Math 2200 Fall 2014, Exam 3 You may use any calculator. You may use a 4 6 inch notecard as a cheat sheet.

1 Math 2200 Fall 2014, Exam 3 You may use any calculator. You may use a 4 6 inch notecard as a cheat sheet. Warning to the Reader! If you are a student for whom this document is a historical artifact, be aware that the definitions and conventions on which some of the questions on this exam are based may differ from those adopted in your course. 1. If X, Y, and W are independent random variables with variances 4, 5, and 6 respectively, then what is the standard deviation of 3 X + 2 Y - W? A) 6.824 B) 7.034 C) 7.244 D) 7.454 E) 7.664 F) 7.874 G) 8.084 H) 8.294 I) 8.504 J) 8.714 Solution. Because Var(3 X + 2 Y - W) = 9 Var(X)+4 Var(y)+Var(W) = 62, we have Sd( 3 X + 2 Y - W ) = 62 = 7.874. Answer: F 2. The weight of fluid in each bottle of Pilsner Aquarelle is normally distributed with mean 12 oz and standard deviation 0.34 oz. Assume that the weights of different bottles are independent. What is the probability that the total weight of the contents of 6 bottles exceeds 73 oz? A) 0.019 B) 0.043 C) 0.067 D) 0.091 E) 0.115 F) 0.139 G) 0.163 H) 0.187 I) 0.211 J) 0.235 Solution. Let W = X 1 + X 2 + + X 6 where X j is the weight of the contents of the j th bottle. Then E(X) = 6 12 = 72, Var(W ) = 6 (0.34) 2 = 0.6936, and Sd(W ) = 0.6936 = 0.83283. We calculate Answer: E P(W > 73) (W 72 > 73 72) W 72 73 72 > 0.83283 0.83283 W E(W) > 1.2007 Sd(W) ( Z > 1.2007) = 1 P ( Z 1.2007) = 1 Φ(1.2007) = 1 0.8851 = 0.1149. 3. Rhys s time for running a 1 mile race is normally distributed with mean 6 minutes and standard deviation 2 minutes. Reese s time for running a 1 mile race is normally distributed with mean 10 minutes and standard deviation 1 minute. If they race each other, what is the probability that Reese will win by at least 30 seconds? (That is, what is the probability that Rhys s time exceeds Reese s time by 30 seconds.) A) 0.0001 B) 0.0040 C) 0.0089 D) 0.0133 E) 0.0177 F) 0.0221 G) 0.0265 H) 0.0309 I) 0.0353 J) 0.0397

2 Solution. Let X be Rhys s time and let Y be Reese s time. Because X and Y are independent normal, the variable X - Y is normal. Moreover, E(X - Y) = 6 10 = 4 and Var(X-Y) = Var(X) + Var(Y) = 2 2 +1 2 = 5. Therefore, Sd(X-Y) = 5 = 2.236. It follows that P(X Y > 1/2) (X Y ( 4) > 1/2 ( 4)) ( X Y ( 4) > 4.5 ) 2.236 2.236 (Z > 2.0125) = 1 Φ(2.0125) = 1 0.9779 = 0.0221. Note, we can obtain Φ(2.0125) from the table as Φ(2.01)+(25/100)(Φ(2.02) Φ(2.01)) = 0.9778+(1/4)(0.0005) = 0.977925. Answer: F 4. Nineteenth century novelist Charles Wutthe was paid for his stories by the word: 1 pence per word by the publishing house Dombey and Son, and two pence per word by The Pickwick Press. Charles found that when Dombey s editor was through with his stories, their lengths were normally distributed with mean 5000 words and standard deviation 200 words. When Pickwick was done editing his stories, their lengths were normally distributed with mean 3000 words and standard deviation 300 words. What was the probability that a story submitted to Dombey would bring more pence than a story submitted to Pickwick? A) 0.057 B) 0.069 C) 0.081 D) 0.093 E) 0.105 F) 0.117 G) 0.129 H) 0.141 I) 0.153 J) 0.165 Solution. Let X and Y be the length random variables of Charles s stories after submission to Dombey and Pickwick respectively. Then E(X 2 Y) = 5000 2 3000 = 1000, Var(X 2 Y) = Var(X) + 2 2 Var(Y) = 200 2 + 4 300 2 = 400000, and Sd(X 2 Y) = 400000 = 632.46. We calculate P (X > 2 Y) (X 2 Y > 0) (X 2 Y ( 1000) > 0 ( 1000)) ( (X 2 Y) ( 1000) > 1000 ) 632.46 632.46 (X 2 Y) E(X 2 Y) > 1.5811 Sd(X 2 Y) (Z > 1.5811) = 1 P (Z 1.5811) = 1 Φ(1.5811) = 1 0.943 = 0.057. Answer: A 5. In 1971 there were 65 highjackings of aircraft. For the purposes of preparedness, it was of interest to know the probability p of more than one highjacking occurring in one day. Assuming that highjackings

