stat202-MT2

stat202-MT2 - Stat 202 Fall 2009 Question Booklet Part 1 -...

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Unformatted text preview: Stat 202 Fall 2009 Question Booklet Part 1 - [8 Marks / 1 Mark each] Multiple Choice: Each question is worth 1 mark. There is no penalty for guessing. Please fill the best correct answer onto the answer sheet provided. 1. The width of a confidence interval estimate of the population mean, with sigma known, tends to widen when the: a level of confidence increases sample size decreases ( (b ( ) ) e) value of the population standard deviation increases (d) All of the above [1} ) (e None of the above 2. Using a 99% confidence interval we wish to estimate the mean of a normal population, whose standard deviation is assumed to be 6, to within $12. This requires a random sample of size (a 166 [1] (b 1 ( CD 5 ,_. CD as. (c d ) ) ) ) 163 ) (e None of the above 3. The 95% confidence interval for a mean is (—4, 12) then (a) The sample mean is between —4 and 12. [1] (b) The population mean is in the interval (—4, 12) (c) The probability that the population mean is in (—4, 12) is 95% (d) This interval is invalid since it goes below zero. ) (e None of the above 4. A 90% confidence interval for the average class midterm grade, based on a sample of 12 students, is (55, 85). This means (a) The width of the interval is 30 (b) If I were to take 100 samples of the class I would find the class average for the midterm in about 90 of the intervals I create (c) I don’t know if the Class average is in the interval (d) All of the above [1] (e) None of the above 5. In a study of plant growth a pilot study indicates that 16 plants have an average height of 32 inches. In a prior study it was determined that the standard deviation was 10 inches. With a 95% level of confidence the authors would like to build an inteval that is no more than 1 inch wide. How many plants should they grow? None of the above. I: I Page 1 6. If the distribution of 7 is ’ 3‘ then the distribution of X, is Nliiogrlm m y (d) We are unable to determine based on the given information. {1] 7. Which of the following could be an appropriate alternative hypothesis? (a) The mean Of a sample is equal to 70 (b) The mean of a population is greater than 70 [1] (c) The mean of a sample is greater than 55 (d) all of the above (e) none of the above 8. If the lower and upper confidence limits of the population proportion p of successes, using a sample of size 1500, are 0.184 and 0.238, respectively, then the lower and upper confidence limits of the total number of successes in the population, given that the population size is 750, 000, are respectively (a) 276 and 357 (b) 137,724 and 178,143 (c) 138,000 and 178,500 [1] (d) 138,276 and 179,857 Page 2 Stat 202 Quantiles for a tn distribution with 72 degrees of {freedom 100 . > 100 0.253 0.8 0.9 6.31 2.92 2.35 2.13 2.02 1.94 1.89 1.86 1.83 1.81 1.80 1.78 1.77 1.76 1.75 1.75 1.74 1.73 1.73 1.72 1.72 1.72 1.71 1.71 1.71 1.71 1.70 1.70 1.70 1.70 ' 1.68 1.68 1.66 1.65 Table 8.2: The quantiles of the t—distribution Page 3 12.7 4.30 3.18 2.78 2.57 2.45 2.36 2.31 