test3_sol - AM8310 PAGE 2-form B PRINT YOUR NAME HERE WRITE...

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Unformatted text preview: AM8310 PAGE 2 --form B PRINT YOUR NAME HERE WRITE CORRECT ANSWER IN THE SPACE PROVIDED BY THE QUESTION NUMBER Information for questions 1-2. Suppose X has a normal p.d.f. with some unknown mean, It and KNOWN standard deviation, 0:81 and we sample n=49 observations of X (X1, X2, X49). Let :Y_ =2 Xi I49 denote the estimator of u. Use the appropriate information to determine the following: Q2. I 1. The maximum error of Y (the sample mean) with probability 0.95. That is, we predict that the error of the sample mean will be no more than in 95% of the samples. Hint: 200251.96 . L9H)!“ L3 2.2.? m 25} 2. The sample size such that the maximum error of X is less than 10 with probability equal to 0.95. . 2 1 a; (ms) I 0L D 3. A research team want to know whether the mean number of breakdowns per month is in fact 2.0 Hence they will test the null hypothesis H0: p=2, They assume that the standard deviation equals to 0.5, La. 0' is known and o=0.5 . The alternative hypothesis to be considered is H1: patZ, 0'=0.5. They will test the null hypothesis at the 0.01 level, using a sample of n=100 days. Under the : 7,92‘ 04‘ «9 I’buucl “I? I +29; ALTERNATIVE hypothesis the averages obtained in samples of n=100, X could have a) Mean equal to 2 and standard deviation of 0.5 b) Mean equal to 3 and standard deviation of 0.5. c) Mean equal to 2 and standard deviation of 0.05 d) Mean equal to 3 and standard deviation of 0.05 ( ( ( ( B 4. Investigator A has 25 subjects and tests Ho;p.=8 vs. Haip $8 at the 0.05 level. Investigator B has 280 subjects and tests H0: u=8 vs. Ha: l1 at 8 at the 0.05 level. Suppose ”=8 (a) Investigator A is less likely than Investigator B to accept Ho and be right. (b) Investigator A and Investigator B are equally likely to accept Ho and be right. (c) Investigator A is more likely than Investigator B to accept Ho and be right. 0315 5. A resear er tests Ho: “=1 vs. H1: p¢1 at the 0.05 level using a sample of n=25. The standard deviation i known to equal 5 (Le, 6:5). He computes the probability of this type II error (B) of his test to be . for u: -1 and for u=3. If 11:3 and n=100 the probability of his type II error is (a) Greater than 0.25. (b) less than 0.25 (0) equal to 0.25 (d) equal to 0.95 C, AMS310 TEST 3 Form B PAGE 3 PRINT YOUR NAME HERE WRITE (2) CORRECT ANSWER IN THE SPACE PROVIDED BY THE QUESTION NUMBER 6. An investigator has two hypotheses (1) ll< 5 and (2) u = 5. Which hypothesis does he test? ,0 2 ‘5 g 7. Suppose we compute a 98% confidence interval estimates of u as being from 3 to 12. Suppose in fact u=10. Which of the following are true? (a) The probability that u is on this interval equals 0.0. (b) The probability that u is on this interval equals 1.0. (c)The probability that p is on this interval equals 0.98. (d) The probability that l1 is on this interval equals 0.02 (e)This demonstrates that we are very good at computing confidence intervals. C 8. Which of the following will increase the width of the 95% confidence interval estimate of u? (a) Decrease the standard deviation of the variable. (b) Increase the mean (c) Increase the confidence value from 95% to 99% (d) Increase the sample size 1 C/ 9. The probability that a random variable having the t distribution with 8 degrees of freedom is less than to.25 equals (a) 0.25 (b) 0.50 (c) 0.75 (d) 1.00 10. A research group tests Ho: “=2 vs. H1: u>2 at the 0.05 level. The P value they obtain is 0.02. (a) Reject H0 at the 0.05 level one sided and conclude that u>2. (b) Reject H0 at the 0.05 level one sided and conclude that u=2. (c) Do not reject H0 at the 0.05 level one sided and conclude u>2. (d) Do not reject H0 at the 0.05 level one sided and conclude u=2. AMS310 TEST 3 FORM B PAGE 4 PRINT YOUR NAME HERE LONGER QUESTIONS. FOR THESE QUESTIONS WHEREVER THERE IS A NUMERICAL ANSWER YOU NEED TO SHOW HOW YOU ARRIVED AT IT. YOUR SCORE WILL BE DETERMINED BY YOUR PRESENTATION OF YOUR METHOD. ”-1. A random sample of 4 aspirin bottles contain on average, 324.4 milligrams (mg.) of aspirin. The sample estimate of standard deviation (s) equals 0.5 mg. Assume that the data may be treated as a random sample from a normal population. (a) What is the point estimate of u, the mean number of mg. of aspirin per bottle? 