MATH 203 Test 2 Solutions: Statistical Inference and Hypothesis Testing - Prof. David K. N, Exams of Statistics

Download MATH 203 Test 2 Solutions: Statistical Inference and Hypothesis Testing - Prof. David K. N and more Exams Statistics in PDF only on Docsity! MATH 203 Test 2 Review Solutions 1. (a) The average lifetime of Type 1 batteries is from 1.55 hours less to 0.86 hours more than average lifetime of Type 2 batteries. (b) It is not conclusive which type of battery has the longer average lifetime. ______________________________________________________________________________ 2. (a) Re-write the interval as 0.024 ≤ p2 – p1 ≤ 0.052 and say The percentage of patients for whom Treatment B worsens the condition is from 2.4 percentage points higher to 5.2 percentage points higher than the percentage of patients under Treatment A. (Without re-writing the interval: The percentage of patients for whom Treatment A worsens the condition is from 5.2 percentage points lower to 2.4 percentage points lower than the percentage of patients under Treatment B.) (b) It is clear that Treatment B has the higher percentage of patients with a worsening of the condition. ______________________________________________________________________________ 3. (a) Test H0 : = 25 vs. Ha : < 25 (b) We can use a Z–Test because the measurement is normally distributed and = 2 is known. (c) z = x − n = 24.2 − 25 2 400 = –8 (d) If = 25 and = 2 were true, then there would be no chance of getting an x of 24.2 or lower with a sample of 400 cars. We can reject H0 . (f) Because we have a large sample, we can still use the Z–Test even if the measurement were not normally distributed. Also, because of the large sample, we can use S as an approximation for and still use the Z–Test. –1.645 N(0, 1) (e) = 0.05 curve Use –1.96 for = 0.025 . ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 4. (a) H0 : = 6 vs. Ha : > 6 (b) With a small sample, we can use the T–Test because the measurement (birth weight) is normally distributed, we do not know , but we have S. (c) If = 6 were true, then there would be an 11.9% chance of getting an x of 6.3 or higher with a sample of size 36. There is not enough evidence to reject H0 . (d) n = 36, use t(35) curve 1.690 = 0.05 (e) t = x − S n = 6.3 − 6 1.5 36 = 1.2 The test stat of 1.2 is not in the rejection region beyond 1.690; so x is not too large and there is not enough evidence to reject H0 . ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 5. Here p = 0.63 (a) Test H0 : p = 0.60 vs. Ha : p > 0.60 (b) z = p − p p × (1 − p) n = 0.63 − 0.60 0.60 × 0.40 1000 = 1.9365 (c) 1.645 N(0, 1) = 0.05 curve Use 1.96 for a 2.5% level of significance. (d) If p = 0.60 were true, then there would be only a 2.64% chance of obtaining a p of 0.63 or higher with a sample of size 1000. We can reject H0 because the P -value is less than 0.05. (Also, the test stat of 1.9365 is in the rejection region beyond 1.645.) (e) For = 0.025, the test statistic of z = 1.9365 is not in the rejection region beyond 1.96 (so the P -value is more than 0.025). Thus, with a 2.5% level of significance, p = 0.63 would not be considered too large and we would not have enough evidence to reject H0 . ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 6. (a) n ≥ z /2 × 0.5 e       2 = 1.96 × 0.5 0.0325       2 = 909.254438 ; a sample of size 910 is required. (b) n ≥ 1800 × 909.254438 1799 + 909.254438 = 604.322 ; so now we only need a sample of size 605. ______________________________________________________________________________ 7. (a) p = 342 950 = 0.36 . Then p ≈ p ± z /2 p (1− p ) n = 0.36 ± 2.576 × 0.36 ×0.64 950 ≈ 0.36 ± 0.04 . That is, 0.32 ≤ p ≤ 0.40 . (Can use the 1–PropZInt to compute the confidence interval with this formula.) With max margin of error: p ≈ p ± z /2 × 0.5 n = 0.36 ± 2.576 ×0.5 950 ≈ 0.36 ± 0.04179