MATH 203 Test 3: Hypothesis Testing and Confidence Intervals - Prof. David K. Neal, Exams of Statistics

Download MATH 203 Test 3: Hypothesis Testing and Confidence Intervals - Prof. David K. Neal and more Exams Statistics in PDF only on Docsity! MATH 203 Test 3 Review Solutions 1. Note that x 1 − x 2 = 39.1 (a) If 1 = 2 were true, then there would be only 0.0027 probability (or a 0.27% chance) of getting an x 1 − x 2 of 39.1 or higher with these sample sizes. We have strong evidence to reject H0 . (b) If 1 − 2 = 40 were true, then there would be a 47.45% chance of getting an x 1 − x 2 of 39.1 or lower with these sample sizes. There is no evidence to reject H0 . (c) z = (x 1 − x 2 ) − M 1 2 n1 + 2 2 n2 = (1079.6 −1040.5) − 40 1802 300 + 1502 250 = –0.06396 → compare on N(0, 1) curve (d) Because we have rather large samples, the S values will approximate the ’s and the test statistic will still be close to N(0, 1), so we can still use a 2Sample Z–Test. ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 2. (a) Test H0 : p1 = p2 vs. Ha : p1 < p2 Note: p 1 – p 2 = –0.03 (b) ˆ p = 110 + 58 200 + 100 = 168 300 = 0.56 (This value is shown on output screen). (Note also: 1 − ˆ p ≈ 0.44) (c) z = p 1 − p 2 ˆ p × (1 − ˆ p ) n1 + ˆ p × (1 − ˆ p ) n2 = 0.55 − 0.58 0.56 × 0.44 200 + 0.56 × 0.44 100 ≈ –0.49346 → N(0, 1). (d) If p1 = p2 were true, then there would be a 31% chance of p 1 – p 2 being –0.03 or lower with samples of these sizes. We cannot reject the claim that p1 = p2 . (e) Ha : p1 – p2 > –0.05; z = p 1 − p 2 − (−0.05) p 1 × (1 − p 1) n1 + p 2 × (1 − p 2 ) n2 = 0.55 − 0.58 − (−0.05) 0.55 × 0.45 200 + 0.58 × 0.42 100 We can reject H0 if z ≥ 1.645 (in the rejection region), but not if z < 1.645. –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 3. (a) Use command 2cdf(lower, upper, degrees) . (i) P(12 ≤ 2(19) ≤ 25) ≈ 0.725 (ii) P( 2 (17) ≤ 20) ≈ 0.726 (iii) P( 2 (22) ≥ 20) ≈ 0.583 0 12 17 25 0 15 20 0 20 (i) (ii) (iii) (b) (i) 90% of the 2 (19) distribution is from 10.12 to 30.14 (ii) 95% of the 2(17) distribution is from 7.564 to 30.19 (iii) 98% of the 2(22) distribution is from 9.542 to 40.29 4. For a normal distribution with = 53 and = 0.9: (a) With n = 40: P(0.81 ≤ S ≤ 0.99) = P 39 × 0.81 2 0.92 ≤ (n −1)S 2 2 ≤ 39 × 0.992 0.92         = P 31.59 ≤ 2 (39) ≤ 47.19( ) ≈ 0.6218 . (b) For n = 30 and S = 0.76, we use the 95% chi-square scores from the 2 (29) curve which are L = 16.05 and R = 45.72. Then, (n − 1) × S2 R ≤ ≤ (n − 1) × S2 L , so we have a 95% confidence interval of 29 × 0.762 45.72 ≤ ≤ 29 × 0.762 16.05 , or 0 .6053 1 .0216 . (c) With S = 0.86 and n = 30 : We test H0 : = 0.90 vs. Ha : < 0.90 . The test stat is (n −1)S2 2 = 29 × 0.862 0.92 = 26.4795 . The P -value is P( 2 (29) ≤ 26.4795) ≈ 0.40 (the left-tail for Ha : < 0.90 ). If = 0.90 were true, then there is a 40% chance of obtaining an S of 0.86 or smaller with a sample of size 30. No evidence to reject H0 . With S = 1.1 and n = 30 : We test H0 : = 0.90 vs. Ha : > 0.90 . The test stat is (n −1)S2 2 = 29 × 1.12 0.9 2 ≈ 43.32 . Now the P -value is P( 2 (29) ≥ 43.32) ≈ 0.0425 (the right-tail for Ha : > 0.90 ). If = 0.90 were true, then there is only a 4.25% chance of obtaining S of 1.1 or larger with a sample of size 30. There is evidence to reject H0 , with a 10% or even 5% level of significance, in favor of > 0.90 –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 5. Obtained frequencies: 1200 responses total Sprint Verizon T-Mobile Cingular AT&T Other 150 210 220 200 70 350 (a) Expected from a sample of 1200 if the given percentages were true: Multiply each stated percentage by 1200 Sprint Verizon T-Mobile Cingular AT&T Other 180 216 240 192 72 300 (b) x = (150 −180)2 180 + (210 − 216)2 216 + (220 − 240)2 240 + (200 −192)2 192 + (70 − 72)2 72 + (350 − 300)2 300 = 30 2 180 + 6 2 216 + 20 2 240 + 8 2 192 + 2 2 72 + 50 2 300 = 15.5 For 6 bins, use the 2 (5) curve: P( 2 (5) ≥ 15. 5 ) ≈ 0.00823 = P -value (right-tail)