7:37.44. is an e/S‘I‘fmcx‘I‘C Fat/VI (b) What is the value of the point estimate of the standard deviation of the sample average? (Kg ‘ g " '3 0.1g 2 S‘e’omcla (I, Est-Ind r” 2 SE \lZ r e“ W ‘ (0) Compute a 95% confidence interval estimate of the mean number of mg. of aspirin per bottle. “ta/owls 2 3. ($2. ._,_> 314.4 i 3. (final? [39.24 9 £25.23 (d) Using your result in (0), determine your conclusions were you to test Ho: p=325 mg. vs. H1zu¢325 mg. at the 0.05 level two sided. A Chi/«AUX? MM. IARVE 3 2'9 . 0 0V1 Im‘I-QJ/VCAQ 5‘0 act-Wt H. 0.1“ a o 9- szj 2. $10qu _ @mflwtt WI ‘2 7’23 525' (e) Test Ho: H = 325 mg. vs. H1: p #fi-with these data at the 0.1 level. Hint: t= -2.4 and Pr (t > 2.4) =0.04 for v = 3 F: 0.09; 5.9 Wfifitgl He al‘ 04 W. AMS310 TEST 3 FORM B PAGE 5 PRINT YOUR NAME HERE “-2. A study is done to compare the levels of dioxin TCDD of 20 Massachusetts Vietnam veterans who were possibly exposed to Agent Orange. The amount of TCDD levels in plasma and fat tissues are to be compared. Thus we have paired data and the test of Ho: lira: - up.asma =0 is reduced to a single sample test of Ho: up: 0 where D= Fat level of TCDD — Plasma level of TCDD in each individual. We obtain the following results: Sample Mean (d = 0.87 n =20 STD DEV(s) =2.98 Standard error = 0.67) Test Ho: u; 0 vs. H1: “$0 at the 0.05 level. In doing so answer the following questions. (a) Give the equation for the test statistic: (b) Give the rejection region. “t 7 m 9 Raw r“ 6+ “’9 W W SW *" “W (0) Give value of value of test statistic. O\%r‘“\ O ________——: [‘30 go “t Ho “‘1‘ 0‘ a; W . 0‘ er} (Law? (d) Give conclusions (reject or accept H0 at the 0.05 level one sided to the right). Acw‘gt H. at 0‘09 W \aerauje (‘30 <lea. (e) Give conclusions of test with respect to whether mean toxin levels are the same in fat as plasma, greater, or less. Write a sentence. COMJtutolfl (PM: fur W3 Mum meow erg/EM 'l‘o Elam; was. (f) Give the P value for these data and indicate whether you would reject H0 at the 0.1 level one- sided. . . F>0\l, “(:>l\32.$ S’o flébept (X'l‘ °~l led/‘91 5 AMS310 TEST 3 FORM B PAGE 6 PRINT YOUR NAME HERE 11-3. A new cure has been developed for a certain type of cement that results in a compressive strength of 5000 kg./cm.2 and a standard deviation of 6:120 kg/cm.2. Suppose we sample n=50 pieces of cement and test them. We will test Ho: u=5000, 0:120 vs. H1: u<5000, o=120. (a)Design a test at the 0.01 level in doing so report the following. (a)(i) The test statistic and your reason for choosing this statistic. 2.3: 3?,“9’000 (D S“ Know/l (lo/r927 @ h lor'SL. (a)(ii) The rejection criterion. 2 < ~23; a (iii) Indicate whether one should accept or reject Ho if the sample mean ; = 4970 kg./m2 Z: ~_~o Xvi-E; : \l'llj [7/0 a (iv) Indicate your conclusions about the cement using the results in a(iii). {\ch Ho 06v om law/t haulage. ‘L‘N >~2x33 (b) Suppose in fact the mean strength is 4960 kg/m. Compute the probability that we will correctly reject Ho and conclude u<5000. That is, compute the power for p: 4960 for a sample of n=50 using a one sided 0.05 level test as described above. The formula on page 8 for the value of [3(u) is likely to be helpful. (DQ07- 1.