STA1510 - Basic Statistics EXAM PACK.

STA1510 - Basic Statistics EXAM PACK. The qualitative variable is “The most frequent use of your microwave oven (reheating, defrosting, warming, other).” Answer: Option 1 Q2. The incorrect statement is “A bar chart is similar to a histogram in the sense that the bases of the rectangles are arbitrary intervals whose centers are the midpoints of the intervals”. Answer: Option 2 Q3. • The number of values larger than 40 is 12. Option 1 is correct. • The median is 2(N+1) 4 position = 2(30 + 1) 4 position = 15.5 position = 3.7 + 0.5(3.8 − 3.7) = 3.7 + 0.5(0.1) = 3.7 + 0.05 = 3.75 Option 2 is incorrect. • The values that lie between 2 and 3 are 6 30 × 100% = 20%. Option 3 is correct. • The mode of the data set is 3.6. Option 4 is correct. page 1 of 11 Page 1 of 174 email: Tel: 9 STA1510 • The sixth smallest value in the data set is 2.8. Option 5 is correct. Answer: Option 2 Q4. n = 11, P x = 82.5 and P x 2 = 638.19. • Sample mean is x = 1 n P x = 1 11 (82.5) = 7.5 Option 1 is correct. • Arranging the data in order gives 5.6 5.9 6.3 6.3 7.0 7.2 8.4 8.5 8.7 8.9 9.7 Median = 2(N+1) 4 position = 2(N+1) 4 = 6 th position = 7.2 Option 2 is correct. • The mode is 6.3. Option 3 is correct. • The range is 9.7 − 5.6 = 4.1 Option 4 is incorrect. • Sample standard deviation, s = s 1 n − 1 Xx 2 i − n Xx 2  = s 1 11 − 1 (638.19 − 11(7.5)2 ) = s 1 10 (638.19 − 618.75) = s 1 10 (19.44) = √ 1.944 ≈ 1.3943 Option 5 is correct. Answer: Option 4 Q5. The correct statement is “Marginal probability is the probability that a given event will occur, with no other event taken into consideration”. Answer: Option 1 page 2 of 11 Page 2 of 174 email: Tel: 9 STA1510 Q6. P(A) = 0.5, P(B) = 0.4 and P(A|B) = 0.6 Note: P(A′ and B′ ) = P(A or B) ′ (Demorgan’s law). P(A|B) = P(A and B) P(B) 0.6 = P(A and B) 0.4 0.6 × 0.4 = P(A and B) 0.24 = P(A and B) P(A or B) = P(A) + P(B) − P(A and B) = 0.5 + 0.4 − 0.24 = 0.66 P(A or B) ′ = 1 − P(A or B) = 1 − 0.66 = 0.34 Answer: Option 5 Q7. The probability is P(HHT H) + P(HT HH) + P(T HHH) =  1 2 × 1 2 × 1 2 × 1 2  +  1 2 × 1 2 × 1 2 × 1 2  +  1 2 × 1 2 × 1 2 × 1 2  = 1 16 + 1 16 + 1 16 = 3 16 No correct answer page 3 of 11 Page 3 of 174 email: Tel: 9 STA1510 Q8. Apartment Size Location One Two Three Total within bedroom bedrooms bedrooms building First floor Second floor Total N = 1 000 Let A be the event its a one bedroom apartment and B be the event its on the second floor. P(A) = 400 1 000 = 0.4, P(B) = 450 1 000 = 0.45, and P(A and B) = 250 1 000 = 0.25 P(A|B) = P(A and B) P(B) = 0.25 0.4 = 0.625 Answer: Option 4 Q9. A is discrete, B is continous, C is continuous and D is discrete. B and C are continous. Answer: Option 2 Q10. Mean, µ = E(X) = Xxip(xi) = 1(0.25) + 2(0.33) + 3(0.17) + 