Chapter 7. Estimates and Sample Sizes

Transcription

1 7- Estimating a Population Proportion Chapter 7 Estimates and Sample Sizes 1. The confidence level was not stated. The most common level of confidence is 95%, and sometimes that level is carelessly assumed without actually being stated.. The margin of error is the maximum likely difference between the point estimate for a parameter and its true value. 3. By including a statement of the maximum likely error, a confidence interval provides information about the accuracy of an estimate. 4. No. A voluntary response sample is not necessarily representative of the population. 5. For 99% confidence, α = = 0.01 and = 0.01/ = For the upper 0.005, A = and z =.575. z = z = For 99.5% confidence, α = = and = 0.005/ = For the upper 0.005, A = and z =.81. z = z = For α = 0.10, = 0.10/ = For the upper 0.05, A = and z = z = z 0.05 = For α = 0.0, = 0.0/ = For the upper 0.01, A = [0.9901] and z =.33. z = z 0.01 = et = the lower confidence limit; U = the upper confidence limit. ˆp = (+U)/ = ( )/ = 0.700/ = E = (U )/ = ( )/ = 0.300/ = The interval can be expressed as ± et = the lower confidence limit; U = the upper confidence limit. ˆp = (+U)/ = ( )/ = 1.500/ = E = (U )/ = ( )/ = 0.060/ = The interval can be expressed as ± et = the lower confidence limit; U = the upper confidence limit. ˆp = (+U)/ = ( )/ = 0.966/ = E = (U )/ = ( )/ = 0.09/ = The interval can be expressed as ± Given that ˆp = 0. and E = 0.044, = ˆp E = = U = ˆp +E = = 0.66 The interval can be expressed as < p < 0.66.

2 198 CHAPTE 7 Estimates and Sample Sizes 13. et = the lower confidence limit; U = the upper confidence limit. ˆp = (+U)/ = ( )/ = 0.740/ = E = (U )/ = ( )/ = 0.100/ = et = the lower confidence limit; U = the upper confidence limit. ˆp = (+U)/ = ( )/ = 1.548/ = E = (U )/ = ( )/ = 0.004/ = et = the lower confidence limit; U = the upper confidence limit. ˆp = (+U)/ = ( )/ = 0.960/ = E = (U )/ = ( )/ = 0.094/ = et = the lower confidence limit; U = the upper confidence limit. ˆp = (+U)/ = ( )/ = 0.338/ = E = (U )/ = ( )/ = 0.134/ = IMPOTANT NOTE: When calculating E = z ˆˆ do not round off in the middle of the problem. This, and the subsequent calculations of U = ˆp + E and = ˆp E may accomplished conveniently on most calculators having a memory as follows. (1) Calculate ˆp = x/n and STOE the value. () Calculate E as 1 ECA = * ECA = n = * z = (3) With the value for E showing on the display, the upper confidence limit U can be calculated by using + ECA =. (4) With the value for U showing on the display, the lower confidence limit can be calculated by using ECA ± + ECA. THE MANUA USES THIS POCEDUE, AND OUNDS THE FINA ANSWE TO 3 SIGNIFICANT DIGITS, EVEN THOUGH IT EPOTS INTEMEDIATE STEPS WITH A FINITE NUMBE OF DECIMA PACES. If the above procedure does not work on your calculator, or to find out if some other procedure would be more efficient on your calculator, ask your instructor for assistance. You must become familiar with your own calculator and be sure to do your homework on the same calculator you will use for the exams. 17. α = 0.05 and z = z 0.05 = 1.96; ˆp = x/n = 400/1000 = 0.40 E = z ˆˆ = 1.96 (0.40)(0.60)/1000 = α = 0.01 and z = z =.75; ˆp = x/n = 0/500 = 0.44 E = z ˆˆ =.575 (0.44)(0.56)/500 = α = 0.0 and z = z 0.01 =.33; ˆp = x/n = [49]/130 = 0.40 E = z ˆˆ =.33 (0.40)(0.60)/130 = NOTE: The value x=[49] was not given. In truth, any 486 x 498 rounds to the given ˆp = x/130 = 40%. For want of a more precise value, ˆp = 0.40 is used in the calculation of E. 0. α = 0.10 and z = z 0.05 = 1.645; ˆp = x/n = [63]/1780 = 0.35 E = z ˆˆ = (0.35)(0.65)/1780 = NOTE: The value x=[63] was not given. In truth, any 615 x 631 rounds to the given ˆp = x/1780 = 35%. For want of a more precise value, ˆp = 0.35 is used in the calculation of E.

