stat12x_f23_practice_test_2b_ubc_sol

STAT 12x Midterm Fall 2018 INSTRUCTOR: Claude Hurtubise COURSE: STAT124 Business Statis- tics Sauder School of Business University of British Columbia Okanagan Date: November 14, 2018; Time: 11:00PM; Duration 80 minutes. This exam has 6 questions for a total of 48 points. Partial marks may be awarded. SPECIAL INSTRUCTIONS Show and explain all your work unless the question directs otherwise. Simplify all answers. Answer the questions in the room provided on the question sheets. If you run out of room for an answer, you may continue on the back page. The use of a business or scientific calculator is permitted. All other electronic devices must be powered off. Question: T/F MC 3 4 5 6 Total Points: 8 6 6 10 10 8 48 Score: STUDENT NAME (print): STUDENT NUMBER: SIGNATURE: This exam consists of 12 pages including this cover page. Check to ensure that it is complete.

STAT 12x (Fall 2018) Midterm (continued) True or False: Total of 10 marks Indicate with a T or F whether each statement is true or false. 1. 8 (a) A statistic is a numeric quantity, usually known, that describes a certain sample characteristic. (a) T (b) The Poisson probability distribution is a continuous probability distribution. (b) F (c) The larger the sample size the smaller the sampling variation. (c) T (d) A 99% confidence interval is wider than a 95% confidence interval. (d) T (e) For a binomial distribution, each trial has a known number of successes. For ex- ample, a 4 question multiple choice test can only have zero, one, two, three or four successes. (e) T (f) The geometric distribution is a special case of the negative binomial distribution. (f) T (g) If you reject the null hypothesis for one-tail test, then you must reject the null hypothesis for a two-tail test as well. (g) F (h) The p-value is based under the assumption that the null hypothesis is true. (h) T Page 2 of 12

STAT 12x (Fall 2018) Midterm (continued) (d) A type I error is committed when A. we don’t reject a null hypothesis that is true. B. we reject a null hypothesis that is false. C. we don’t reject a null hypothesis that is false. D. we reject a null hypothesis that is true. E. none of these (e) If the p-value is 0.034, which of the following is false ? A. you will reject H 0 at α = 0 . 05. B. you will retain H 0 at α = 0 . 01. C. you will retain H 0 at α = 0 . 02. D. you will reject H 0 at α = 0 . 01 . E. you will reject H 0 at α = 0 . 10. (f) Which of the following describes what the property of unbiasedness means? A. The shape of the sampling distribution is approximately normally dis- tributed. B. The center of the sampling distribution is found at the population stan- dard deviation. C. The center of the sampling distribution is found at the popula- tion parameter that is being estimated. D. The sampling distribution in question has the smallest variation of all possible sampling distributions. E. None of these describe the property of unbiasedness. Page 4 of 12

STAT 12x (Fall 2018) Midterm (continued) Long Answer: Marks indicated as shown All calculations must be shown to receive full marks. 3. The length of human pregnancies from conception to birth approximates a normal dis- tribution with a mean of 266 days and a standard deviation of 16 days. (a) 4 What proportion of all pregnancies will last between 240 and 270 days (roughly between 8 and 9 months)? Solution: X ∼ N (266 , 16). z = 240 − 16 112 ≈ − 1 . 63 z = 270 − 16 112 ≈ 0 . 25 P (240 < X < 270) = P ( − 1 . 63 < Z < 0 . 25) = P ( Z < 0 . 25) − P ( Z < − 1 . 63) = 0 . 5987 − 0 . 0516 = 0 . 5471 (b) 2 What length of time marks the shortest 70% of all pregnancies? Solution: X ∼ N (527 , 112). P ( X < ?) = 0 . 70 ⇒ P ( z < ?) = 0 . 70 ⇒ z = 0 . 52 X = 266 + 0 . 52(16) = 274 . 32 Page 5 of 12