4(0.15) + 5(0.1) = 0.25 + 0.66 + 0.51 + 0.6 + 0.5 = 2.52 Variance, σ2 = X(xi − µ) 2 p(xi) = (1 − 2.52)2 × 0.25 + (2 − 2.52)2 × 0.33 + . . . + (5 − 2.52)2 × 0.1 = (−1.52)2 × 0.25 + (−0.52)2 × 0.33 + . . . + (2.48)2 × 0.1 = 0.5776 + 0. + 0. + 0.32856 + 0.61504 = 1.6496 page 4 of 11 Page 4 of 174 email: Tel: 9 STA1510 or Variance, σ2 = E(X 2 ) − µ 2 = Xx 2 i p(xi) − µ 2 = 12 (0.25) + 22 (0.33) + . . . + 52 (0.1) − (2.52)2 = 0.25 + 1.32 + 1.53 + 2.4 + 2.5 − 6.3504 = 8 − 6.3504 = 1.6496 =⇒, standard deviation, σ = √ 1.6496 ≈ 1.2844 Answer: Option 4 Q11. More than 60% of the next 5 patients is 0.6 × 5 = 3 =⇒ more than 3 X = Binomial, n = 5 and π = 0.8 P(X = x) = n x ! (π) n (1 − π) n−x The probability is P(X 3) = P(X = 4) + P(X = 5 = 5 4 ! (0.8)4 (1 − 0.8)5−4 + 5 5 ! (0.8)5 (1 − 0.8)5−5 = 5 4 ! (0.8)4 (0.2)1 + 5 5 ! (0.8)5 (0.2)0 = 0.4096 + 0.32768 ≈ 0.7373 Using table 4, n = 5 and π = 0.8 P(X 3) = P(X = 4) + P(X = 5) = 0.4096 + 0.3277 = 0.7373 Answer: Option 5 Q12. λ = 1.8. X ∼ Poisson(1.8) P(X = x) = e−λ(λ) x x! P(X = 0) = e −1.8 (1.8)0 0! = e −1.8 ≈ 0.1653 Using table 6, λ = 1.8 =⇒ P(X = 0) = 0.1653 Answer: Option 1 page 5 of 11 Page 5 of 174 email: Tel: 9 STA1510 Q13. The statement which is incorrect “In the normal distribution, the mean, median, mode and the variance are all at the same position on the horizontal axis since the distribution is symmetric”. The variance is not in the same position as the mean, median and mode. Answer: Option 3 Q14. X ∼ Exponential =⇒ P(X = x) = λe−λx σ = 200 = 1 λ =⇒ λ = 1 200 Thus, P(X = x) = 1 200 e − 1 200 x P(X ≤ k) = e −λk Now the probability that the bulb will last at least 10 hours is P(X ≥ 0) = e − 1 200 ×10 = e −0.05 ≈ 0.9512 Answer: Option 5 Q15. µ = 0.1, σ = 2.1 and n = 900. If mean is negative then X is less than 0. =⇒ P(X 0) = P  Z 0 − µ √σ n   = P  Z 0 − 0.1 √ 2.1 900   = P  Z −0.1 0.07  = P(Z −1.9) = P(Z −1.43) = 0.0764 Answer: Option 3 Q16. π = 0.6 and n = 200 P(p ≥ 0.58) = P  Z ≥ 0.58 − π qπ(1−π) n   = P  Z ≥ 0.58 − 0.6 q0.6(1−0.6) 200   = P  Z ≥ −0.02 q0.6×0.4 200   = P Z ≥ −0.02 √ 0.0012! page 6 of 11 Page 6 of 174 email: Tel: 9 STA1510 = P  Z ≥ −0.02 0. = P(Z ≥ −0.) = P(Z ≥ −0.58) = 1 − P(Z −0.58) = 1 − 0.281 = 0.719 Answer: Option 3 Q17. n = 15, x = 23.4 and s = 8.5. Using z α 2 = 1.96 Thus, the 95% confidence interval for µ is x ± z α 2 × s √ n 23.4 ± 1.96 × 8.5 √ 15 23.4 ± 4. 