3 1. α = 0.05 and z = z 0.05 = 1.96; ˆp = x/n = 40/00 = ± 1.96 (0.000)(0.8000)/ ± < p < α = 0.05 and z = z 0.05 = 1.96; ˆp = x/n = 400/000 = ± 1.96 (0.000)(0.8000)/ ± < p < α = 0.01 and z = z =.575; ˆp = x/n = 109/136 = ±.575 (0.088)(0.9118)/ ± < p < α = 0.01 and z = z =.575; ˆp = x/n = 481/500 = ±.575 (0.971)(0.079)/ ± < p < Estimating a Population Proportion SECTION α = 0.05, z = z 0.05 = 1.96 and E = 0.045; ˆp unknown, use ˆp =0.5 n = [(z = [(1.96) (0.5)(0.5)]/(0.045) = 474.7, rounded up to 475 ˆˆ ) pq]/e 6. α = 0.01, z = z =.575 and E = 0.005; ˆp unknown, use ˆp =0.5 n = [(z = [(.575) (0.5)(0.5)]/(0.005) = , rounded up to 66,307 ˆˆ ) pq]/e 7. α = 0.01, z = z =.575 and E = 0.0; ˆp estimated to be 0.14 n = [(z = [(.575) (0.14)(0.86)]/(0.0) = , rounded up to 1996 ˆˆ ) pq]/e 8. α = 0.05, z = z 0.05 = 1.96 and E = 0.03; ˆp estimated to be 0.87 n = [(z = [(1.96) (0.87)(0.13)]/(0.03) = 48.76, rounded up to 483 ˆˆ ) pq]/e 9. et x = the number of girls born using the method. a. ˆp = x/n = 55/574 = , rounded to 0.915

4 00 CHAPTE 7 Estimates and Sample Sizes b. α = 0.05, z = z 0.05 = ± 1.96 (0.9146)(0.0854)/ ± < p < c. Yes. Since 0.5 is not within the confidence interval, and below the interval, we can be 95% certain that the method is effective. 30. et x = the number of boys born using the method. a. ˆp = x/n = 17/15 = , rounded to b. α = 0.01, z = z = ±.575 (0.8355)(0.1645)/ ± < p < c. Yes. Since 0.5 is not within the confidence interval, and below the interval, we can be 99% certain that the method is effective. 31. et x = the number of deaths in the week before Thanksgiving. a. ˆp = x/n = 606/1000 = 0.505, rounded to b. α = 0.05, z = z 0.05 = ± 1.96 (0.505)(0.4948)/ ± < p < c. No. Since 0.5 is within the confidence interval, there is no evidence that people can temporarily postpone their death in such circumstances. 3. et x = the number of suits dropped or dismissed. a. ˆp = x/n = 856/18 = , rounded to b. α = 0.01, z = z = ±.575 (0.6971)(0.309)/ ± < p < c. Yes. Since 0.5 is not within the confidence interval, and below the interval, we can be 99% certain that more than half the suits are dropped or dismissed. 33. et x = the number of yellow peas a. α = 0.05, z = z 0.05 = 1.96 and ˆp = x/n = 15/(48+15) = 15/580 = ± 1.96 (0.61)(0.7379)/ ± < p < 0.98 or.6% < p < 9.8% b. No. Since 0.5 is within the confidence interval, it is a reasonable possibility for the true population proportion. The results do not contradict the theory.

5 Estimating a Population Proportion SECTION et x = the number who say they voted. a. α = 0.01, z = z =.575 and ˆp = x/n = 701/100 = ±.575 (0.6996)(0.3004)/ ± < p < b. No. Since 0.61 is not within the confidence interval, the results are not consistent with the actual voter turnout. It appears that people do not tell the truth about whether they voted. 35. et x = the number that develop those types of cancer. a. α = 0.05, z = z 0.05 = 1.96 and ˆp = x/n = 135/40095 = ± 1.96 ( )( )/ ± < p < or 0.067% < p < % b. No. Since % = is within the confidence interval, it is a reasonable possibility for the true population value. The results do not provide evidence that cell phone users have a different cancer rate than the general population. 36. et x = the number who think global warming demands priority attention. a. ˆp = x/n = 939/1708 = , rounded to or 55.0% b. α = 0.01, z = z = ±.575 (0.5498)(0.450)/ ± < p < c. Yes. Since 0.5 is not within the confidence interval, and below the interval, we can be 99% certain that more than half the population thinks global warming demands such attention. 37. et x = the number who say they use the Internet. α = 0.05, z = z 0.05 = 1.96 and ˆp = x/n = [198]/3011 = 0.73 NOTE: The value x=[198] was not given. In truth, any 183 x 13 rounds to the given ˆp = x/3011 = 73%. For want of a more precise value, ˆp = 0.73 is used in the calculations. Technically, this should limit the exercise to two significant digit accuracy 0.73 ± 1.96 (0.73)(0.7)/ ± < p < No. Since 0.75 is not within the confidence interval, it is not likely to be the correct value of the population proportion and should not be reported as such. In this particular exercise, however, the above NOTE indicates that the third significant digit in the confidence interval endpoints is not reliable and if ˆp is really 13/3011 = , for example, the confidence interval is < p < and 75% is acceptable.

