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CHAPTER 11 REVIEW 1. Find the critical value of t required to construct a 99% confidence interval for a population mean based on a sample of size 15. ANS: Looking in the row for df = 15 – 1 = 14, and the column for 99%, we find t* = 2.977. 2. The following data were collected as part of a study. Construct a 90% confidence interval for the true difference between the means (μ1 – μ2). Does it seem likely that, despite the sample differences, that there is not real difference between the population means? The samples were SRSs from independent, approximately normal, populations. Population 1 2 n 20 18 x-bar 9.87 7.13 s 4.7 4.2 ANS: The relatively small values of n tell us that we need to use a 2-sample t interval. The conditions necessary for using this interval are given in the problem: SRSs from independent, approximately normal, proportions. Using the “conservative” method of choosing the degrees of freedom: df = min {n1 – 1, n2 – 1} = min {19, 17} = 17 t* = 1.740 4.7 2 4.2 2 2.74 1.740(1.444) (0.227, 5.25) 20 18 We are 90% confident that the true difference between the means lies in the interval from 0.227 to 5.25. If the true difference between the means is zero, we would expect to find 0 in the interval. (9.87 7.13) 1.740 3. The local farmers association in Cass County wants to estimate the mean number of bushels of corn produced per acre in the county. A random sample of 13 1-acre plots produced the following results (in number of bushels per acre): 98, 103, 95, 99, 92, 106, 101, 91, 99, 101, 97, 95, 98. Construct a 95% confidence interval for the mean number of bushels per acre in the entire county. The local association has been advertising that the mean yield per acre is 100 bushels. Do you think they are justified in this claim? ANS: The population standard deviation is unknown, and the sample size is small (13), so we need to use a t procedure. Now, x-bar = 98.1, s = 4.21, df = 13 – 1 = 12 t* = 2.179. The 95% confidence interval is 4.21 98.1 2.179 (95.56, 100.64) 13 Because 100 is contained in this interval, we do not have strong evidence against the association’s claim that the mean number of bushels per acre is 100, even though the sample mean is only 98.1. 4. A study done to determine if male and female 10th graders differ in performance in mathematics was conducted. Twenty-three randomly selected males and 26 randomly selected females were each given a 50-question multiple-choice test as part of the study. The scores were approximately normally distributed. The results of the study were as follows: Sample size Mean St. Deviation Males 23 40.3 8.3 Females 26 39.2 7.6 Construct a 99% confidence interval for the true difference between the mean score for males and the mean score for females. Does the interval suggest that there is a difference between the true means for males and females? ANS: The problem states that the samples were randomly selected and that the scores were approximately normally distributed, so we can construct a two-sample t interval. df = min {23 – 1, 26 – 1} = 22 t* = 2.819 8.32 7.6 2 (40.3 39.2) 2.819 (-5.34, 7.54) 23 26 0 is a likely value for the true difference between the means because it is in the interval. Hence, we do not have evidence that there is a difference between the true means for males and females. 5. A company president believes that there are more absences on Monday than on other days of the week. The company has 45 workers. The following table gives the number of worker absences on Mondays and Wednesdays for an 8-week period. Do the data provide evidence that there are more absences on Mondays? Monday Wednesday Week 1 5 2 Week 2 9 5 Week 3 2 4 Week 4 3 0 Week 5 2 3 Week 6 6 1 Week 7 4 2 Week 8 1 0 ANS: Because the data are paired on a weekly basis, the data we use for this problem are the difference between the days of the week for each of the 8 weeks. Adding a row to the table gives the differences (absences on Monday minus absences on Wednesday), we have: Monday Wednesday Difference Week 1 5 2 3 Week 2 9 5 4 Week 3 2 4 -2 Week 4 3 0 3 Week 5 2 3 -1 Week 6 6 1 5 Week 7 4 2 2 Week 8 1 0 1 Let μ = the true mean difference between the number of absences on Monday and absences on Wednesday. H0: μ = 0 and Ha: μ > 0 We will use a one-sample t test for the difference scores. 