Solutions for STAT101 Fall 2004 Hypothesis Testing and Confidence Intervals, Exams of Statistics

Download Solutions for STAT101 Fall 2004 Hypothesis Testing and Confidence Intervals and more Exams Statistics in PDF only on Docsity! SOLUTIONS FINAL STAT101 FALL 2004 1. A. Plainly we take H0 : µ ≤ 60 and Ha : µ > 60. Now t = x̄− µ s/ √ n = 62.6− 60 7.4/ √ 334 = 6.42. Since n is large it matters not if we compare with z∗1 or t ∗ 1. In any case t = 6.42 is off the scale and t >> z∗1 = 2.33 so we reject H0. B. (i) Here we want the probability of getting a type II error. The acceptance region is given by x̄ ≤ µ0 + z∗1 σ√n = 60 + 2.33 7.4√ 334 = 60.9434. And so we want P (x̄ ≤ 60.9434|µ = 61) = P (Z ≤ 60.9434−61 7.4/ √ 334 ) = 0.4443. (ii) Using the formula we gave in class n = (zα + zβ)2σ2 (µa − µ0)2 = (2.33 + 1.645)2(7.4)2 61.60 = 865.24, so 866 will do. C. Here we take H0 : µd = 0 and Ha : µd 6= 0. Comparing t = 3.25− 0 7.15/ √ 20 = 2.033 with t∗ = 2.093 (taking 19 degrees of freedom) we see that we must retain H0. 2. A. (i) Of course we take a normal with the same mean and standard deviation: N(2.77, 2.1082). (ii) As the saying goes, all roads lead to Roma: it is plain from the histogram or the normal quantile plot that the data is right leaning. Comparing the mean and the median supports this. It may be the case that a normal approximation is sufficiently accurate in the middle hunk of data. B. (i) This is just x̄± z∗ σ√ n or 2.77± 1.96 2.108√ 400 = 2.77± 0.2068. Thus the confidence interval runs from 2.563 to 2.977, which we could just read off the JMP output to a sufficiently high degree of accuracy! (ii) Using the formula we discussed in class with E = 0.2 we have n = z∗2σ2 E2 = 1.9622.112 0.22 = 427.58, so that 428 will suffice. C. Again with wanton abandon we apply the formula we learn’t in class. We have p̂ = 66400 = 0.165, and so the confidence interval interval is estimated by 1 2 SOLUTIONS FINAL STAT101 FALL 2004 p̂± z∗ √ p̂(1− p̂√ n = 0.16± 1.96 √ (0.165)(0.835)√ 400 = 0.165± 0.036. Alternatively we could use p̃ = 66+2400+4 = 0.1683 and obtain 0.1683 ± 0.0365. D. (i) Here we look at the contingency table and extract conditional probabilities. For example, the probability of not getting a job given that one is a finance major is 9.52%. The others are around 20%; specifically, management is the highest at 21.74%, market- ing at 19.44% and ‘other’ at 18.9970. (ii) Literally the P -value in this context says that if the fraction of students without jobs were uniform across the majors the probability that we would observe the discrepancy in the data in one more extreme (i.e. fractions unequal) happens about 3.6% of the time. If, for example, α = 0.05 we would reject the null hypothesis in favour of a relationship between major and students not getting a job offer. 3. A. We have µ = 0(0.15) + 1(0.15) + · · · + 10(0.1) = 2.57 which one can check against the JMP output. B. Without thinking we write down H0 : p = 0.15 and Ha : p 6= 0.15. We compute z = 0.165− 0.15√ 0.15(0.85)/400 = 0.84 and since |z| < z∗ = 1.96 we regretfully retain the null hypothesis. C. (i) Of course we have an instance of our old friend the central limit theorem, and just as a week is a long time in politics, a thou- sand is a big number in probability; anyhow a cursory perusal of the normal quantile plot confirms once again our faith in this marvellous result. (ii) This is what one might term a ‘derived’ hypothesis test. A moment’s reflection tells us that we take H0 : p = 0.95 and Ha : p 6= 0.95. Here we have p̂ = 961/1000 = 0.961 and z = p̂− p0√ p0(1− p0)/n = 0.961− 0.95√ 0.95(0.05)/1000 = 1.59, and since |z| = 1.59 < z∗ = 1.96 we retain H0. 4 A. One could begin by inspecting the equations for each fit, and note that the gradients in particular are vastly different. One could also compare specific instances of the two predictors: take, for example, X = 5. Then Ŷ1 = 29.89 + 9.4378(5) = 74 and Ŷ2 = 56.73 + 2.069(5) = 67.1. We could be more quantitative and compare RMSE: in the first case we have 17.85 and the second 6.71. What’s happening here is the associating 0 values for 0 jobs inflates the variance.