6 0 CHAPTE 7 Estimates and Sample Sizes 38. et x = the number who display little or no knowledge of the company. α = 0.01, z = z =.575 and ˆp = x/n = [71]/150 = 0.47 NOTE: The value x=[71] was not given. In truth, any 70 x 71 rounds to the given ˆp = x/150 = 47%. For want of a more precise value, ˆp = 0.47 is used in the calculations. Technically, this should limit the exercise to two significant digit accuracy ±.575 (0.47)(0.53)/ ± < p < Yes. Since 0.50 is within the confidence interval, it is a likely value for the true population proportion. 39. et x = the number who indicate the outbreak would deter them from taking a cruise. α = 0.05, z = z 0.05 = 1.96 and ˆp = x/n = [130]/34358 = 0.6 NOTE: The value x=[130] was not given. In truth, any 1131 x 1473 rounds to the given ˆp = x/34358 = 6%. For want of a more precise value, ˆp = 0.6 is used in the calculations. Technically, this should limit the exercise to two significant digit accuracy. 0.6 ± 1.96 (0.6)(0.38)/ ± < p < 0.65 No. Since the sample is a voluntary response sample, the respondents are not likely to be representative of the population. 40. et x = the number who correctly identified which hand Emily selected. a. If the touch therapists made random guesses, one would expect the proportion of correct responses to be 0.50 regardless of how Emily chose which hand to use even if, for example, she always used the right hand. b. ˆp = x/n = 13/80 = , rounded to c. α = 0.01, z = z = ±.575 (0.4393)(0.5607)/ ± < p < d. Since the confidence interval includes 0.50, the therapists success rate is consistent with chance guessing. There is no evidence that the professional touch therapists have any special ability in this area. 41. α = 0.01, z = z =.575 and E = 0.0 a. ˆp unknown, use ˆp =0.5 n = [(z ) pq]/e = [(.575) (0.5)(0.5)]/(0.0) = , rounded up to 4145 b. ˆp estimated to be 0.73 ˆˆ n = [(z ˆˆ ) pq]/e = [(.575) (0.73)(0.7)]/(0.0) = 367.4, rounded up to 368

7 4. α = 0.10, z = z 0.05 = and E = 0.04 a. ˆp unknown, use ˆp =0.5 ˆˆ Estimating a Population Proportion SECTION 7-03 n = [(z ) pq]/e = [(1.645) (0.5)(0.5)]/(0.04) = 4.8, rounded up to 43 b. ˆp estimated to be 0.08 n = [(z ˆˆ ) pq]/e = [(1.645) (0.08)(0.9)]/(0.04) = 14.48, rounded up to α = 0.05, z = z 0.05 = 1.96 and E = 0.03; ˆp unknown, use ˆp =0.5 n = [(z ˆˆ ) pq]/e = [(1.96) (0.5)(0.5)]/(0.03) = , rounded up to α = 0.05, z = z 0.05 = 1.96 and E = 0.0; ˆp estimated to be 0.10 n = [(z ˆˆ ) pq]/e = [(1.96) (0.10)(0.90)]/(0.0) = , rounded up to et x = the number of green M&M s. α = 0.05, z = z 0.05 = 1.96 and ˆp = x/n = 19/100 = ± 1.96 (0.19)(0.81)/ ± < p < 0.67 or 11.3% < p < 6.7% Yes. Since is within the confidence interval, this result is consistent with the claim that the true population proportion is 16%. 46. et x = the number students who gained weight during their freshman year. Of the 67 students, 17 lost weight and 5 stayed the same and 45 gained weight. a. ˆp = x/n = 45/67 = , rounded to 0.67 or 67.% b. α = 0.05, z = z 0.05 = ± 1.96 (0.6716)(0.384)/ ± < p < or 55.9% < p < 78.4% c. It is estimated that 67% of U.S. college students gain weight during their freshman year. This result comes from a study of 67 men and women published in the Journal of American College Health, Vol. 55, No, 1. That estimate has an 11% margin of error with a 95% confidence level. In other words, 95% of all such studies can be expected to produce estimates that are within 11% of the true proportion of all college students who gain weight during their freshman year. 47. et x = the number of days with precipitation. α = 0.05, z = z 0.05 = 1.96 Wednesdays: ˆp = x/n = 16/53 = Sundays: ˆp = x/n = 15/5 = ± 1.96 (0.3019)(0.6981)/ ± 1.96 (0.885)(0.7115)/ ± ± < p < < p < 0.41 The confidence intervals are similar. It does not appear to rain more on either day.

8 04 CHAPTE 7 Estimates and Sample Sizes 48. et x = the number of movies with ratings. α = 0.05, z = z 0.05 = 1.96 and ˆp = x/n = 1/35 = ± 1.96 (0.349)(0.6571)/ ± < p < No. To more significant digits, the upper limit of the confidence limit is Since the confidence interval includes values higher than 0.50, it is not an unreasonable possibility that the proportion of movies rated is greater than ½. We cannot conclude that most movies have a rating different from. 49. α = 0.05, z = z 0.05 = 1.96 and E = 0.03; ˆp unknown, use ˆp =0.5 Npq[z ˆˆ ] n = pq[z ˆˆ ] + (N-1)E (1784)(0.5)(0.5)[1.96] = = = , rounded up to 985 (0.5)(0.5)[1.96] + (1783)(0.03) No. The sample size is not too much lower than the n=1068 required for a population of millions of people. 50. α = 0.05, z α = z 0.05 = and ˆp = x/n = 630/750 = p ˆ - z ˆˆ α (0.8400)(0.1600)/ < p The interval is expressed as p > The desired figure is 81.8%. 51. α = 0.05, z = z 0.05 = 1.96 and ˆp = x/n = 3/8 = ± 1.96 (0.3750)(0.650)/ ± < p < Yes. The results are reasonably close being shifted down 4.5% from the correct interval < p < But depending on the context, such an error could be serious. 5. α = 0.01, z = z =.575 and ˆp = x/n = 95/100 = ±.575 (0.9500)(0.0500)/ ± < p < This interval is noteworthy because the upper limit is greater than 1, the maximum possible value for p in any problem. This occurs because the normal distribution used is only an approximation to the binomial. In this case the approximation is barely appropriate since nq 100(0.05) = 5, the minimum acceptable value to use the normal to approximate the binomial. In such cases the interval should be reported as 0.894<p<1. NOTE: Do not use 0.894<p 1, because the presence of 5 tails indicates that p=1 is not true.