3 were a Poisson process, what was p? A) 0.0031 B) 0.0141 C) 0.0250 D) 0.0361 E) 0.0471 F) 0.0581 G) 0.0691 H) 0.0801 I) 0.0911 J) 0.1021 Solution. Set λ = 65/365 = 0.1780822, the average number of highjackings per day. Let X be the number of highjackings on any given day. Then P (X = k) = exp( λ) λ k /k! for k = 0, 1, 2,.... We calculate p = 1 ( P (X = 0) + P (X = 1) ) Answer: B λ λ0 λ1 = 1 e e λ 0! 1! = 1 e λ e λ λ = 1 0.83687 0.14903 = 0.0141. 6. If X and Y are independent standard normal random variables, what is P ( 2.4 < X 2 + Y 2 < 4.6 )? A) 0.05 B) 0.10 C) 0.15 D) 0.20 E) 0.25 F) 0.30 G) 0.35 H) 0.40 I) 0.45 J) 0.50 Solution. ( ( Let F be the cumulative distribution function of χ 2 2. We are asked for F(4.6) - F(2.4), or 1 P χ 2 2 > 4.6 )) ( 1 P ( χ 2 2 > 2.4 )) = (1 0.1) (1 0.3) = 0.2 Answer: D 7. A random sample X 1, X 2,..., X 25 is drawn from a distribution. The variance of the sampling distribution of (X 1 + X 2 + + X 10 ) /10 is 60. What is the variance of the sampling distribution of (X 11 + X 12 + + X 25 ) /15? A) 5 B) 10 C) 15 D) 20 E) 25 F) 30 G) 35 H) 40 I) 45 J) 50 Solution. Let σ be the population standard deviation. Then σ 2 10 = Var X1 + X 2 + + X 10 = 60, 10 so σ 2 = 600. It follows that X11 + X 12 + + X 25 Var = σ2 15 15 = 600 15 = 40. Answer: H 8. A random sample X 1, X 2,... X 34 is drawn from a distribution. The variance of the sampling distribution of the average of the first 18 of these variables is exactly three times the standard deviation of the sampling distribution of the average of the remaining 16 of these variables. What is the population standard deviation? A) 9.5 B) 10 C) 10.5 D) 11 E) 11.5 F) 12 G) 12.5 H) 13 I) 13.5 J) 14

4 Solution. Let σ be the population standard deviation. The given information can be expressed as the following equation: σ 2 /18 = 3 σ/ 16. The solution is 13.5. Answer: I 9. It has been estimated that employers verify the asserted educational qualifications of job applicants 75% of the time. Use the normal approximation without the continuity correction to approximate the probability that the educational credentials of at least 70 of 100 applicants will be checked. (It might be useful to read the next question before answering this one.) A) 0.76 B) 0.78 C) 0.80 D) 0.82 E) 0.84 F) 0.86 G) 0.88 H) 0.90 I) 0.92 J) 0.94 Solution. Let X j = 1 if the educational qualifications of the j th applicant are verified, and X j = 0 if not. Let Y = X 1 + X 2 + + X 100. Then E(Y) = 100 (0.75) = 75, Var(Y) = 100 (0.75) (0.25) = 18.75, and Sd(Y) = 18.75 = 4.33. Thus, P(Y 70) (Y 75 70 75) Y 75 70 75 4.33 4.33 (Z 1.1547) (Z 1.1547) = Φ(1.1547) = 0.8759. Answer: G 10. Answer the preceding problem once again, but this time use the correction for continuity. A) 0.76 B) 0.78 C) 0.80 D) 0.82 E) 0.84 F) 0.86 G) 0.88 H) 0.90 I) 0.92 J) 0.94 Solution. The solution follows that of the preceding problem, but we use 60.5 instead of 70: P(Y 69.5) (Y 75 70 69.5) Y 75 69.5 75 4.33 4.33 (Z 1.2702) (Z 1.2702) = Φ(1.2702) = 0.898. Using the binomial distribution, the exact probability is 100 k=70 ( 100 k Answer: H ) (0.75) k (0.25) 100 k = 180019797683981972115291244561520585282085418955028516121773 200867255532373784442745261542645325315275374222849104412672 0.89621276