1.75 0 x 1.75, s 2.49, t 1.99, df 8 1 7 0.025 P 0.05 (from Table C). [Using the 2.49 8 TI-83, P = 0.043) The P-value is small. This provides us with evidence that there are more absences on Mondays than on Wednesdays. 6. Twenty-six pairs of identical twins are enrolled in a study to determine the impact of training on ability to memorize a string of letters. Two programs (A and B) are being studied. One member of each pair is randomly assigned to one of the two groups and the other twin goes into the other group. Each group undergoes the appropriate training program, and then the scores for pair of twins is compared. The means and standard deviations for groups A and B are determined as well as the mean and standard deviation for the difference between each twins score. Is this study a one-sample or two-sample situation, and how many degrees of freedom are involved in determining the t value? ANS: This is a paired study because the scores for each pair of twins is compared. Hence, it is a onesample situation, and there are 26 pieces of data to be analyzed, which are the 26 differences scores between the twins. Hence, df = 26 – 1 = 25. 7. An avid reader, Booker Worm, claims that he reads books that average more than 375 pages in length. A random sample of 13 books on his shelf had the following number of pages: 595, 353, 434, 382, 420, 225, 408, 422, 315, 502, 503, 384, 420. Do the data support Booker’s claim? Test at the 0.05 level of significance. ANS: Let μ = the true average number of pages in the books Booker reads. H0: μ = 375 and Ha: μ > 375. We are going to use a one-sample t test at α = 0.05. n 13, x 412.5, s 91.35, df 13 1 12 412.5 375 1.48 91.35 13 0.05 < P < 0.10 (from Table C) or P = 0.082 from TI-83. Because P > 0.05, we cannot reject H0. We do not have strong evidence to back up Booker’s claim that the books he reads actually average more than 375 pages in length. t 8. The statistics teacher, Dr. Tukey, gave a 50-point quiz to his class of 10 students and they did not do very well, at least by Dr. Tukey’s standards. Rather than continuing to the next chapter, he spent some time reviewing the material and then gave another quiz. The quizzes were comparable in length and difficulty. The results of the two quizzes were as follows. Student Quiz 1 Quiz 2 1 45 42 2 40 38 3 36 34 4 38 37 5 34 36 6 28 26 7 44 44 8 35 32 9 42 38 10 30 31 Do the data indicate that the review was successful, at the 0.05 level, of improving the performance of the students on this material? Give good statistical evidence for your conclusion. ANS: The data are paired by individual students, so we need to test the difference scores for the students rather than the means for each quiz. The differences are given in the following table. Student Quiz 1 Quiz 2 Difference (Q1 – Q2) 1 45 42 3 2 40 38 2 3 36 34 2 4 38 37 1 5 34 36 -2 6 28 26 2 7 44 44 0 8 35 32 3 9 42 38 4 10 30 31 -1 Let μ = the mean of the differences between the scores of students on Quiz 2 and Quiz 1. H0: μ = 0 and Ha: μ > 0 This is a matched pairs t test. That is, it is a one-sample t test for a population mean. n 10, x 1.4, s 1.90, df 10 1 9 1.4 0 2.33 1.90 10 0.02 < P < 0.025 (from Table C) or P = 0.022 (from TI-83) Because P < 0.05, we reject the null. The data provide evidence at the 0.05 level that the review was successful at improving student performance on the material. t 9. A company uses two different models, call them model A and model B, of a machine to produce electronic locks for hotels. The company has several hundred of each machine in use in its various factories. The machines are not perfect, and the company would like to phase out of service the one that produces the most