Chapter 10 1. Online testing: Do you prefer taking tests on paper or online? A college instructor gave identical tests to two randomly sampled groups of 35 students. One group took the test on paper and the other took it online. Following are the test scores. Page 486 Paper 79 75 49 78 73 81 70 63 79 65 60 74 64 64 74 69 57 65 61 58 92 71 69 66 71 68 67 81 78 80 76 67 49 56 45 Online 79 75 71 81 56 72 49 75 63 81 74 72 71 73 83 59 78 65 53 47 63 82 81 76 65 82 78 76 65 72 85 84 81 50 79 a. Construct a 95% confidence interval for the difference in mean scores between paper and online tests. b. The instructor claims that the mean scores are the same for both the paper and the online versions of the test. Does the confidence interval contradict this claim? 2. Pain after surgery: In a random sample of 50 patients undergoing a standard surgical procedure, 15 required medication for postoperative pain. In a random sample of 90 patients undergoing a new procedure, only 16 required pain medication. 1.Construct a 95% confidence interval for the difference in the proportions of patients needing pain medication between the old and new procedures. 2.A physician claims that the proportion of patients who need pain medication is the same for both procedures. Does the confidence interval contradict this claim? 3. High cholesterol: A group of eight individuals with high cholesterol levels were given a new drug that was designed to lower cholesterol levels. Cholesterol levels, in milligrams per deciliter, were measured before and after treatment for each individual, with the following results: Individual Before After 1 283 215 2 299 206 3 274 187 4 284 212 5 248 178 6 275 212 7 293 192 8 277 196 a.Construct a 90% confidence interval for the mean reduction in cholesterol level. b.A physician claims that the mean reduction in cholesterol level is more than 80 milligrams per deciliter. Does the confidence interval contradict this claim? Chapter 11 1. Empathy: The Interpersonal Reactivity Index is a survey designed to assess four different types of empathy. One type of empathy, called Empathetic Concern, measures the tendency to feel sympathy and compassion for people who are less fortunate. The index ranges from 0 (less empathetic) to 28 (more empathetic). The following data, representing random samples of 16 males and 16 females, are consistent with results reported in psychological studies. Boxplots show that it is reasonable to assume that the populations are approximately normal. Males 13 20 12 16 13 26 21 23 8 15 18 25 15 23 17 22 Females 22 20 26 25 28 24 16 19 20 23 21 23 15 26 19 25 Can you conclude that there is a difference in mean empathy score between men and women? Use the = 0.05 level. 2. Computer chips: A computer manufacturer has a choice of two machines, a less expensive one and a more expensive one, to manufacture a particular computer chip. Out of 500 chips manufactured on the less expensive machine, 70 were defective. Out of 400 chips manufactured on the more expensive machine, only 20 were defective. The manufacturer will buy the more expensive machine if he is convinced that the proportion of defectives is more than 5% less than on the less expensive machine. Let p1 represent the proportion of defectives produced by the less expensive machine, and let p2 represent the proportion of defectives produced by the more expensive machine. a. State appropriate null and alternate hypotheses. b. Compute the value of the test statistic. c. Can you reject H0 at the = 0.05 level? d. Which machine should the manufacturer buy? 3. Strength of concrete: The compressive strength, in kilopascals, was measured for concrete blocks from five different batches of concrete, both three and six days after pouring. The data are as follows: Block 1 2 3 4 5 After 3 days 1341 1316 1352 1355 1327 After 6 days 1376 1373 1366 1384 1358 Can you conclude that the mean strength after three days differs from the mean strength after six days? a. State the null and alternate hypotheses. b. Compute the test statistic. c. State a conclusion. Use the = 0.05 level of significance. Question1 Paper mean=68.4 standard deviation= 10.20 Online mean=71.31 