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Review I Key MA 223 1. E(3X − Y ) = E(3X) + E(−Y ) = 3E(X) − E(Y ) and V (3X − Y ) = V (3X) + V (−Y ) = 32 V (X) + (−1)2 V (Y ) = 9V (X) + V (Y ). 2. P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = 0.45 (Venn diagram!). If A and B are independent P (A ∩ B) = P (A)P (B), which isn’t quite true. 3. (a) f ≥ 0 and integrates to 1. (b) Find ∫ E(X) = ∫ 6 6 xf (x) dx = 9/2, V (X) = 0 0 (x − 9/2)2 f (x) dx = 27/20. 4. (a) Find x̄ = 4 and s2 = 11/2. The sorted data is 1, 2, 2, 3, 4, 4, 5, 7, 8, so the median is 4. The first quartile is the median of 1, 2, 2, 3, 4, which is 2, and the third quartile is 5. The interquartile range is 5 − 2 = 3. (b) Draw a box going from 2 to 5, whiskers out to 1 and 8. 5. (a) Do a PAIRED t-test. The appropriate hypotheses are H0 : µD = 0 versus Ha : µD < 0, where µD is the mean reduction in SENS (assumed the same over all patients). Note, we’re looking for solid evidence of a REDUCTION, so this Ha is appropriate. (b) Using Minitab we find a p-value of 0.009, so we reject H0 . Alternatively, let di = xi − yi ¯ for the before √ and after data points. You can compute that d = 7.87, sd = 5.61. Let ¯ d / 6), which should follow a t distribution with 5 df. We reject H0 if T is too T = d/(s large—at 5 percent significance this means T > 2.015. We actually obtain T = 3.44, significant at the 0.009 level. 6. Do a two-sample test. The appropriate test statistic is x̄1 − x̄2 Z=√ s21 /n1 + s22 /n2 which under H0 is standard normal (or t with lots of degrees of freedom—69, actually, so basically normal). We will reject H0 at α = 0.05 if |Z| ≥ 1.96. In this case you can compute that Z = 0.267, so we do not reject H0 . In fact this Z value corresponds to p ≈ 0.79. 7. (a) Do a one-way ANOVA. In Minitab the ANOVA table comes out as Source Factor Error Total df 2 15 17 SS MS 3.648 1.824 7.630 0.509 11.278 F p-value 3.59 0.053 We just barely fail to reject H0 : τ1 = τ2 = τ3 = 0 (or H0 : µ1 = µ2 = µ3 ). 1 (b) The 160 flow rate has the highest mean, 4.417. The confidence interval is √ √ 4.417 − t0.025,15 M SE /6 ≤ µ + τ2 ≤ 4.417 − t0.025,15 M SE /6 which works out to 3.796 ≤ µ + τ2 ≤ 5.038. 8. (a) The scatterplot looks like good support for a linear regression; everything lies close to a line. (b) The regression in Minitab gives slope β1 = 0.827 and intercept β0 = −1.13. (c) This is just (0.827)(50) − 1.13 = 40.22. (d) Here r2 = 0.975. (e) It appears that the residuals increase in variability as the rainfall volume increases. This doesn’t mess up anything above (since we haven’t done any statistics in this problem) but it would mess up the usual statistical computations one does for confidence intervals on the slope and intercept estimates (which we haven’t done in this course). 2