5 11. In a large class, scores on an exam have the distribution N(18.8, 4.2). (Recall that, in our notation, the second parameter is the population standard deviation and not anything else.) If a random sample of size 10 is drawn, what is the probability that the sample variance is greater than 5.292? A) 0.400 B) 0.500 C) 0.600 D) 0.700 E) 0.750 F) 0.800 G) 0.900 H) 0.950 I) 0.975 J) 0.990 Solution. Let S 2 (10 1) be the sample variance. Then (4.2) 2 S 2 χ 2 10 1. We have P ( S 2 > 5.292 ) (10 1) (4.2) 2 S 2 (10 1) > (4.2) 2 5.292 ( χ 2 9 > 2.7 ) = 0.975. Answer: I 12. The heights of women in a large statistics class have the distribution N(µ, σ) with unknown population mean µ and unknown standard deviation σ. The heights of three randomly chosen women in the class are 64, 67, and 67 inches. What is the probability that the population mean µ is less than 65 inches A) 0.01 B) 0.11 C) 0.21 D) 0.31 E) 0.41 F) 0.51 G) 0.61 H) 0.71 I) 0.81 J) 0.91 Solution. Let X 1, X 2, X 4 be the three heights. The sample mean is 66, the sample variance is 3, and the sample standard deviation is 3. Furthermore ( X µ ) /(S/ 3) t 3 1. Thus, P(µ < 65) ( µ > 65) (X µ > X 65) ( X µ S/ 3 > X 65 ) S/ 3 ( t 2 > X 65 ) S/ 3 ( ) 66 65 t 2 > 3/ 3 (t 2 > 1). From the table of t-values, we find P (t 2 > 0.8165) = 0.25 and P (t 2 > 1.0607) = 0.2. We will do a linear interpolation. The line through the points (0.8165,0.25) and (1.0607,0.2) has slope (0.2-0.25)/(1.0607-0.8165), or -0.20475. Therefore it has equation y = 0.20475 (x 1.0607) + 0.2. Substituting x = 1, we obtain y = 0.212428325. That is our approximation of P (t 2 > 1). Answer: C 13. In a poll of 400 registered voters, [0.30792, 0.41208] was the confidence interval for the proportion of voters who supported candidate Todd Crude. What was the confidence level? A) 90% B) 91% C) 92% D) 93% E) 94% F) 95% G) 96% H) 97% I) 98% J) 99%

6 Solution. First we obtain p = (0.30792 + 0.41208)/2 = 0.36. Next we obtain ME ( p) = 0.41208 0.36 = 0.05208. We also can find SE ( p) = 0.36(1 0.36)/400 = 0.024. If z is the critical value of the confidence interval, we have 0.05208 = ME ( p) = z SE ( p) = z 0.024, or z = 0.05208/0.024 = 2.17. From the Phi table we have Φ(2.17) = 0.985. Thus, α/2 = 1 0.985 = 0.015 and α = 0.03. The interval is a 100(1 0.03)%, or 97% confidence interval. Answer: H 14. A charity wishes to estimate the proportion p of individuals on its mailing list who are willing to contribute $50 or more. To do so, a random survey of size n is conducted. In order to have a margin of error no greater than 0.05 and a 95% confidence level, what is the smallest sample size n that meets the requirements? A) 355 B) 360 C) 365 D) 370 E) 375 F) 380 G) 385 H) 390 I) 395 J) 400 Solution. Here z 0.025 = 1.96 and 2 ME 0 = 0.1. The required sample size is n = (1.96/0.10) 2 = 384.16 rounded up to the nest whole number. Answer: G 15. In a study of 6,000 diabetic patients treated by primary care physicians in Alabama, Iowa, and Maryland, 84% of the subjects were not receiving specialized blood work recommended by the American Diabetes Association. What is the upper bound for a 90% confidence interval for the proportion of all diabetic patients in those states who receive the recommended blood work? A) 0.1632 B) 0.1655 C) 0.1678 D) 0.1701 E) 0.1724 F) 0.1747 G) 0.1770 H) 0.1793 I) 0.1816 J) 0.1839 Solution. The requested upper bound is 0.16 + z 0.10/2 (0.84)(0.16) 6000 = 0.84 + 1.6449 0.004733 = 0.1678. Answer: C 16. Lars Birdski, a star of European basketball, made 46 baskets (successes) in 50 free throws (trials). Find the lower bound for a 95% confidence interval for his success rate p. A) 0.805 B) 0.818 C) 0.831 D) 0.843 E) 0.856 F) 0.869 G) 0.882 H) 0.895 I) 0.907 J) 0.920 Solution. The 10-10 Success-Failure Condition is not satisfied. Therefore, the Agresti-Coull adjustment is needed. With the four phony trials included, we have n = 54, p = 48/54 = 0.8889, 1 p = 6/54 = 0.1111, SE ( p) = (0.8889)(0.1111)/54 = 0.0427648, ME ( p) = 1.96 SE ( p) = (1.96)(0.0427648) = 0.0838, and p ME ( p) = 0.8889 0.0838 = 0.8051 is the lower bound for a 95% confidence interval for Lars Birdski s success rate p. Answer: A 17. The Acme Oat Foods Company packages its delicious oatmeal crisps in 18 oz cereal boxes. Of course, the reality is that there is some variability. Thanks to a long history of production, Acme knows that