9 Estimating a Population Proportion SECTION a. If ˆp = x/n = 0/n = 0, then (1) np 0 < 5, and the normal approximation to the binomial does not apply. () E = z ˆˆ = 0, and there is no meaningful interval. b. Since ˆp = x/n = 0/0 = 0, use 3/n = 3/0 = 0.15 as the 95% upper bound for p. NOTE: The corresponding interval would be 0 p<0.15. Do not use 0<p<0.15, because the failure to observe any successes in the sample does not rule out p=0 as the true population proportion. 54. Since 19 cases out of 0 implies 19/0 = 0.95 = 95% confidence, use α = Since ˆp is unknown, use ˆp = 0.5. n = [(z = [(1.96) (0.5)(0.5)]/(0.01) = 9604 ˆˆ ) pq]/e 7-3 Estimating a Population Mean: σ Known 1. A point estimate is a single value used to estimate a population parameter. If the parameter in question is the mean of a population, the best point estimate is the mean of a random sample from that population.. No. The list of the employees at her facility from which she obtained her simple random sample is itself a convenience sample. Those employees are likely not representative of the population by age, gender, ethnicity, or other factors that may affect leg length. 3. It is estimated that the mean height of U.S. women is inches. This result comes from the Third National Health and Nutrition Examination Survey of the U.S. Department of Health and Human Services. It is based on an in-depth study of 40 women and assumes a population standard deviation of.5 inches. The estimate has a margin of error of inches with a 95% level of confidence. In other words, 95% of all such studies can be expected to produce estimates that are within inches of the true population mean height of all U.S. women. 4. While any one particular estimate for a population parameter may not be correct, a statistic is an unbiased estimator of a population parameter if its long-run expected value (i.e., its mean value) is equal to the true value of the parameter being estimated. 5. For 90% confidence, α = = 0.10 and = 0.10/ = For the upper 0.05, A = and z = z = z 0.05 = For 98% confidence, α = = 0.0 and = 0.0/ = For the upper 0.01, A = [0.9901] and z =.33. z = z 0.01 = For α = 0.0, = 0.0/ = For the upper 0.10, A = and z = 1.8. z = z 0.10 = For α = 0.04, = 0.0/ = 0.0. For the upper 0.0, A = [0.9798] and z =.05. z = z 0.0 =.05

10 06 CHAPTE 7 Estimates and Sample Sizes 9. Since σ is known and n>30, the methods of this section may be used. α = 0.05, z = z 0.05 = 1.96 E = zσ/ n = 1.96(68)/ 50 = 18.8 FICO units x ± E ± < μ < (FICO units) NOTE: The above interval assumes x = Technically, the failure to report x to tenths limits the endpoints of the confidence interval to whole number accuracy. 10. Since σ is known and n>30, the methods of this section may be used. α = 0.05, z = z 0.05 = 1.96 E = zσ/ n = 1.96(7)/ 3 =.4 feet x ± E ± < μ < (feet) NOTE: The above interval assumes x = Technically, the failure to report x to tenths limits the endpoints of the confidence interval to whole number accuracy. 11. Since n<30 and the population is far from normal, the methods of this section may not be used. 1. Since n<30 and the population is far from normal, the methods of this section may not be used. 13. α = 0.05, z = z 0.05 = 1.96 n = [z σ/e] = [(1.96)(68)/(3)] = , rounded up to α = 0.01, z = z =.575 n = [z σ/e] = [(.575)(7)/()] = 81.3, rounded up to α = 0.01, z = z =.575 n = [z σ/e] = [(.575)(0.1)/(0.010)] = , rounded up to α = 0.05, z = z 0.05 = 1.96 n = [z σ/e] = [(1.96)(18.6)/()] = 33.6, rounded up to x = 1.1 mg < μ <.387 (mg) 19. E = (U )/ = ( )/ = ± 1.67 (mg) 0. We are 95% confident that the interval from mg to.387 mg contains the true mean amount of tar in all king-size, non-filtered, non-menthol, and non-light cigarettes. 1. a. x = 146. lbs b. α = 0.05, z = z 0.05 = 1.96 x ± z σ / n 146. ± 1.96(30.86)/ ± < μ < (lbs)