23.4 ± 4.3016 (23.4 − 4.3016 ; 23.4 + 4.3016) (19.0984 , 27.7016) Answer: Option 1 Q18. n = 500, π = 0.85 P(0.83 ≤ p ≤ 0.88) = P   0.83 − π qπ(1−π) n ≤ Z ≤ 0.88 − π qπ(1−π) n   = P   0.83 − 0.85 q0.85(1−0.85) 500 ≤ Z ≤ 0.88 − 0.85 q0.85(1−0.85) 500   = P   −0.02 q0.85×0.15 500 ≤ Z ≤ 0.03 q0.85×0.15 500   = P −0.02 √ 0. ≤ Z ≤ 0.03 √ 0.! = P  −0.02 0. ≤ Z ≤ 0.03 0. = P(−1. ≤ Z ≤ 1.) = P(−1.25 ≤ Z ≤ 1.88) = 0.9699 − 0.1056 = 0.8643 Answer: Option 1 page 7 of 11 Page 7 of 174 email: Tel: 9 STA1510 Q19. The incorrect statement “A type I error is the probability of not rejecting a false null hypothesis.” A type I error is when you reject H0 when in actual fact it is true. Answer: Option 2 Q20. This is a two tailed test. Thus, p − value = P(Z −1.59) + P(Z 1.59) = 0.0559 + 1 − 0.9441 = 0.0559 + 0.0559 = 0.1118 Answer: Option 1 Q21. The standardised test statistic is equal to Z = x − µ √σ n = 53.5 − 50 √ 10 64 = 3.5 1.25 = 2.8 Answer: Option 3 Q22. µ = 880C, x = 910C, s = 40C and n = 16 • H0 : µ = 88 H1 : µ 88 Option 1 is correct. • Test statistic is T = x − µ √s n = 91 − 88 √ 4 16 = 3 1 = 3 Option 2 is incorrect. • Test is one-tailed, α = 0.05 tα;n−1 = t0.05;16−1 = t0.05;15 = 1.753 Option 3 is correct. page 8 of 11 Page 8 of 174 email: Tel: 9 STA1510 • The confidence interval is x ± t α 2 ;n−1 × √s n α = 0.05, α 2 = 0.025 and t0.025;15 = 2.131 The 95% confidence interval is x ± t α 2 ;n−1 × s √ n 91 ± 2.131 × 4 √ 16 91 ± 2.131 (91 − 2.131 , 91 + 2.131) (88.869 , 93.131) Option 4 is correct. • We reject H0 if T 1.753. Since 3 1.753, the null hypothesis is rejected. Option 5 is correct. Answer: Option 2 Q23. Absence Weather Yes No Total Rainy Non-rainy Total The expected values are Row total × Column total Grand total Absence Weather Yes No Total Rainy 6.3 113.7 120 Non-rainy 14.7 265.3 280 Total The test statistic is χ 2 calc = X (Oij − Eij ) 2 Eij = (10 − 6.3)2 6.3 + (110 − 113.7)2 113.7 + (11 − 14.7)2 14.7 + (269 − 265.3)2 265.3 = (3.7)2 6.3 + (−3.7)2 113.7 + (−3.7)2 14.7 + (3.7)2 265.3 = 2.1730 + 0.1204 + 0.9313 + 0.0516 = 3.2763 Answer: Option 3 page 9 of 11 Page 9 of 174 email: Tel: 9 STA1510 Q24. n = 7, P xi = 210, P x 2 i = 7 308, P xiyi = 1 590.48, P yi = 49.56 and P y 2 i = 362.1628 • x = 1 n P x = 1 7 (210) = 30 y = 1 n P y = 1 7 (49.56) = 7.08 Option 1 is correct. • b = P xiyi − nxy P x 2 i − n(x) 2 = 1 590.48 − 7(30)(7.08) 7 308 − 7(30)2 = 1 590.48 − 1 486.8 7 308 − 6 300 = 103.68 1 008 = 0. ≈ 0.1029 Option 2 is correct. • b0 = y − b1x = 7.08 − 0.(30) = 7.08 − 3.0857 = 3.9943 Option 3 is correct. • yˆ = b0 + b1x = 3.9943 + 0.1029X Option 4 is incorrect. • Estimated yield when x = 50 is yˆ = 3.9943 + 0.1029(50) = 3.9943 + 5.145 = 9.1393 ≈ 9.14 Option 5 is correct. Answer: Option 4 page 10 of 11 Page 10 of 174 email: Tel: 9 STA1510 Q25. r = n P xiyi − ( P xi)(P yi) q [n P x 2 i − ( P xi) 2 ][n P y 2 i − ( P yi) 2 ] = 7(1 590.48) − (210)(49.56) q [7(7 308) − (210)2 ][7(362.1628) − (49.56)2 ] = 11 133.36 − 10 407.6 q [51 156 − 44 100][2 535.1396 − 2 456.1936] = 725.76 q [7 056][78.946] = 725.76 q [557 042.976] = 725.76 746. ≈ 0.9724 Answer: Option 3 Wish You Good Luck!!!!! Muchengetwa 2011 for Polokwane Group