defects in the locks. A random sample of 13 model A machines and 11 model B machines are tested and the data for the average number of defects per week are given in the following table: Model A Model B n 13 11 x-bar 11.5 13.1 s 2.3 2.9 Dot plots of the data indicate that there are no outliers or strong skewness in the data and that there are no strong departures from normal. Do these data provide statistically convincing evidence that the two machines differ in terms of the number of defects produced? ANS: Let μ1 = the true mean number of defects produced by machine A. Let μ2 = the true mean number of defects produced by machine B. H0: μ1 – μ2 = 0 and Ha: μ1 – μ2 ≠ 0 We use a two-sample t test for the difference between means. The conditions for this procedure are given in the problem: both samples are simple random samples from independent, approximately normal populations. df = min {13 – 1, 11 – 1} = 10 11.5 13.1 t 1.48 0.05 P 0.10 (from Table C). However, since it is a two-sided test, 2.3 2 2.9 2 13 11 we have to multiply P-value by 2. Therefore, it is 0.10 < P < 0.20. The P-value is too large to be strong evidence against the null hypothesis that there is no difference between the machines. We do not have strong evidence that the types of machines actually differ in the number of defects produced. 10. You must estimate the number of students who will attend the annual Math Meet at your university this year. Over the past 8 years, the attendance has been rather consistent, with a mean of 122 and a standard deviation of 12. Estimate the mean number of attendees with 95% confidence. ANS: x-bar = 122, s = 12, n = 8, and C = 95% The sample is a simple random sample. The population is normal and σ is unknown. Therefore, we can use one-sample t interval. df = 8 – 1 = 7 t* = 2.365 12 Confidence interval = 122 2.365 (112, 132) 8 We are 95% confident that the number of students who will attend this year’s annual Math Meet will be between 112 and 132. 11. Two independent random samples were taken to compare the means of the two populations. The sample statistics are summarized in the table below. Construct a 99% confidence interval for the estimation of the difference in the means of the two samples. Sample A B n 35 50 x-bar 63.5 50.4 s 5.2 9.6 ANS: x1 63.5, x2 50.4, n1 35, n2 50, C 99% s1 5.2, s 2 9.6 df = min {35 – 1, 50 – 1} = 34 t* = 2.750 (Since our table C does not have df = 34, we err on the conservative side and take the next lowest value the table contains. In this case, it would be df = 30.) 5.2 2 9.6 2 (8.8, 17.4) 35 50 We are 99% confident that the difference between means of these two populations is between 11.495 and 14.705. Confidence interval = (63.5 50.4) 2.750 12. When 10 cars of a new model were tested for gas mileage, the results showed a mean of 27.2 miles per gallon with a standard deviation of 1.8 miles per gallon. What is a 95% confidence interval estimate for the gas mileage achieved by this model? ANS: The population standard deviation is unknown, and so we use the t-distribution. The standard 1.8 deviation of the sample means is s 0.569. With 10 – 1 = 9 degrees of freedom t* = 10 2.262. Thus, we can be 95% certain that the gas mileage of the new model is in the range of 1.8 27.2 2.262 (25.9, 28.5) miles per gallon. 10 13. A new process for producing synthetic gems yielded six stones weighing 0.43, 0.52, 0.46, 0.49, 0.60, and 0.56 carats, respectively, in its first run. Find a 90% confidence interval estimate for the mean carat weight from this process. ANS: Using 1-Var Stats, we get x-bar = 0.51 and s = 0.0632. With df = 6 – 1 = 5 t* = 2.015. Thus, we can be 90% sure that the new process will yield stones weighing 0.0632 0.51 2.015 (0.458, 0.562) carats. 