Standard deviation=10.71 ( n 11 ) s 21+ ( n21 ) s 22 ( 351 ) 10.202+ ( 351 ) 71.312 s p= = =50.94 n 1+n 22 35+352 ( 2 1 t ( s ) n11 + n12 )=(71.3168.4 1.96(50.94) 351 + 351 )=2.91 23.866899 p The 95% confident that in the data, the mean difference between online and paper is between -20.95 and 26.77. Question 2 Letting n1=50; the number of standard surgical procedure patients X1=15; the number of patients requiring medication for prospective pain Proportion; p1=15/50=0.3 Patients undergoing new procedure; n2=90 The new procedure patients requiring medication for prospective pain is x2=12 Proportion; p2=16/90=0.178 a.) Calculating 95% interval; ( p1 p2 ) z / 2 p1 (1p 1) p2 (1 p 2) + n1 n2 (0.30.178) z 0.05/ 2 and substituting in the formula, we obtain; 0.3(10.3) 0.178(10.178) + 50 90 (0.122 0.1496)=(0.027,0 .272) as the interval. b.) Stating the null and alternative hypotheses; Null hypothesis: Proportion of pain is the same in both methods. ^ H0 : ^ p1 p 2 Alternative hypothesis: proportion of pain is not the same for the two methods. H 0 : P1 P2 The test statistic; Z p1 p 2 0.3 0.178 0.3(1 0.3) 0.178(1 0.178) p1q1 p2 q2 50 90 n1 n2 0.122 1.602 0.07632 Critical value at 5% significance level = 1.96 hence the test statistic < critical value; 1.602 < 1.96 Therefore, the null hypothesis cannot be rejected. Conclusion Proportion of pain medication is the same for the two methods. Confidence interval does not contradict with the test result since zero exists in both the statistical test result and in the confidence interval. Thus, it can be concluded that proportion of pain is the same for both methods. Question 3 a.) Looking at the assumptions, sample size is small. We therefore, construct a differences boxplot. There are no outliers but just a slight skew to the right. b.) It is difference may Formulating meaningful in the believe that the be as big as 88.3 hypotheses; H 0 : u1 u2 0 H1 : u1 u2 0 Significance level = 0.05 This is a two- tailed test and the means are; x1 17.9375 x2 22 and the s1 5.17002579 s2 3.687817783 Therefore, the standard error of the difference, s12 s2 2 sD sqrt ( ) n1 n2 standard deviations are; and; n1= sample size of group 1 = 16 n2 = sample size of group 2 = 16 Hence; df = n1 + n2 -2 = 30 and sD 1.587631228 The test statistic is; [ X X 2 uD ] t 1 2.558843595 If uD = hypothesized difference = 0 sD tcritical 2.042272456 Since /t/ > 2.0422, we reject H0. By use of the p-values; P = 0.015787775. P < 0.05, therefore, we reject H0. Conclusion: the population mean for males is not the same as that of female. 2. Computer chips a.) b.) c.) S i n c e test statistic (2.11) critical value (1.645), H0 is rejected. d.) The more expensive machine should be bought. 3. a.) b.) c.) since test statistic (-4.790) is not in the interval ( 2.776) of the critical values, we reject H0. the Conclusion: the mean strength after 3 days is not same as the mean strength after 6 days. Question1 Paper mean=68.4 standard deviation= 10.20 Online mean=71.31 Standard deviation=10.71 ( n 11 ) s 21+ ( n21 ) s 22 ( 351 ) 10.202+ ( 351 ) 71.312 s p= = =50.94 n 1+n 22 35+352 ( 2 1 t ( s ) n11 + n12 )=(71.3168.4 1.96(50.94) 351 + 351 )=2.91 23.866899 p The 95% confident that in the data, the mean difference between online and paper is between -20.95 and 26.77. Question 2 Letting n1=50; the number of standard surgical procedure patients X1=15; the number of patients requiring medication for prospective pain Proportion; p1=15/50=0.3 Patients undergoing new procedure; n2=90 The new procedure patients requiring medication for prospective pain is x2=12 Proportion; p2=16/90=0.178 a.) Calculating 95% interval; ( p1 p2 ) z / 2 p1 (1p 1) p2 (1 p 2) + n1 n2 (0.30.178) z 0.05/ 2 and substituting in the formula, we obtain; 0.3(10.3) 0.178(10.178) + 50 90 (0.122 0.1496)=(0.027,0 .272) as the interval. b.) Stating the null and alternative hypotheses; Null hypothesis: Proportion of pain is the same in both methods. H 0 : P 1=P 2 Alternative hypothesis: proportion of pain is not the same for the two methods. H0: P 1 P 2 The test statistic Z= ; Z= p1 p2 p1q1 p2q2 + n1 n2 0.30.178 0.3(10.3) 0.178(10.178) + 50 90 Solve to get the answer