7 the standard deviation of the weight of the contents is 0.66 oz. As part of its customary quality control, Acme randomly sampled 36 boxes and found the sample mean to be 17.88 oz. What is the upper bound of a 99% confidence interval for the weight of the cereal? A) 17.89 B) 17.98 C) 18.07 D) 18.16 E) 18.25 F) 18.34 G) 18.43 H) 18.52 I) 18.61 J) 18.70 Solution. Because z 0.005 = 2.5758, the required upper bound is 17.88 + 2.5758 0.66/ 36, or 18.16. Answer: D 18. A random sample of size 64 from a distribution has sample standard deviation equal to 10. What is the length of a 92.16% confidence interval for the mean of the distribution? A) 4.40 B) 4.54 C) 4.68 D) 4.82 E) 4.96 F) 5.10 G) 5.24 H) 5.38 I) 5.52 J) 5.66 Solution. Solving the equation 100(1 α) = 92.16, we obtain α = 0.0784. Therefore, α/2 = 0.0392. From the table, we find Φ(z) = 1 0.0392, or Φ(z) = 0.9608, has solution z = 1.76. Therefore, the length of the described confidence interval is 2 ME = 2 1.76 SE = 2 1.76 10/ 64, or 4.4. Answer: A 19. The sample 5, 7, 10, 14 was drawn from a normal distribution. What is the margin of error for an 80% confidence interval for the mean? A) 2.750 B) 2.864 C) 2.978 D) 3.092 E) 3.206 F) 3.320 G) 3.434 H) 3.548 I) 3.662 J) 3.776 Solution. For this small sample (n = 4) from a normal distribution, we must use a Student-t distribution with n 1, or 3, degrees of freedom. From the Student-t table, we have t 0.1,3 = 1.6377. The sample mean is 9 and the sample standard deviation is S = 3.91578. The margin of error is 1.6377.91578/ 4, or 3.206. Answer: E 20. A random sample of size 11 was drawn from a normal distribution with unknown population mean and unknown population variance σ 2. The sample variance was 20. Which of the following is a 95% confidence interval for σ 2? A) [8.3, 59.6] B) [8.6, 60.0] C) [8.9, 60.4] D) [9.2, 60.8] E) [9.5, 61.2] F) [9.8,61.6] G) [10.1, 62.0] H) [10.4, 62.4] I) [10.7, 62.8] J) [11.0, 63.2] Solution. Unlike confidence intervals for means, which are based on symmetric normal and Student-t random variables, there is no natural center for a confidence interval of σ 2. That is because such confidence intervals are based on asymmetric χ 2 df random variables. Still, there is a standard procedure: if S 2 is the sample variance, then S 2 11 1 [ 11 1, S 2] is a 95% confidence interval for σ 2. Substituting the observed value χ 2 0.025,11 1 χ 2 0.975,11 1 [ ] of S 2 and the tabulated values χ 2 0.025,10 = 20.4832 and χ 2 10 10 0.975,10 = 3.2470, we obtain 20, 20, χ 2 0.025,10 χ 2 0.975,10 or [ 10 20.4832 20, 10 3.2470 20], or [9.8, 61.6] Answer: F