-—fl = l' g 20. - m WI?) :1, “flies—2.59) “2 EMU : «1%. AMS310 TEST 3 FORM B PAGE 7 PRINT YOUR NAME HERE WRITE CORRECT ANSWER IN THE SPACE PROVIDED BY THE QUESTION NUMBER ”-4. Prove: If we accept Ho: p= H0 at the a level then Ho is on the (1- CL) 100% confidence interval foru. . II- 5. (Extra Credit). We wish to test Ho: p= 0.3 vs. H1: p < 0.30 using a sample of n=20 observations. (a) Suppose we sample n=20 individuals and observe that X=3. Use test statistic for large samples: Z= (X- npm/ np0(1-po) and test the null hypothesis at the 0.05 level. (+2 points) (b) Indicate your conclusions as to whether or not p<0.30 based (a) (+‘Lpoint). (c) Find the exact P value for this result using the binomial distribution (+4 points). Indicate your conclusions about whether or not one major earns more or less than the other or whether the evidence here points to an equal mean salary for these two groups. w ' H “3 ._ \l 037m» >42» ' 1‘05 “l- 46. AMS310 PAGE 2 --form A PRINT YOUR NAME HERE WRITE CORRECT ANSWER IN THE SPACE PROVIDED BY THE QUESTION NUMBER Information for questions 1-2. Suppose X has a normal p.d.f. with some unknown mean, p. and KNOWN standard deviation, o=49and we sample n=81 observations of X (X1, X2, X31). Let 3? =2 Xi I81 denote the estimator of p. Use the appropriate information to determine the following: Il- Fl 1. The maximum error of 3(— (the sample mean) with probability 0.98. That is, we predict that the error of the sample mean will be no more than in 98% of the samples. Hint: zom= 2.33. 2.721;;«31 7.. 2.5“7xfl 2 (2.6% W W I 3‘ 2. The sample size such that the maximum error of f is less than 10 with probability equal to 0.98. /h 7/ (2%.‘6‘>L at 1‘33[X4QJL: [34%4‘ E o 3. A research team want to know whether the mean number of breakdowns per month is in fact 2.0 Hence they will test the null hypothesis H0: p=2, They assume that the standard deviation equals to 0.5, La. 0 is known and c=0.5 . The alternative hypothesis to be considered is H1: (#2, o=0.5. They will test the null hypothesis at the 0.01 level, using a sample of n=100 days. Under the null hypothesis the averages obtained in samples of n=100, X could have L (a) Mean equal to 2 and standard deviation of 0.5 (b) Mean equal to 3 and standard deviation of 0.5. (0) Mean equal to 2 and standard deviation of 0.05 (d) Mean equal to 3 and standard deviation of 0.05 B 4. Investigator A has 25 subjects and tests H01p=8 vs. Hazp $8 at the 0.05 level. Investigator B has 280 subjects and tests Ho: p=8 vs. Ha: p ¢ 8 at the 0.05 level. Suppose p=8 (a) Investigator A is less likely than Investigator B to accept Ho and be right. (b) Investigator A and Investigator B are equally likely to accept Ho and be right. (c) Investigator A is more likely than Investigator B to accept Ho and be right. _I_3_ 0‘25. 5. A researc er tests Ho: p=1 vs. H1: p¢1 at the 0.05 level using a sample of n=25. The standard deviation is own to equal 5 (Le, 0:5). He computes the probability of this type II error ([3) of his test to be .5 for p: -1 and for p=3. If p=10 and n=25 the probability of his type II error is (a) Greater than 0.25. (b) less than 0.25 (0) equal to 0.25 (d) equal to 0.95 AMS310 TEST 3 Form A PAGE 3 PRINT YOUR NAME HERE WRITE CORRECT ANSWER IN THE SPACE PROVIDED BY THE QUESTION NUMBER (17 6. An investigator has two hypotheses (1) p. at 5 and (2) l1 = 5. Which hypothesis does he test? .14; 9 7. Suppose we compute a 98% confidence interval estimates of u as being from 5 to 8. Suppose in fact u=10. Which of the following are true? (a) The probability that p is on this interval equals 0.0. (b) The probability that H is on this interval equals 1.0. (c)The probability that u is on this interval equals 0.98. (d) The probability that u is on this interval equals 0.02 (e)This demonstrates that we are not very good at computing confidence intervals. 8. Which of the following will increase the width of the 95% confidence interval estimate of u? (a) Increase the sample size (b) Decrease the confidence value from 95% to 90% (c) increase the mean (d) Increase the confidence value from 95% to 99% A 9. The probability that a random variable having the t distribution with 8 degrees of freedom is less than tom equals (a) 0.99 (b) 0.01 (c) 0.98 (d) 0.02 12 10. A research group tests Ho: u=2 vs. H1: u>2 at the 0.05 level. The P value they obtain is 0.07. (a) Reject H0 at the 0.05 level one sided and conclude that u>2. (b) Reject H0 at the 0.05 level one sided and conclude that u=2. (c) Do not reject H0 at the 0.05 level one sided and conclude u>2. (d) Do not reject H0 at the 0.05 level one sided and conclude u=2. ...
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