11 Estimating a Population Mean: σ Known SECTION a. x = $415,953 b. α = 0.05, z = z 0.05 = 1.96 x ± z σ / n 415,953 ± 1.96(463,364)/ ,953 ± 143,598 7,355 < μ < 559,551 (dollars) c. Yes. In this case the confidence interval includes the true population mean. 3. a. x = 58.3 seconds b. α = 0.05, z = z 0.05 = 1.96 x ± z σ / n 58.3 ± 1.96(9.5)/ ± < μ < 61. (seconds) c. Yes. Since the confidence interval contains 60 seconds, it is reasonable to assume that the sample mean was reasonably close to 60 seconds and it was, in fact, 58.3 seconds. 4. a. x = 4.63 cells/microliter b. α = 0.01, z = z =.575 x ± z σ / n 4.63 ±.575(0.54)/ ± < μ < 4.83 (cells/microliter) c. The intervals are not directly comparable, since the two given in part (c) are normal ranges for individual counts and the one calculated in part (b) is a confidence interval for mean counts. One would expect the confidence interval for mean counts to be well within the normal ranges for individual counts. The fact that the point estimate and the lower confidence interval limit for the mean are so close to the lower limit of the normal ranges for individuals suggests that the sample may consist of persons with lower red blood cell counts. 5. a. α = 0.05, z = z 0.05 = 1.96 b. α = 0.01, z = z =.575 x ± z σ / n x ± z σ / n 15 ± 1.96(333)/ ±.575(333)/ ± ± < μ < < μ < 1599 c. The 99% confidence interval in part (b) is wider than the 95% confidence interval in part (a). For an interval to have more confidence associated with it, it must be wider to allow for more possibilities. 6. a. α = 0.05, z = z 0.05 = 1.96 b. α = 0.05, z = z 0.05 = 1.96 x ± z σ / n x ± z σ / n 3433 ± 1.96(495)/ ± 1.96(495)/ ± ± < μ < 3545 (grams) 349 < μ < 3437 (grams) c. The n=75 confidence interval in part (a) is wider than the n=75,000 confidence interval in part (b). There is less accuracy associated with smaller samples.

12 08 CHAPTE 7 Estimates and Sample Sizes 7. summary statistics: n = 14 Σx = 1875 x = α = 0.05, z = z 0.05 = 1.96 x ± z σ / n ± 1.96(10)/ ± < μ < 139. (mmhg) Ideally, there is a sense in which all the measurements should be the same and in that case there would be no need for a confidence interval. It is unclear what the given σ = 10 represents in this situation. Is it the true standard deviation in the values of all people in the population (in which case it would not be appropriate in this context where only a single person is involved)? Is it the true standard deviation in momentary readings on a single person (due to constant biological fluctuations)? Is it the true standard deviation in readings from evaluator to evaluator (when they are supposedly evaluating the same thing)? Using the methods of this section and assuming σ = 10, the confidence interval would be 18.7 < μ < 139. as given above even if all the readings were the same. 8. a. summary statistics: n = 10 Σx = 39 x = 3.9 α = 0.05, z = z 0.05 = 1.96 x ± z σ / n 3.9 ± 1.96(.87)/ ± < μ < 5.7 b. No; since n<30 and the population distribution is not normal, the methods of this section do not apply. No; since the methods of this section do not apply, the confidence interval does not provide a good estimate. Even though the confidence interval may include the true mean, the endpoints of the confidence limits do not carry the supposed level of confidence. 9. summary statistics: n = 35 Σx = 4305 x = α = 0.05, z = z 0.05 = 1.96 x ± z σ / n ± 1.96(100)/ ± < μ < (million dollars) 30. summary statistics: n = 100 Σx = x = α = 0.01, z = z =.575 x ± z σ / n ±.575(9.)/ ± < μ < 76.9 (FICO units) 31. α = 0.05, z = z 0.05 = 1.96 n = [z σ/e] = [(1.96)(15)/(5)] = 34.57, rounded up to α = 0.01, z = z =.575 n = [z σ/e] = [(.575)(.5)/(0.)] = , rounded up to α = 0.05, z = z 0.05 = 1.96 n = [z σ/e] = [(1.96)(10.6)/(0.5)] = , rounded up to 6907 The sample size is too large to be practical.

13 Estimating a Population Mean: σ Known SECTION α = 0.10, z = z 0.05 = n = [z σ/e] = [(1.645)(0.88)/(0.1)] = 09.55, rounded up to α = 0.05, z = z 0.05 = 1.96 Using the range rule of thumb: = 40,000 0 = 40,000, and σ /4 = 40,000/4 = 10,000. n = [z σ/e] = [(1.96)(10,000)/(100)] = 38416, rounded up to α = 0.05, z = z 0.05 = 1.96 a. Using the range rule of thumb: = = 40, and σ /4 = 40/4 = 10. n = [z σ/e] = [(1.96)(10)/()] = 96.04, rounded up to 97 b. Using the sample standard deviation: σ s = n = [z σ/e] = [(1.96)(11.97)/()] = 1.56, rounded up to 13 c. The two values are relatively close. Since s (which considers all the data) is a better estimator for σ than /4 (which is based entirely on the extreme values), the sample size of 13 should be preferred. 37. Since n/n = 15/00 = 0.65 > 0.05, use the finite population correction factor. α = 0.05, z = z 0.05 = 1.96 x ± [zσ/ n ] (N-n)/(N-1) 15 ± [1.96(333)/ 15] (00-15)/(00-1) 15 ± [ ] [0.6139] 15 ± < μ < 1558 The confidence interval becomes narrower because the sample is a larger portion of the population. As n approaches N, the length of the confidence interval shrinks to 0 because when n=n the true mean μ can be determined with certainty. 38. From Exercise 3: α = 0.01, z = z =.575 and σ =.5 and E = 0.. Nσ (z ) 500(.5) (.575) n = = = =337.48, rounded up to 338 (N-1)E + σ (z ) (500-1)(0.) + (.5) (.575) Yes; the information about the population size has a significant effect, dropping the required sample size from 1037 to Estimating a Population Mean: σ Not Known 1. According to the point estimate ( average ), the parameter of interest is a population mean. But according to the margin of error ( percentage points ), the parameter of interest is a population proportion. It is possible that the margin of error the paper intended to communicate was 1% of $483 (or $4.83, which in a 95% confidence interval would correspond to a sample standard deviation of $6.57) but the proper units for the margin of error in a situation like this are dollars and not percentage points.. obust against departures from normality mans that that the requirement that the original population be approximately normal is not a strong requirement, and that the methods of this section still give good results if the departure from normality is not too extreme. The methods of this section are not robust against poor sampling methods, as poor sampling methods can yield data that are entirely useless.