is Z=1.602 Critical value at 5% significance level = 1.96 hence the test statistic < critical value; 1.602 < 1.96 Therefore, the null hypothesis cannot be rejected. Conclusion Proportion of pain medication is the same for the two methods. Confidence interval does not contradict with the test result since zero exists in both the statistical test result and in the confidence interval. Thus, it can be concluded that proportion of pain is the same for both methods. Question 3 a.) Looking at the assumptions, sample size is small. We therefore, construct a differences boxplot. There are no outliers but just a slight skew to the right. dt 2 sd s < d +t / 2 d n n 79.375 1.896501( 13.3837588 ) 8 79.375 8.9650 1 70.4< d 88.3 b.) It is meaningful in the believe that the difference may be as big as 88.3 Formulating hypotheses; H 0 : 12=0 H 1 : 1 2=0 Significance level = 0.05 This is a two-tailed test and the means are; x 1=17.937 5 x 2=22 and the standard deviations are; s 1=5.1 7 s 2=3.687 8 Therefore, the standard error of the difference, 2 2 s1 s2 sD= ( + ) n1 n 2 and; n1= sample size of group 1 = 16 n2 = sample size of group 2 = 16 Hence; df = n1 + n2 -2 = 30 and sD The test statistic is; t= [ X 1 X 2 uD ] =2.558 8 sD If uD = hypothesized difference = 0 tcritical 2.042272456 Since /t/ > 2.0422, we reject H0. 1.587631228 By use of the p-values; P = 0.015787775. P < 0.05, therefore, we reject H0. Conclusion: the population mean for males is not the same as that of female. 2. Computer chips a.) H 0 : p1 p 2=0 .05 H 1 : pp 2=0 .05 b) b.) ; Z= Z= ( p 1 p2 ) p d p1 q1 p2q2 + n1 n2 70 20 0.05 500 400 70 430 20 3 8 0 ( ) ( ) 500 500 4 00 4 00 + 50 0 4 00 c.) Since test statistic (2.11) = 2.11 critical value (1.645), H0 is rejected. d.) The more expensive machine should be bought. 3. a.) H 0 : d=0 H 1 : d 0 t= d0 33.20 = sd 15.4984 =-4.79 5 n b.) c.) since test statistic (-4.790) is not in the interval ( 2.776) of the critical values, we reject H0. Conclusion: the mean strength after 3 days is not the same as the mean strength after 6 days. Question1 Paper mean=68.4 standard deviation= 10.20 Online mean=71.31 Standard deviation=10.71 ( n 11 ) s 21+ ( n21 ) s 22 ( 351 ) 10.202+ ( 351 ) 71.312 s p= = =50.94 n 1+n 22 35+352 ( 2 1 t ( s ) n11 + n12 )=(71.3168.4 1.96(50.94) 351 + 351 )=2.91 23.866899 p The 95% confident that in the data, the mean difference between online and paper is between -20.95 and 26.77. Question 2 Letting n1=50; the number of standard surgical procedure patients X1=15; the number of patients requiring medication for prospective pain Proportion; p1=15/50=0.3 Patients undergoing new procedure; n2=90 The new procedure patients requiring medication for prospective pain is x2=12 Proportion; p2=16/90=0.178 a.) Calculating 95% interval; ( p1 p2 ) z / 2 p1 (1p 1) p2 (1 p 2) + n1 n2 (0.30.178) z 0.05/ 2 and substituting in the formula, we obtain; 0.3(10.3) 0.178(10.178) + 50 90 (0.122 0.1496)=(0.027,0 .272) as the interval. b.) Stating the null and alternative hypotheses; Null hypothesis: Proportion of pain is the same in both methods. ^ H0 : ^ p1 p 2 Alternative hypothesis: proportion of pain is not the same for the two methods. H 0 : P1 P2 The test statistic; Z p1 p 2 0.3 0.178 0.3(1 0.3) 0.178(1 0.178) p1q1 p2 q2 50 90 n1 n2 0.122 1.602 0.07632 Critical value at 5% significance level = 1.96 hence the test statistic < critical value; 1.602 < 1.96 Therefore, the null hypothesis cannot be rejected. Conclusion Proportion of pain medication is the same for the two methods. Confidence interval does not contradict with the test result since zero exists in both the statistical test result and in the confidence interval. Thus, it can be concluded that proportion of pain is the same for both methods. Question 3 a.) Looking at the assumptions, sample size is small. We therefore, construct a differences boxplot. There are no outliers but just a slight skew to the right. b.) It is difference may Formulating meaningful in the believe that the be as big as 88.3 hypotheses; H 0 : u1 u2 0 H1 : u1 u2 0 Significance level = 0.05 This is a two- tailed test and the means are; x1 17.9375 x2 22 and the s1 5.17002579 s2 3.687817783 Therefore, the standard error of the difference, s12 s2 2 sD sqrt ( ) n1 n2 standard deviations are; and; n1= sample size of group 1 = 16 n2 = sample size of group 2 = 16 Hence; df = n1 + n2 -2 = 30 and sD 1.587631228 The test statistic is; [ X X 2 uD ] t 1 2.558843595 If uD = hypothesized difference = 0 sD tcritical 2.042272456 Since /t/ > 2.0422, we reject H0. By use of the p-values; P = 0.015787775. P < 0.05, therefore, we reject H0. Conclusion: the population mean for males is not the same as that of female. 