Dinov's Stat 10 1 Dinov's Stat 10 Final Exam Review - SolutionsProbabilities and Proportions 1. (a) Females Males Total Stat10 0.3003 0.2949 0.5952 Stat11 0.1924 0.2123 0.4048 Total 0.4928 0.5072 1 (b) (i) 2949.02011593= (ii) 4928.02011991= (iii) 1924.02011387= (c) 5246.0814427= (d) 4186.01020427= 2. (a) 40 or under Over 40 Total Wearing a seat belt 0.484 0.369 0.853 Not wearing seat belt 0.066 0.081 0.147 Total 0.55 0.45 1 (b) 0.147 (c) 551.0147.0081.0= (d) 0.55 3. (a) Under 40 40 or over Total Mild cases 16 20 36 Serious cases 15 35 50 Total 31 55 86 (b) (i) 1860.08616= (ii) 8140.086153520=++ (iii) 4070.08635= (c) 3.05015= (d) 3636.05520= Dinov's Stat 10 2 4. (a) High-risk Low-risk Total In default 40% of 0.05 = 0.02 0.03 0.05 Not in default 0.2185 77% of 0.95 = 0.7315 0.95 Total 0.2385 0.7615 1 (b) 23.85% (c) 0839.02385.002.0= 5. (a) Female Male Total Business degree 27% of 0.175 = 0.04725 0.12775 0.175 Other degree 0.42281 48.75% of 0.825 = 0.40219 0.825 Total 0.47006 0.52994 1 (b) 0.52994 (c) 0.12775 (d) 1005.047006.004725.0= 6. (2)Stage I Statistics Tutorial Answers – Chapter 6 3 7. Let M be the morning travel time and N be the evening travel time. (a) Want pr(M + N > 60) E(M + N) = E(M) + E(N) = 31 + 35.5 = 66.5 minutes sd(M + N) = 22)(sd)(sd NM + =225.33 + = 4.60977, assuming M and N are independent. M + N ~ approx. Normal (µ = 66.5, σ = 4.60977) pr(M + N > 60) = 1 – pr(M + N ≤ 60) = 1 – 0.0793 = 0.9207 (b) Want pr(M > N) = pr(M – N > 0) E(M – N) = E(M) – E(N) = 31 – 35.5 = –4.5 minutes sd(M – N) = 22)(sd)(sd NM + =225.33 + = 4.60977, assuming M and N are independent. M – N ~ approx. Normal (µ = –4.5, σ = 4.60977) pr(M – N > 0) = 1 – pr(M – N < 0) = 1 – 0.8355 = 0.1645 (c) Want pr(N – M > 5) E(N – M) = E(N) – E(M) = 35.5 – 31 = 4.5 minutes sd(N – M) = 22)(sd)(sd MN + =2235.3 + = 4.60977, assuming N and M are independent. N – M ~ approx. Normal (µ = 4.5, σ = 4.60977) pr(N – M > 5) = 1 – pr(N – M < 5) = 1 – 0.5432 = 0.4568 (d) (i) Let TM = M1 + M2 + M3 + M4 + M5, where Mi ~ approx. Normal (µ = 31, σ = 3) E(TM) = E(M1 + M2 + M3 + M4 + M5) = 5E(M) = 5 x 31 = 155 minutes sd(TM) = sd(M1 + M2 + M3 + M4 + M5) = )(sd5 M× = 35 × = 6.7082 minutes, assuming independence of the morning travel times. TM ~ approx. Normal (µ = 155min, σ = 6.7082min) (ii) Let TN = N1 + N2 + N3 + N4 + N5, where Ni ~ approx. Normal (µ = 35.5, σ = 3.5) E(TN) = E(N1 + N2 + N3 + N4 + N5) = 5E(N) = 5 x 35.5 = 177.5 minutes sd(TN) = sd(N1 + N2 + N3 + N4 + N5) = )(sd5 N× = 35 × .5 = 7.8262 minutes, assuming independence of the morning travel times. TN ~ approx. Normal (µ = 177.5min, σ = 7.8262min) (iii) Let T = TM + TN E(T) = E(TM + TN) = E(TM) + E(TN) = 155 + 177.5 = 332.5 minutes sd(T) = sd(TM + TN) = 22)(sd)(sdNMTT + = 228262.77082.6 + = 10.3078 minutes, assuming independence of ten travel times in the week. Dinov's Stat 10, UCLA 1 Dinov's Stat 10 Final Exam Review- Solutions Sampling Distributions of Estimates, CLT 1. (a) (i) µµ=X (ii) nXσσ= (b) X is exactly Normally distributed. (c) (i) X is approximately Normally distributed. (ii) Central limit theorem. 