2.26 2.23 2.20 2.18 2.16 2.14 2.13 2.12 2.11 2.10 2.09 2.09 2.08 2.07 ' 2.07 2.06 2.06 2.06 2.05 2.05 2.05 2.04 2.02 2.01 1.98 1.96 31.8 6.96 4.54 3.75 3.36 3.14 3,00 2.90 2.82 2.76 2.72 2.68 2.65 2.62 2.60 2.58 2.57 2.55 2.54 2.53 2.52 2.51 2.50 2.49 2.49 2.48 2.47 2.47 2.46 2.46 2.42 2.40 2.36 2.33 0.95 0.975 0.99 0.995 0.9995 63.7 9.92 5.84 4.60 4.03 3.71 ' 3.50 3.36 43.25 3.17 3.11 3.05 39.1 2.98 2.95 2.92 2.90 2.88 2.86 2.85 2.83 2.82 2.81 2.80 2.79 2.78 2.77 i 2.76 2.76 2.75 2.70 2.68 2.63 2.58 Fall 2009 Stat 202 0.0 0.1 0.2 0.3 0.4 0.5 0. 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 7 2.4 - 2.5 2.6 2.7 2.8 2.9 3.0, Cumulative distx‘ibutian for a C(01) random variable 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.5000 0.5398 0.5793 0.6179 0.6554 0.6915 0.7257 0.7580 0.7881 0.8159 0.8413 0.8643 0.8849 0.9032 0.9192 0.9332 0.9452 0.9554 0.9641 0.9713 0.9772 0.9821 0.9861 . 0.9893 0.9918 0.9938 0.9953 0.9965 0.9974 . 0.9981 0.9987 0.5040 0.5438 0.5832 0.6217 0.6591 0.6950 0.7291 0.7611 0.7910 0.8186 0.8438 0.8665 0.8869 0.9049 0.9207 0.9345 0.9463 0.9564 0.9649 0.9719 0.9778 0.9826 0.9864 0.9896 0.9920 0.9940 0.9955 0.9966 0.9975 0.9982 0.9987 0.5080 0.5478 0.5871 0.6255 0.6628 0.6985 0.7324 0.7642 0.7939 0.8212 0.8461 0.8686 0.8888 0.9066 0.9222 0.9357 0.9474 0.9573 0.9656 0.9726 0.9783 0.9830 0.9868 0.9898 0.9922 0.9941 0.9956 0.9967 0.9976 0.9982 0.9987 0.5120 0.5517 0.5910 0.6293 0.6664 0.7019 0.7357 0.7673 0.7967 0.8238 0.8485 0.8708 0.8907 0.9082 0.9236 0.9370 0.9484 0.9582 0.9664 0.9732 0.9788 0.9834 0.9871 0.9901 0.9925 0.9943 0.9957 0.9968 0.9977 0.9983 0.9988 0.5160 0.5557 0.5948 0.6331 0.6700 0.7054 0.7389 0.7704 0.7995 0.8264 0.8508 0.8729 0.8925 0.9099 0.9251 0.9382 0.9495 0.9591 0.9671 0.9738 0.9793 0.9838 0.9875 0.9904 0.9927 0.9945 0.9959 0.9969 0.9977 0.9984 0.9988 "0.8944 0.5199 0.5596 0.5987 0.6368 0.6736 0.7088 0.7422 0.7734 0.8023 0.8289 0.8531 0.8749 0.5239 0.5636 0.6026 0.6406 , 0.6772 0.7123 0.7454 0.7764 0.8051 0.8315 0.8554 0.8770 0.8962 0.9131 0.9279 0.9406 0.9515 0.9608 0.9686 0.9115 0.9265 0.9394 0.9505 0.9599 0.9678 0.9744 0.9798 0.9842 0.9878 0.9906 0.9929 0.9946 0.9960 0.9970 0.9803 0.9846 0.9881 0.9909 0.9931 0.9948 0.9961 0.9971 0.9978 0.9979 0.9984 0.9985 0.9989 0.9989 0.9750 - 0.5279 0.5675 0.6064 {0.6443 0.6808 0.7157 0.7486 0.7794 0.8078 0.8340 0.8577 0.8790 “0.8980 0.9147 0.9292 0.9418 0.9525 0.9616 0.9693 0.9756 0.9808 0.9850 0.9884 0.9911 0.9932 0.9949 0.9962 0.9972 0.9979 0.9985 0.9989 Fall 2009 0.08 0.5319 0.5714 0.6103 0.6480 0.6844 0.7190 0.7517 0.7823 0.8106 0.8365 0.8599 0.8810 0.8997 0.9162 0.9306 0.9429 0.9535 0.9625 0.9699 0.9761 0.9812 0.9854 0.9887 0.9913 0.9934 0.9951 0.9963 0.9973 0.9980 0.9986 0.9990 0.09 0.5359 0.5753 0.6141 0.6517 0.6879 0.7224 0.7549 7 0.7852 0.8133 0.8389 0.8621 0.8830 0.9015 0.9177 ‘ 0.9319 0.9441 0.9545 0.9633 0.9706 0.9767 0.9817 0.9857 0.9890 0.9916 0.9936 0.9952 0.9964 0.9974 0.9981 0.9986 0.9990 T able 8.1: The Cumulative Distribution Function F : P(X S for 4 G10. 