6 14. A survey is run to determine the difference in the cost of groceries in suburban stores versus inner city stores. A preselected group of items is purchased in a sample of 45 suburban and 35 inner city stores, and the following data are obtained. Suburban stores Inner city stores n1 = 45 n2 = 35 x-bar1 = $36.52 x-bar2 = $39.40 s1 = $1.10 s2 = $1.23 Find a 90% confidence interval estimate for the difference in the cost of groceries. ANS: df = min {45 – 1, 35 – 1} = 34 t* = 1.697 (1.23) 2 (1.10) 2 (2.44, 3.32) Confidence interval = (39.40 36.52) 1.697 35 45 Thus, we are 90% certain that the selected group of items costs between $2.44 and $3.32 more in suburban stores that in inner city stores. 15. Two varieties of corn are being compared as to difference in maturation time. Ten plots of the first variety reach maturity in an average of 95 days with a standard deviation of 5.3 days, while eight plots of the second variety reach maturity in an average of 74 days with a standard deviation of 4.8 days. Determine a 95% confidence interval estimate for the difference in maturation time. ANS: n1 = 10 n2 = 8 x-bar1 = 95 x-bar2 = 74 s1 = 5.3 s2 = 4.8 With df = min {10 – 1, 8 – 1} = 7 t* = 2.365 (5.3) 2 (4.8) 2 (15.36, 26.64) 10 8 Thus, we can be 95% certain that the first variety of corn will take between 15.36 and 26.64 more days to read maturity than the second variety. Confidence interval = (95 74) 2.365 16. In a test for acid rain, an SRS of 49 water samples showed a mean pH level of 4.4 with a standard deviation of 0.35. Find a 90% confidence interval estimate for the mean pH level. (a) 4.4 ± 0.01 (b) 4.4 ± 0.08 (c) 4.4 ± 0.32 (d) 4.4 ± 0.35 (e) 4.4 ± 0.58 ANS: B – Using t-scores: df = 49 – 1 = 48 t* = 1.676 0.35 Confidence interval = 4.4 1.676 4.4 0.084 49 17. One gallon of gasoline is put in each of 30 test autos, and the resulting mileage figures are tabulated with x-bar = 28.5 and s = 1.2. Determine a 95% confidence interval estimate of the mean mileage. (a) (28.46, 28.54) (b) (28.42, 28.58) (c) (28.1, 28.9) (d) (27.36, 29.64) (e) (27.3, 29.7) ANS: C – Using t-scores: df = 30 – 1 = 29 t* = 2.045 1.2 Confidence interval = 28.5 2.045 (28.1, 28.9) 30 18. A company owns 335 trucks. For an SRS of 30 of these trucks, the average yearly road tax paid is $9540 with a standard deviation of $1205. What is a 99% confidence interval estimate for the total yearly road taxes paid for the 335 trucks? (a) $9540 ± $103 (b) $9540 ± 567 (c) $3,196,000 ± $567 (d) $3,196,000 ± $35,000 (e) $3,196,000 ± $203,000 ANS: E – Using t-scores: df = 30 – 1 = 29 t* = 2.576 1205 Confidence interval = 9540 2.756 9540 606.3 30 335 (9540 ± 606.3) = $3,196,000 ± $203,000 19. In a study aimed at reducing developmental problems in low-birth-weight (under 2500 grams) babies, 347 infants were exposed to a special educational curriculum while 561 did not receive any special help. After 3 years the children exposed to the special curriculum showed a mean IQ of 93.5 with a standard deviation of 19.1; the other children had a mean IQ of 84.5 with a standard deviation of 19.9. Find a 95% confidence interval estimate for the difference in mean IQs of low-birth-weight babies who receive special intervention and those who do not. (a) 9.0 ± 2.60 (b) 9.0 ± 4.42 (c) 9.0 ± 6.24 (d) 89.0 ± 19.5 (e) 89.0 ± 39.0 ANS: A – With df = min {347 – 1, 561 – 1} = 346 t* = 1.984 n1 = 347, x-bar1 = 93.5, s1 = 19.1 n2 = 561, x-bar2 = 84.5, s2 = 19.9 Confidence interval = (93.5 84.5) 1.984 19.12 19.9 2 9.0 2.60 347 561 20. Does socioeconomic status relate to age at time of HIV infection? For 274 high-income HIVpositive individuals the average of infection was 33.0 years with a standard deviation of 6.3, while for 90 low-income individuals the average was 28.6 years with a standard deviation of 6.3. Find a 90% confidence interval estimate for the difference in ages of high- and low-income people at the time of HIV infection. (a) 4.4 ± 0.963 (b) 4.4 ± 1.26 (c) 4.4 ± 2.51 (d) 30.8 ± 2.51 (e) 30.8 ± 6.3 ANS: B – With df = min {274 – 1, 90 – 1} = 89 t* = 1.664 n1 = 274, x-bar1 = 33, s1 = 6.3 n2 = 90, x-bar2 = 28.6, s2 = 6.3 Confidence interval = (33 28.6) 1.664 6.3 2 6.3 2 4.4 ± 1.26 274 90 21. Acute renal graft rejection can occur years after the graft. In one study, 21 patients showed such late acute rejection