14 10 CHAPTE 7 Estimates and Sample Sizes 3. No; the estimate will not be good for at least two reasons. First, the sample is a convenience sample using the state of California, and California residents may not be representative of then entire country. Secondly, any survey that involves self-reporting (especially of financial information) is suspect because people tend to report favorable rather than accurate data. 4. The degrees of freedom in this survey is 4. In general, the degrees of freedom in a problem is the number of data values that are free to vary without changing the estimate of the parameter of interest. The estimate of a mean is determined by Σx, and n-1 of the data values are free to vary so long as the n th value is the one necessary to produce the required Σx. 5. σ unknown, normal population, n=3: use t with df = α = 0.05, t df, = t,0.05 =.074 IMPOTANT NOTE: This manual uses the following conventions. (1) The designation df stands for degrees of freedom. () Since the t value depends on the degrees of freedom, a subscript may be used to clarify which t distribution is being used. For df =15 and =0.05, for example, one may indicate t 15, =.13. As with the z distribution, it is also acceptable to use the actual numerical value within the subscript and indicate t 15,.05 =.13. (3) Always use the closest entry in Table A-3. When the desired df is exactly halfway between the two nearest tabled values, be conservative and choose the one with the lower df. (4) As the degrees of freedom increase, the t distribution approaches the standard normal distribution and the large row of the t table actually gives z values. Consequently the z score for certain popular α and values may be found by reading Table A-3 frontwards instead of Table A- backwards. This is not only easier but also more accurate since Table A-3 includes one more decimal place. Note the following examples. For large df and = 0.05, t = = z (as found in the z table). For large df and = 0.01, t =.36 = z (more accurate than the.33 in the z table). This manual uses this technique from this point on. [For df = large and = 0.005, t = = z (as found in the z table). This is a discrepancy caused by using different mathematical approximation techniques to construct the tables, and not a true difference. While.576 is the more standard value, his manual will continue to use.575.] 6. σ known, normal population: use z α = 0.01, z = z = σ unknown, population not normal, n=6: neither normal nor t applies 8. σ unknown, population not normal, n=40: use t with df =39 α = 0.05, t df, = t 39,0.05 = σ known, population not normal, n=00: use z α = 0.10, z = z 0.05 = σ unknown, population not normal, n=9: neither normal nor t applies 11. σ unknown, population normal, n=1: use t with df = 11 α = 0.01, t df, = t 11,0.005 = σ unknown, population not normal, n=38: use t with df =37 α = 0.05, t df, = t 39,0.05 =.06

15 Estimating a Population Mean: σ Not Known SECTION σ unknown, normal distribution: use t with df = 19 α = 0.05, t df, = t 19,0.05 =.093 a. E = t s/ n b. x ± E =.093(569)/ ± 66 = 66 dollars 8738 < μ < 970 (dollars) 14. σ unknown, normal distribution: use t with df = 6 α = 0.01, t df, = t 6,0.005 = a. E = t s/ n b. x ± E = 3.707(0.04)/ ± 0.06 = 0.06 grams/mile 0.06 < μ < 0.18 (grams/mile) 15. From the SPSS display: < μ< (grams) There is 95% confidence that the interval from grams to grams contains the true mean weight of all U.S. dollar coins in circulations. 16. From the TI-83/84 Plus display: < μ <.706 (lbs) There is 99% confidence that the interval from lbs to.706 lbs contains the true mean annual weight of the plastic discarded by U.S. households. 17. a. x = 3. mg/d b. σ unknown, n > 30: use t with df=46 [45] α = 0.05, t df, = t 46,0.05 =.014 x ± t s/ n 3. ±.014(18.6)/ ± < μ < 8.7 (mg/dl) Since the confidence interval includes 0, there is a reasonable possibility that the true value is zero i.e., that the Garlicin treatment has no effect on D cholesterol levels. 18. a. x = 3103 grams b. σ unknown, n > 30: use t with df=185 [00] α = 0.05, t df, = t 185,0.05 = 1.97 x ± t s/ n 3103 ± 1.97(696)/ ± < μ < 304 (grams) c. Yes. Since the confidence interval for the mean birth weight for mothers who used cocaine is entirely below the confidence interval in part (b), it appears that cocaine use is associated with lower birth rates. 19. a. x = 98.0 F b. σ unknown, n > 30: use t with df=105 [100] α = 0.01, t df, = t 105,0.005 =.66 x ± t s/ n 98.0 ±.66(0.6)/ ± < μ < ( F) c. No, the confidence interval does not contain the value 98.6 F. This suggests that the common belief that 98.6 F is the normal body temperature may not be correct.