8 21. A dietician created two low calorie diets, A and B. Of the two, diet B was also low in fat. The dietician performed an experiment to determine if the diets reduced obesity during a fixed length of time. Let µ A and µ B be the true average loss in pounds for the entire population. The dietician had 100 subjects follow diet A and 86 subjects follow diet B. The sample variances were S 2 A = 22.4 and S 2 B = 15.7. A 95% confidence interval for µ B µ A was found. What was the margin of error? A) 1.25 B) 1.30 C) 1.35 D) 1.40 E) 1.45 F) 1.50 G) 1.55 H) 1.60 I) 1.65 J) 1.70 Solution. The standard deviation is S = 22.4 100 + 15.7 86 = 0.6376. The margin of error is z 0.025 S, or 1.96 0.6376, or 1.249696. Remark: The authorized alternative is to use t 0.025,85, or 1.988267907, instead of z 0.025. Doing so results in t 0.025,85 S, or 1.988267907 0.6376, or 1.2677, which leads to the same answer choice. Answer: A 22. The mean time µ for mice to find their way through a particular maze is 18 seconds. Psychologists conjectured that if the mice were to hear a loud, startling noise at the outset, then they might be motivated to navigate the maze more quickly. They trialed 49 mice, calculated the sample mean, which was 17 seconds, and sample standard deviation, which was 4 seconds, and tested the null hypothesis µ = 18 versus the alternative µ < 18 at the 5% significance level. Which of the following numbers is an endpoint of the critical region? A) 17.060 B) 17.164 C) 17.268 D) 17.372 E) 17.476 F) 17.580 G) 17.684 H) 17.788 I) 17.892 J) 17.996 Solution. The critical region is µ < µ 0 t 0.05,48 4 49 If z 0.05 = 1.645 is used instead of the t-value, then the answer is 18 1.645 4/7, or 17.06. If instead, the t-table is used to approximate t 0.05,48, we interpolate. Values are t 0.05,40 = 1.6839 and t 0.05,50 = 1.6759. The line segment between these two points has equation t = (1.6759 1.6839) 50 40 (df 40) + 1.6839. If we substitute df = 48 in this equation, we obtain t = 1.6775. With this t-value, the endpoint of the critical region is 18 1.6775 4/7, or 17.041. Answer: A 23. A certain management task was believed to fall to senior management 50% of the time. But, because of a perceived trend, the alternative hypothesis that the task fell to senior management more than 50% of the time was considered possible. A survey of 153 companies was conducted. The particular task in question was handled by senior management at 57% of the companies. What was the p-value? A) 0.01 B) 0.02 C) 0.03 D) 0.04 E) 0.05 F) 0.06 G) 0.07 H) 0.08 I) 0.09 J) 0.10

9 Solution. If the null hypothesis is true, then the standard deviation of the sample mean of a random sample of size 153 is (0.5)(0.5)/153, or 0.0404226. The requested p-value is p 0.5 0.57 0.5 P ( p 0.57 p = 0.5) 0.0404226 0.0404226 p = 0.5 p 0.5 1.731704542 p = 0.5 0.0404226 (Z 1.731704542) = 1 P (Z 1.731704542) = 1 Φ(1.7317) = 0.04166, Answer: D 24. An article in the British Medical Journal in 2004 described how dogs can be trained to detect human bladder cancer by sniffing urine. It s a dog s life, isn t it. An experiment was set up so that the probability of correctly guessing the presence of bladder cancer was 1/7. In 81 trials, dogs were successful 33 times. Let p be the true canine success rate. Test the hypothesis H 0 : p = 1/7 versus H a : p 1/7 at 5% significance level. If the critical region is symmetric about 1/7, which of these numbers is an endpoint of the critical region? A) 0.206 B) 0.219 C) 0.232 D) 0.245 E) 0.258 F) 0.271 G) 0.284 H) 0.297 I) 0.310 J) 0.323 Solution. The answer choices are all greater than 1/7, which is about 0.14, so we are looking for the left endpoint p of an interval [p, 1]. The endpoint we seek is 1 (1/7)(6/7) 7 + 1.96, or 0.219. 81 Answer: B 25. Muzzle velocities of eight shells have a sample mean X = 2959 feet per second and sample standard deviation S = 38.68. The gunpowder manufacturer s claim is that the mean muzzle velocity is 3000 feet per second. Test that null hypothesis against the alternative that the mean muzzle velocity is less than A) 0.005 B) 0.010 C) 0.015 D) 0.020 E) 0.025 3000 feet per second. What is the p-value? F) 0.050 G) 0.100 H) 0.200 I) 0.300 J) 0.400

10 Solution. We calculate the p-value as follows: Answer: B P ( X 2959 µ = 3000 ) ( X 3000 2959 3000 µ = 3000 ) ( X 3000 S/ 8 ( X 3000 38.68/ 8 ( X µ 2959 3000 S/ 8 ) µ = 3000 2959 3000 38.68/ µ = 3000 8 ) S/ 2959 3000 8 38.68/ µ = 3000 8 ( ) 2959 3000 t 7 38.68/ 8 (t 7 2.9980) (t 7 2.9980) = 0.010. )

11

12 Chi-Squared Values Left Tails.

Chi-Squared Values Central Hump + Left Tails. 13

14 Student-t Values Left Tails α = 0.45, 0.40, 0.35, 0.30, 0.25, 0.20.

Student-t Values Left Tails α = 0.15, 0.10, 0.05, 0.025, 0.010, 0.005. 15