when the ages of their grafts (in years) were 9, 2, 7, 1, 4, 7, 9, 6, 2, 3, 7, 6, 2, 3, 1, 2, 3, 1, 1, 2, and 7, respectively. Establish a 90% confidence interval estimate for the ages of renal grafts that undergo late acute rejection. (a) 2.024 ± 0.799 (b) 2.024 ± 1.725 (c) 4.048 ± 0.799 (d) 4.048 + 1.041 (e) 4.048 ± 1.725 ANS: D – x-bar = 4.048, s = 2.765, df = 21 – 1 = 20 t* = 1.725 2.765 Confidence interval = 4.048 1.725 4.048 ± 1.041 21 22. In a sample of ten basketball players the mean income was $196,000 with a standard deviation of $315,000. a) Assuming all necessary assumptions are met, find a 95% confidence interval estimate of the mean salary of basketball players. b) What assumptions are necessary for the above estimate? Do they seem reasonable here? ANS: a) With df = 10 – 1 = 9 t* = 2.262 315000 Confidence interval = 196000 2.262 196000 ± 225000 10 b) We muse assume the sample is an SRS and that basketball salaries are normally distributed. This does not seem reasonable—the salaries are probably strongly skewed to the right by a few high ones. 23. A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reported does a quick check on 15 cigarettes representing a cross section of the market. a) What conclusion is reached if the sample mean is 5.63 milligrams of tar with a standard deviation of 1.61? Assume a 10% significance level. b) What is the conclusion at the 5% significance level? c) What is the P-value? ANS: a) We have H0: μ < 5 and Ha: μ > 5, α = 0.10, and with df = 15 – 1 = 14 t* = 1.345. 5.63 5 1.52 T-statistic = 1.61 15 Since 1.52 > 1.345, the industry spokesperson’s remarks should be rejected at the 10% significance level. b) α = 0.05, df = 14, t* = 1.761 Since 1.52 < 1.761, the remarks cannot be rejected at the 5% significance level. c) Since 1.52 is between 1.345 and 1.761, we can conclude that P-value is between 0.05 and 0.10. [On the TI-83, P = 0.0759] 24. A local chamber of commerce claims that the mean sale price for homes in the city is $90,000. A real estate salesperson notes eight recent sales of $75,000, $102,000, $80,000, $85,000, $79,000, $95,000, $98,000, and $62,000. How strong is the evidence to reject the chamber of commerce claim? ANS: H0: μ = 90000 and Ha: μ ≠ 90000 x-bar = 84,500 and s = 13,341.70. 84500 90000 1.17 t-statistic = 13341.7 8 With df = 8 – 1 = 7, we note that 1.17 is between 1.119 and 1.415, corresponding to 0.15 and 0.10, respectively. Doubling these values because this is a two-sided test, we see that the P-value is between 0.20 and 0.30. With such a high P, we conclude that there is no evidence to reject the chamber of commerce claim. [The TI-83 gives P = 0.2818] 25. A store manager wishes to determine whether there is a significant difference between two trucking firms with regard to the handling of egg cartons. In a simple random sample of 200 cartons on one firm’s truck there was an average of 0.7 broken eggs per carton with a standard deviation of 0.31, while a sample of 300 cartons on the second firm’s truck showed an average of 0.775 broken eggs per carton with a standard deviation of 0.42. Is the difference between the averages significant at a significance level of 5%? At a significance level of 1%? ANS: H0: μ1 = μ2 and Ha: μ1 ≠ μ2 (where μ1 = mean number of broken eggs per carton for the first firm and μ2 = mean number of broken eggs per carton for the second firm) 0.7 0.775 2.29 t-statistic = 0.312 0.42 2 200 300 Using Table C, with df = min {200 – 1, 300 – 1} = 199, we find that -2.29 is between 2.081 and 2.364, corresponding to 0.01 and 0.02. The test is two-sided, and so the P-value is between 0.02 and 0.04. So the observed difference is statistically significant at the 5% level but not at the 1% level. [On the TI-83, P = 0.0222.] 