1) “‘41de 1vn ARIA" Page 4 Stat 202 Fall 2009 Answer Sheet — Test 2 Stat 202 /40 Last (Family) Name: First (Given) Name: Numeric ID: User ID: Part 1 [8 Marks / 1 mark each, NO penalty for guessing] Please write your multiple choice answers in part 1 below: Part 2 Short Answer. Most of the grade is dependent on how clearly you answer the problem. In this test, perform a continuity correction wherever necessary. 1. In your local newspaper there was a table that shows the high temperatures recorded for cities in Canada on August fifth. You select a random sample of 5 cities in Canada and record the following temperatures: 29, 16, 22, 24 and 31. (a) [5 Marks] Build a 99% confidence interval for the mean temperature of Canadian cities on August fifth. 29+16+22+24+31 r = —5——————— = 24.4 [1] 82 _ :56? ~ 71(5)2 n — 1 _ 292 + + 312 —- (5) (24.4)2 5 — 1 = 5.942 [1] c : 4.604 [1] r :t 95 = 24.4 i —4'60E:/(55"94) [1] => (12.15, 36.65) [1] (b) [2 Marks] Why is it important that the sample is random? Refer back to class examples. As we saw in class with our age experiment, if the sample is not random we will not find the mean in 99% of them... i.e. we will not have a 99% confidence interval. (c) [4 Marks] Approximate the probability that 5 randomly selected cities’ average temperature is more than 22 if it is known that the population mean temperature is 24 and the standard deviation is 6. Pr (Y > 22) = Pr (Z > 23/924) : [2] Pr (2 > —O.745356) : Pr (Z < 0.75) [1] => 77.34% [1] Page 5 Stat 202 Fall 2009 2. A scientist is capturing tadpoles in a pond. It is known that 25% of the ponds’ tandpoles are bullfrogs, 50% are green frogs and the remainder are of other breeds. (a) [4 Marks] In a bucket of 300 captured tadpoles write a formula for the exact probability that there is more than 60 but at most 85 bullfrogs. Pr (60 < X g 85) = f(61) + f(62) + + f(85) 85 : 2(330) (.25y’ (.75)300_i [4] i=61 (b) [10 Marks] In a bucket of 300 captured tadpoles, approximate the probability that there are exactly 75 bullfrog tadpoles. E(X) = np 2 300 (0.25) = 75 [1] Va7’(X) = np(1 —p) = 30 (0.75) (0.25) = 56.25 [1] Pr (X = 75) = Pr (74.5 < X < 75.5) (Continuity Correction) [2] : Pr < Z < [1] = Pr (—0.07 < Z < 0.07) = Pr (Z < 0.07) — Pr(Z < —0.07) [1] = Pr (2 < 0.07) — Pr(Z > 0.07) [1] Pr (2 < 0.07) — (1 — Pr(Z < 0.07)) [1] H 2 (0.5279) — 1 [1] ll 0.055 8 [1] (c) [5 Marks] Build a 90% confidence interval for the number of bullfrogs in the pond if 60 bullfrog tadbpoles are found in our bucket from "b". p : 1.654154%?2 = 3.51 1.645 2 [2] = 20% :: 0.04 [1] => 16% to 24% [1] (d) [2 Marks] Is the population proportion of bullfrog tadpoles in your interval? Yes or no and should we be concerned if the answer is yes or no? Why or why not? No. The population proportion is 25%. It is not in the interval. We should not be concerned as a confidence interval will only contain 25%, 90% of the time. Page 6 ...
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This note was uploaded on 02/26/2011 for the course STAT 202 taught by Professor Springer during the Spring '09 term at Waterloo.

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stat202-MT2 - Stat 202 Fall 2009 Question Booklet Part 1 -...

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