16 1 CHAPTE 7 Estimates and Sample Sizes 0. a. x =.1 lbs b. σ unknown, n > 30: use t with df=39 α = 0.01, t df, = t 39,0.005 =.708 x ± t s/ n.1 ±.708(4.8)/ 40.1 ±.1 0 < μ < 4. (lbs) c. Yes; since the confidence interval does not include 0, the diet appears to be effective. No; since the amount of weight loss is so small, the diet does not appear to be practical. 1. a. σ unknown, n > 30: use t with df=336 [300] b. σ unknown, n > 30: use t with df=369 [400] α = 0.05, t df, = t 336,0.05 = α = 0.05, t df, = t 369,0.05 = x ± t s / n x ± t s / n 6.0 ± 1.968(.3)/ ± 1.966(.4)/ ± ± < μ < 6. (days) 5.9 < μ < 6.3 (days) c. The two confidence intervals are very similar and overlap considerably. There is no evidence that the echinacea treatment is effective.. a. σ unknown, n > 30: use t with df=141 [100] b. σ unknown, n > 30: use t with df=79 [80] α = 0.05, t df, = t 141,0.05 = α = 0.05, t df, = t 79,0.05 = x ± t s / n x ± t s / n 1.8 ± 1.984(1.4)/ ± 1.990(1.)/ ± ± < μ <.0 (headaches) 1.3 < μ < 1.9 (headaches) c. The two confidence intervals are very similar and overlap considerably. There is no evidence that the acupuncture treatment is effective. 3. a. σ unknown, n 30: if approximately b. σ unknown, n 30: if approximately normal distribution, use t with df=19 normal distribution, use t with df=19 α = 0.05, t df, = t 19,0.05 =.093 α = 0.05, t df, = t 99,0.05 =.093 x ± t s / n x ± t s / n 5.0 ±.093(.4)/ ±.093(.9)/ ± ± < μ < 6.1 (VAS units) 3.4 < μ < 6.1 (VAS units) c. The two confidence intervals are very similar and overlap considerably. There is no evidence that the magnet treatment is effective. 4. a. σ unknown, n > 30: use t with df=78 [80] b. σ unknown, n > 30: use t with df=78 [80] α = 0.01, t df, = t 78,0.005 =.639 α = 0.01, t df, = t 78,0.005 =.639 x ± t s / n x ± t s / n 35.8 ±.639(11.3)/ ±.639(8.9)/ ± ± < μ < 39. (years) 41. < μ < 46.4 (years) c. The two confidence intervals differ considerably and do not overlap at all. Women Oscar winners are considerably younger than their male counterparts. Either women and men reach their peak acting ability at different years, or the standards for judging women and men are not really the same.

17 Estimating a Population Mean: σ Not Known SECTION preliminary values: n = 6, Σx = 9.3, Σx = x = (Σx)/n = (9.3)/6 = s = [n(σx ) (Σx) ]/[n(n-1)] = [6(3.5197) (9.3) ]/[6(5)] = s = σ unknown (and assuming the distribution is approximately normal), use t with df=5 α = 0.05, t df, = t 5,0.05 =.571 x ± t s/ n ±.571(1.914)/ ± < μ < [which should be adjusted, since negative values are not possible] 0 < μ < (micrograms/cubic meter) Yes. The fact that 5 of the 6 sample values are below x raises a question about whether the data meet the requirement that the underlying distribution is normal. 6. preliminary values: n = 7, Σx = 0.85, Σx = x = (Σx)/n = (0.85)/7 = 0.11 s = [n(σx ) (Σx) ]/[n(n-1)] = [7(0.113) (0.85) ]/[7(6)] = s = σ unknown (and assuming the distribution is approximately normal), use t with df=6 α = 0.0, t df, = t 6,0.01 = x ± t s/ n 0.11 ± 3.143(0.0389)/ ± < μ < (grams/mile) No. Since the confidence interval includes values greater than 0.165, there is a reasonable possibility that the true mean emission amount is greater than that. NOTE: This is a two-sided 98% confidence interval, and the requirement is one-sided (i.e., that μ < 0.165). This means that the level of significance associated with the interval may not be the same level of significance associated with a conclusion about the requirement. 7. preliminary values: n = 10, Σx = 04.0, Σx = x = (Σx)/n = (04.0)/10 = 0.40 s = [n(σx ) (Σx) ]/[n(n-1)] = [10(5494.7) (04.0) ]/[10(9)] = s = a. σ unknown (and assuming the distribution is approximately normal), use t with df=9 α = 0.05, t df, = t 9,0.05 =.6 x ± t s/ n 0.40 ±.6(1.171)/ ± < μ < 9.1 (million dollars) b. No. Since the data are the top 10 salaries, they are not a random sample. c. There is a sense in which the data are the population (i.e., the top ten salaries) and are not a sample of any population. Possible populations from which the data could be considered a sample (but not a representative sample appropriate for any statistical inference) would be the salaries of all TV personalities, the salaries of the top 10 salaries of TV personality for different years. d. No. Since no population can be identified from which these data are a random sample, the confidence interval has no context and makes no sense.

