- August 19, 2020

1Solutions to Practice Exam A 1. [15 marks] (a) Define µB and µA as the true mean pre- and post-treatment testosterone levels. i. H0 : µB − µA = 0 (or H0 : µB = µA) versus Ha : µB − µA > 0 (or Ha : µB > µA). ii. This is a paired t-test, so the test statistic is: T = X¯d Sd/ √ n ∼ t(7). The observed test statistic is: tobs = x¯d sd/ √ n = 762 211/ √ 8 ‘ 10.21. iii. Need to use the correct given output (pt(10.21,7) = 0.9999): P -value = P (T ≥ tobs) = P (T ≥ 10.21) = 1− P (T < 10.21) = 1− 0.9999 < 0.05. iv. The P -value is very small, so we have strong evidence to reject H0. We can conclude that Depo-Provera is successful in lowering testos- terone levels in males. (b) We require that: • The observations are independent. ⇒ Examine the sampling design and hope for a random sample or a simple random sample (SRS) (however no info given in the question about this). • The distribution of the observed (mean) differences, i.e., X¯d is ap- proximately normal. ⇒ Sample size is a bit small; a normal quantile plot of differences might be useful to check the approximation. (c) i. A 95% CI for before-treatment testosterone level is (need to use t∗ = qt(0.975,7) = 2.3646):( x¯± t∗ s√ n ) = ( 834.25± 2.365× 226.04√ 8 ) = (645.28, 1023.22). 2ii. 679 is within the bounds of the above confidence interval. So, there is no evidence the before-treatment testosterone levels are higher on average than military men. 32. [15 marks] (a) Position Meniscal Tear MCL Tear ACL/Other Total Forward 13 14 11 38 Back 12 9 17 38 Total 25 23 28 76 (b) Expected count = row total× column total n = 38× 25 76 ' 12.5. Position Meniscal Tear MCL Tear ACL/Other Forward 12.5 11.5 14 Back 12.5 11.5 14 (c) [6 marks] i. H0: Type of knee injury and position are independent (or no associ- ation) Ha: Type of knee injury and position are not independent (or asso- ciation). ii. X2 = ∑ (observed− expected)2 expected = (13− 12.5)2 12.5 +· · ·+(17− 14) 2 14 = 2.41. iii. X2 is χ2 with (2− 1)× (3− 1) = 2 degrees of freedom. iv. Need to use the correct given output (pchisq(2.41,2) = 0.7003): P -value = P (χ22 ≥ 2.41) = 1− P (χ22 < 2.41) = 1− 0.7003 = 0.2997. v. The P -value is large so we fail to reject H0 in favour of Ha. “There is no evidence that knee injury type and position are not independent” (or there is no evident association between knee injury type and position). (d) • The 76 data points are assumed to be independent. ⇒ This can’t be verified from the question since there is not enough info, but it can be achieved through (say) a random sample or a SRS of birth records. • All expected frequencies are assumed to be > 10. ⇒ This is easily verified from the expected counts table. 43. [15 marks] (a) Let X be the number of children born to females 15–24 years. i. P (X = 0) = 1− 0.09− 0.04− 0.01 = 0.86. ii. µX = ∑ all x xkpk = 0× 0.86 + 1× 0.09 + 2× 0.04 + 3× 0.01 = 0.20. iii. Use σX = √∑ (xk − µx)2pk. (b) Let Y be number of children born to females over 24 years. i. µX+Y = µX + µY = 2.02 + 0.20 = 2.22. ii. σX+Y = √ σ2X + σ 2 Y = √ 0.54772 + 1.482 = √ 2.490 ‘ 1.578. (c) i. A random sample or an SRS. ii. • To ensure observations are independent. • To remove selection bias/ensure sample is representative of pop- ulation. (d) Assign each listing a random number, sort, and select the first 600 phone numbers in the sorted list. Picking 600 numbers out of a hat is also OK, although a logistic night- mare! (e) Undercoverage is where some subjects in the population are not included in the sample process. For example, some ACT residents might not have a phone number listed in the ACT White Pages. 54. [15 marks] (a) pˆ ∼ N ( p, √ p(1−p) n ) . (b) The sample proportion is pˆ = 354/600, and it’s standard error is se(pˆ) = √ pˆ(1− pˆ) n = √ (354/600)(1− (354/600)) 600 = 0.0201. (c) i. We assume that: (1) the sample is a random sample or simple random sample (in other words, all observations are independent). (2) the normal approximation is valid for pˆ. ii. (1) We can’t check the independence assumption, as it’s not men- tioned anywhere in the question if this is a random sample or an SRS. (2) Check using npˆ = 354 > 10 and n(1 − pˆ) = 246 > 10, so the normality assumption holds. (Note that ). (d) A 95% CI for p is (using the give output):( pˆ± z∗ √ pˆ(1− pˆ) n ) = (0.59± 1.96× 0.0201) = (0.551, 0.629). (e) We want to solve for n such that: 1.96 √ p(1− p) n = 0.02. The maximum standard error is achieved at p = 1 2 , so: 1.96 √ 1 4n = 0.02 1 4n = ( 0.02 1.96 )2 n = 1 4 ( 1.96 0.02 )2 = 2401, so 2401 NSW residents need to be included in the sample to ensure that the margin of error is no more than 0.02 at 95% confidence.