26. An automotive company executive claims that a mean of 48.3 cars per dealership are being sold each month. A major stockholder believes this claim is high and runs a test by sampling 30 dealerships. What conclusion is reached if the sample mean is 45.4 cars with a standard deviation of 15.4? (a) There is sufficient evidence to prove the executive’s claim is true. (b) There is sufficient evidence to prove the executive’s claim is false. (c) The stockholder has sufficient evidence to reject the executive’s claim. (d) The stockholder does not have sufficient evidence to reject the executive’s claim. (e) There is not sufficient data to reach any conclusion. ANS: D We have H0: μ = 48.3 and Ha: μ < 48.3. 45.4 48.3 1.03 t-statistic = 15.4 30 Using Table C, with df = 30 – 1 = 29, we find that 1.03 < 1.055, corresponding to 0.15. [On the T83, P = 0.1554.] With such a large P-value, there is little evidence against H0 and the stockholder should not reject the executive’s claim. 27. A pharmaceutical company claims that a medication will produce a desired effect for a mean time of 58.4 minutes. A government researcher runs a hypothesis test of 250 patients and calculates a mean of x-bar = 59.5 with a standard deviation of s = 8.3. In which of the following intervals is the P-value located? (a) P < 0.01 (b) 0.01 < P < 0.02 (c) 0.02 < P < 0.05 (d) 0.05 < P < 0.10 (e) P > 0.10 ANS: C We have H0: μ = 58.4 and Ha: μ ≠ 58.4 59.5 58.4 2.10 t-statistic = 8.3 250 Using Table C, with df = 250 – 1 = 249, we find the corresponding probability to be between 0.01 and 0.02. Doubling this value because the test is two-sided results in a P-value between 0.02 and 0.04. [On the TI-83, P = 0.0371.] 28. In a one-sided hypothesis test for the mean (>), in a random sample of size 10 the t-score of the sample mean is 2.79. Is this significant at the 5% level? At the 1% level? (a) Significant at the 1% level but not at the 5% level. (b) Significant at the 5% level but not at the 1% level. (c) Significant at both the 1% and 5% levels. (d) Significant at neither the 1% nor 5% level. (e) Cannot be determined from the given information. ANS: B With df = 10 – 1 = 9, the critical t-scores for the 0.05 and 0.01 tail probabilities are 1.833 and 2.821, respectively. We have 2.79 > 1.833, but 2.79 < 2.821, and so 2.79 is significant at the 5% level but not at the 1% level. 29. A recent study of health service costs for coronary angioplasty versus coronary artery bypass surgery at a London hospital showed an average cost of £6176 with a standard deviation of £329 for 231 angioplasties and an average cost of £8164 with a standard deviation of £264 for 221 bypass surgeries. Is this sufficient evidence to say that the average cost of angioplasty is less than the average cost of bypass surgery? (a) P < 0.001, so this is a very strong evidence that angioplasty costs less. (b) P is between 0.001 and 0.01, so this is strong evidence that angioplasty costs less. (c) P is between 0.01 and 0.05, so this is moderate evidence that angioplasty costs less. (d) P is between 0.05 and 0.10, so there is some evidence that angioplasty costs less. (e) P > 0.10, so there is little evidence that angioplasty costs less. ANS: A 6176 8164 71, and the P-value is 0.0000. This is very strong evidence that t-statistic = 2 2 329 264 231 221 angioplasty costs less than bypass surgery. 30. The 26 contestants in the pentathlon event of the 1992 Olympics ran the 200-meter dash in the following times (in seconds): 25.44, 24.39, 25.66, 23.93, 23.34, 25.01, 24.27, 24.54, 24.86, 23.95, 23.31, 24.60, 23.12, 25.29, 24.40, 25.24, 24.43, 23.70, 25.20, 25.09, 24.48, 26.13, 25.28, 24.18, and 23.83. In testing the null hypothesis that the mean time required for pentathlon contestants to run this race is 25 seconds against the alternative hypothesis that the mean time required is less than 25 seconds, in which of the following intervals is the P-value located? (a) P < 0.005 (b) 0.005 < P < 0.01 (c) 0.01 < P < 0.05 (d) 0.05 < P < 0.10 (e) 0.10 < P ANS: B We have H0: μ = 25 and Ha: μ < 25 x-bar = 24.581 and s = 0.7793 24.581 25 2.74 t-statistic = 0.7793 26 In looking at the df = 26 – 1 = 25 row in Table C, note that 2.74 is between 2.485 and 2.787, so P is between 0.005 and 0.01. [On the TI-83, P = 0.0056.]