2. Computer chips a.) b.) c.) S i n c e test statistic (2.11) critical value (1.645), H0 is rejected. d.) The more expensive machine should be bought. 3. a.) b.) c.) since test statistic (-4.790) is not in the interval ( 2.776) of the critical values, we reject H0. the Conclusion: the mean strength after 3 days is not same as the mean strength after 6 days. Question1 Paper mean=68.4 standard deviation= 10.20 Online mean=71.31 Standard deviation=10.71 ( n 11 ) s 21+ ( n21 ) s 22 ( 351 ) 10.202+ ( 351 ) 71.312 s p= = =50.94 n 1+n 22 35+352 ( 2 1 t ( s ) n11 + n12 )=(71.3168.4 1.96(50.94) 351 + 351 )=2.91 23.866899 p The 95% confident that in the data, the mean difference between online and paper is between -20.95 and 26.77. Question 2 Letting n1=50; the number of standard surgical procedure patients X1=15; the number of patients requiring medication for prospective pain Proportion; p1=15/50=0.3 Patients undergoing new procedure; n2=90 The new procedure patients requiring medication for prospective pain is x2=12 Proportion; p2=16/90=0.178 a.) Calculating 95% interval; ( p1 p2 ) z / 2 p1 (1p 1) p2 (1 p 2) + n1 n2 (0.30.178) z 0.05/ 2 and substituting in the formula, we obtain; 0.3(10.3) 0.178(10.178) + 50 90 (0.122 0.1496)=(0.027,0 .272) as the interval. b.) Stating the null and alternative hypotheses; Null hypothesis: Proportion of pain is the same in both methods. H 0 : P 1=P 2 Alternative hypothesis: proportion of pain is not the same for the two methods. H0: P 1 P 2 The test statistic Z= ; Z= p1 p2 p1q1 p2q2 + n1 n2 0.30.178 0.3(10.3) 0.178(10.178) + 50 90 Solve to get the answer is Z=1.602 Critical value at 5% significance level = 1.96 hence the test statistic < critical value; 1.602 < 1.96 Therefore, the null hypothesis cannot be rejected. Conclusion Proportion of pain medication is the same for the two methods. Confidence interval does not contradict with the test result since zero exists in both the statistical test result and in the confidence interval. Thus, it can be concluded that proportion of pain is the same for both methods. Question 3 a.) Looking at the assumptions, sample size is small. We therefore, construct a differences boxplot. There are no outliers but just a slight skew to the right. dt 2 sd s < d +t / 2 d n n 79.375 1.896501( 13.3837588 ) 8 79.375 8.9650 1 70.4< d 88.3 b.) It is meaningful in the believe that the difference may be as big as 88.3 Formulating hypotheses; H 0 : 12=0 H 1 : 1 2=0 Significance level = 0.05 This is a two-tailed test and the means are; x 1=17.937 5 x 2=22 and the standard deviations are; s 1=5.1 7 s 2=3.687 8 Therefore, the standard error of the difference, 2 2 s1 s2 sD= ( + ) n1 n 2 and; n1= sample size of group 1 = 16 n2 = sample size of group 2 = 16 Hence; df = n1 + n2 -2 = 30 and sD The test statistic is; t= [ X 1 X 2 uD ] =2.558 8 sD If uD = hypothesized difference = 0 tcritical 2.042272456 Since /t/ > 2.0422, we reject H0. 1.587631228 By use of the p-values; P = 0.015787775. P < 0.05, therefore, we reject H0. Conclusion: the population mean for males is not the same as that of female. 2. Computer chips a.) H 0 : p1 p 2=0 .05 H 1 : pp 2=0 .05 b) b.) ; Z= Z= ( p 1 p2 ) p d p1 q1 p2q2 + n1 n2 70 20 0.05 500 400 70 430 20 3 8 0 ( ) ( ) 500 500 4 00 4 00 + 50 0 4 00 c.) Since test statistic (2.11) = 2.11 critical value (1.645), H0 is rejected. d.) The more expensive machine should be bought. 3. a.) H 0 : d=0 H 1 : d 0 t= d0 33.20 = sd 15.4984 =-4.79 5 n b.) c.) since test statistic (-4.790) is not in the interval ( 2.776) of the critical values, we reject H0. Conclusion: the mean strength after 3 days is not the same as the mean strength after 6 days