2. (a) (i) pP=ˆµ (ii) nppP)1(ˆ−=σ (b) For large samples Pˆ is approximately Normally distributed. 3. A parameter is a numerical characteristic of a population. 4. An estimate is a known quantity calculated from data in order to estimate an unknown parameter. 5. (4) 6. (a) Xµ = 280 seconds Xσ = 1660 = 15 seconds X ~ approximately Normal (µ = 280s, σ = 15s) (b) pr(X > 240) = 1 – pr( X < 240) = 1 – 0.0038 = 0.9962 7. (a) (i) 15.7=Xµ litres 2.112.1===nXXσσ litres (ii) 15.7=Xµ litres 6.042.1===nXXσσ litres (iii) 15.7=Xµ litres 3.0162.1===nXXσσ litres (b) The standard deviation differs. This is because as the sample size increases there is a decrease in the variability of the sample mean.Dinov' Stat 10, UCLA 2 8. (a) The proportion of university students who belong to the student loan scheme. (b) 65.0ˆ=Pµ ()0675.05065.0165.0ˆ=−=Pσ Pˆ ~ approx Normal (µ = 0.65, σ = 0.0675) (c) pr( Pˆ > 0.7) = 1 – 0.7707 = 0.2293 (d) pr(0.45 < Pˆ < 0.55) = pr( Pˆ < 0.55) - pr( Pˆ < 0.45) = 0.0691 – 0.0015 = 0.0676 9. (a) 9477.1,125.10 == sx (b) nsx ×± 2 = 10.125 ± 2 x 89477.1 = (8.75, 11.50) (c) (i) wider (ii) nothing (iii) narrower 10. 3.012036ˆ==p nppp)ˆ1(ˆ2ˆ−×± = 0.3 ± 1207.03.02×× = (0.216, 0.384) 11. (2) Dinov's Stat 10, UCLA 1 Dinov's Stat 10, Final Exam Review- Solutions Confidence Intervals Section A: Confidence intervals for a mean, proportion and difference between means 1. (a) θ = µ, the population mean grade(mark) for the exam. (b) x=θ ˆ = 38.20, the mean mark of the sample of 30 marks. (c) se(θˆ) = se(x ) = 9809.13085.10==ns (d) df = 30 – 1 = 29 (e) t-multiplier = 2.045 (f) 95% c.i. is: )(xsetx ×± = 38.20 ± 2.045 x 1.9809 = 38.20 ± 4.0509 = (34.15, 42.25) (g) There are many ways of interpreting a confidence interval. Three different ways follow. (1) With 95% confidence, we estimate that the population mean mark is somewhere between 34.15 and 42.25 marks. (2) We estimate that the population mean mark is somewhere between 34.15 and 42.25 marks. A statement such as this is correct, on average, 19 times out of every 20 times we take such a sample. (3) We estimate the population mean mark to be 38.20 with a margin or error of 4.05. A statement such as this is correct, on average, 19 times out of every 20 times such a sample is taken. (h) We don’t know. The population mean mark is not known so we don’t know whether this particular 95% confidence interval contains the population mean. However, in the long run, the population mean will be contained in 95% of the 95% confidence intervals calculated from such samples. 2. (a) θ = p, the proportion of female Spanish prisoners in 1995 who had tuberculosis. (b) 4.09036ˆˆ=== pθ, the proportion in the sample of female Spanish prisoners who had tuberculosis. (c) se(θˆ) = se( pˆ) = 051640.0906.04.0)ˆ1(ˆ=×=−npp (d) z-multiplier = 1.96 (e) 95% c.i. is: )ˆ(ˆpsetp ×± = 0.4 ± 1.96 x 0.051640 = 0.4 ± 0.1012 = (0.299, 0.501) (f) We estimate that