STAT 12x (Fall 2018) Midterm (continued) (v) Find the p-value: Solution: The p-value is smaller than the smallest value in the table, so p < 0 . 0001. (vi) State your decision: Solution: Since p < α , we reject H 0 . (vii) State your conclusion: Solution: There is sufficient evidence to conclude boys are more common than girls in the entire population at the 5% significance level. (viii) Suppose you made an incorrect decision in part ( a ). What type of error did you make? Solution: Since you rejected H 0 , you have made a type I error. (b) 2 Construct 95% confidence interval for the true proportion of newborn babies likely to be boys. (Keep 4 decimal places.) Solution: 0 . 5172 ± 1 . 96 · s 0 . 5172 (1 − 0 . 5172) ⇒ 0 . 5172 ± 1 . 96 · 0 . 003131 ⇒ 0 . 5172 ± 0 . 006137 ⇒ (0 . 5111 , 0 . 5233) Page 7 of 12

STAT 12x (Fall 2018) Midterm (continued) 5. 10 About 8% of males are colour blind. A researcher needs subjects for an experiment and begins sampling from a very large university with over 10,000 male students. (a) If he samples 10 male students, what is the probability that exactly 3 of them will be colour blind? Solution: P ( X = 3) = 10 3 (0 . 08) 3 (0 . 92) 7 = 120(0 . 000512)(0 . 557847) = 0 . 034274 (b) If he samples 10 male students, what is the probability that his 3rd colour blind subject will occur on the 10th male student? Solution: P ( X = 3 , on 10th male student) = 10 − 1 3 − 1 (0 . 08) 3 (0 . 92) 7 = 36(0 . 000512)(0 . 557847) = 0 . 010282 (c) How many male students must he sample before he gets 10 colour blind subjects? Solution: E( X ) = k p = 10 0 . 08 = 125 Page 8 of 12

STAT 12x (Fall 2018) Midterm (continued) 6. A local newspaper reported the result of a telephone poll of 800 adult Canadians. The question posed of the Canadians who were surveyed was: “Should the federal tax on e-cigarettes be raised to pay for health care reform?” The results of the survey were: Non-smokers Smokers n 605 195 Counts 351 said yes 41 said yes ˆ p (a) 8 Is there sufficient evidence at the α = 0 . 05, to conclude that the two populations, smokers and non-smokers, differ significantly with respect to their opinions? (i) State the hypotheses: Solution: H 0 : p 1 = p 2 H A : p 1 ̸ = p 2 (ii) Check the CLT conditions: Solution: np = 25 , 468(0 . 5) = 12 , 734 n (1 − p ) = 25 , 468(1 − 0 . 5) = 12 , 734 np = 25 , 468(0 . 5) = 12 , 734 n (1 − p ) = 25 , 468(1 − 0 . 5) = 12 , 734 (iii) Compute the pooled proportion and standard error: Solution: ˆ p = Y 1 + Y 2 n 1 + n 2 = 41 + 351 195 + 605 = 392 800 = 0 . 49 SE ˆ p 1 − ˆ p 2 = s 0 . 49(1 − 0 . 49) 1 195 + 1 605 = 0 . 0412 Page 10 of 12

STAT 12x (Fall 2018) Midterm (continued) (iv) Calculate z -statistic: Solution: z = 0 . 58 − 0 . 21 0 . 0412 ≈ 8 . 79 (v) Find the p-value: Solution: The p-value is smaller than the smallest value in the table, so at 0 . 0001. Since the alternative hypothesis is 2-sided, we double it. Hence, p < 0 . 0002 (vi) State your decision: Solution: Since p < α , we reject H 0 . (vii) State your conclusion: Solution: There is sufficient evidence to conclude that the proportion of adults who think the federal tax on e-cigarettes should be raised to pay for health care reform is significantly different at the 5% significance level. (viii) Construct a 95% confidence interval for the true difference, but do not interpret it. Solution: ⇒ 0 . 58 − 0 . 21 ± 1 . 96 · r (0 . 58)(0 . 42) 605 + (0 . 21)(0 . 79) 195 ⇒ 0 . 37 ± 1 . 96 · √ 0 . 00040264 + 0 . 00085077 ⇒ 0 . 37 ± 0 . 0354 ⇒ (0 . 3346 , 0 . 4054) Page 11 of 12