0 Chance isn't a good explanation, so it's probably something about the wording of the question. 7. (a) 592 assigned to treatment, 48.3% re-arrested 154 assigned to control group, 49.4% re-arrested null hypothesis: treatment group no better than control alternative hypothesis: treatment group does better (less arrests) SE%(T): [(0.483)(0.517)]^0.5/(592^0.5) x 100% = 2.05% SE%(C): [(0.494)(0.506)]^0.5/(154^0.5) x 100% = 4.03% z-score: [(48.3 - 49.4) - 0]/[(2.05)(2.05) + (4.03)(4.03)]^0.5 = -0.24 p-value = (100-19.74)/2 = 40.13 -- p-value = 0.4 So, there is no significant difference. (b) null hypotheis: no difference in the averages for the 2 groups alternative hypothesis: income support group has lower average SE for treatment average: 0.7 wks SE for control average: 1.4 wks SE for the difference: 1.6 wks observed difference: -7.5 z-score: -7.5/1.6 is about -4.7 -- p-value almost 0 So, we reject the null hypothesis in favor of the idea that income support makes the released prisoners work less.
Additional Problem
1. You should have used JMP-IN to do the calculations. (See my posting on the newsgroup if you need the instructions for later reference.) You should have gotten a t-score of 1.08 and a p-value that was greater than 0.2. Based on the sample, there is no reason to believe that the average number of disputes is higher or lower on Mondays than on any other day of the week.