18 14 CHAPTE 7 Estimates and Sample Sizes 8. preliminary values: n = 1, Σx = 1461, Σx = 18,435 x = (Σx)/n = (1461)/1 = s = [n(σx ) (Σx) ]/[n(n-1)] = [1(18435) (1461) ]/[1(11)] = s = a. σ unknown (and assuming the distribution is approximately normal), use t with df=11 α = 0.01, t df, = t 11,0.005 = x ± t s/ n ± 3.106(0.356)/ ± < μ < (minutes) b. While it is tempting to add 30 minutes to the upper confidence interval limit associated with the mean times, it is not appropriate to make a decision about individual times based on the distribution of the means. Without knowing the distribution of the lengths of the individual films, and without assuming they are normally distributed, it is still possible to give the manager some guidance. Defining an outlier to be any values more than two standard deviations from the mean the usual maximum and minimum film lengths are: usual min: (0.356) = 81.0 minutes usual max: (0.356) = 16.5 minutes If the manager allowed = 19.5 minutes between showings, he would accommodate all but the unusually long films. In practice, in order to use round numbers and err slightly on the conservative side, he should consider a regular schedule of 195 minutes (i.e., 3 hours and 15 minutes) between feature showings. 9. preliminary values: n = 1, Σx = 5118, Σx = 8,07,688 x = (Σx)/n = (5118)/1 = s = [n(σx ) (Σx) ]/[n(n-1)] = [1(807688) (5118) ]/[1(11)] = s = σ unknown (and assuming the distribution is approximately normal), use t with df=11 α = 0.05, t df, = t 1,0.05 =.01 x ± t s/ n ±.01(394.91)/ ± < μ < (seconds) 30. preliminary values: n = 43, Σx = 358, Σx = 130,930 x = (Σx)/n = (358)/43 = s = [n(σx ) (Σx) ]/[n(n-1)] = [43(130930) (358) ]/[43(4)] = s = 6.18 σ unknown and n>30, use t with df=4 [40] α = 0.01, t df, = t 4,0.005 =.704 x ± t s/ n ±.704(6.18)/ ± < μ < 57.4 (years) There is a sense in which the data are the population (i.e., the ages at inauguration of all US Presidents) and are not a sample of any population. Possible populations from which the data could be considered a sample (but not a representative sample appropriate for any statistical inference) would be the ages of all US adults, the ages upon taking office of word heads of states, the ages at inauguration of all past-present-future US presidents.

19 Estimating a Population Mean: σ Not Known SECTION a. preliminary values: n = 5, Σx = 31.4, Σx = x = (Σx)/n = (31.4)/5 = 1.56 s = [n(σx ) (Σx) ]/[n(n-1)] = [5(40.74) (31.4) ]/[5(4)] = 3.54/600 = s = 0.39 σ unknown (and assuming the distribution is approximately normal), use t with df=4 α = 0.05, t df, = t 4,0.05 =.064 x ± t s/ n 1.56 ±.064(0.39)/ ± < μ < 1.35 (mg) NOTE: The Minitab output for this exercise is given below. Variable N Mean StDev SE Mean 95% CI nicotine ( , ) b. preliminary values: n = 5, Σx =.9, Σx =.45 x = (Σx)/n = (.9)/5 = s = [n(σx ) (Σx) ]/[n(n-1)] = [5(.45) (.9) ]/[5(4)] = 36.84/600 = s = σ unknown (and assuming the distribution is approximately normal), use t with df=4 α = 0.05, t df, = t 4,0.05 =.064 x ± t s/ n ±.064(0.478)/ ± < μ < 1.0 (mg) NOTE: The Minitab output for this exercise is given below. Variable N Mean StDev SE Mean 95% CI nicotine ( , ) c. There is no overlap in the confidence intervals. Yes; since the CI for the filtered cigarettes is completely below the CI for the unfiltered cigarettes, the filters appear to be effective in reducing the amounts of nicotine. 3. a. preliminary values: n = 40, Σx = 776, Σx = x = (Σx)/n = (776)/40 = 69.4 s = [n(σx ) (Σx) ]/[n(n-1)] = [40(19763) (776) ]/[40(39)] = s = σ unknown and n>30, use t with df=39 α = 0.05, t df, = t 39,0.05 =.04 x ± t s/ n 69.4 ±.04(11.97)/ ± < μ < 73.0 (beats/min) NOTE: The Minitab output for this exercise is given below. Variable N Mean StDev SE Mean 95% CI PUSE ( , ) b. preliminary values: n = 40, Σx = 305, Σx = x = (Σx)/n = (305)/40 = 76.3 s = [n(σx ) (Σx) ]/[n(n-1)] = [40(38960) (305) ]/[40(39)] = s = 1.499

20 16 CHAPTE 7 Estimates and Sample Sizes σ unknown and n>30, use t with df=39 α = 0.05, t df, = t 39,0.05 =.04 x ± t s/ n 76.3 ±.04(1.499)/ ± < μ < 80.3 (beats/min) NOTE: The Minitab output for this exercise is given below. Variable N Mean StDev SE Mean 95% CI PUSE (7.307, ) c. Since the two confidence intervals overlap, we cannot conclude that the two population means are different. But recall the CAUTION in this section that the overlapping of confidence intervals should not be used for making formal and final conclusions about equality of means. 33. preliminary values: n = 43, Σx = 738, Σx = 307,50 x = (Σx)/n = (738)/43 = s = [n(σx ) (Σx) ]/[n(n-1)] = [43(30750) (738) ]/[43(4)] = s = σ unknown and n>30, use t with df=4 [40] α = 0.01, t df, = t 4,0.005 =.704 x ± t s/ n ±.704(56.54)/ ± < μ < 86.9 (years) Yes, the confidence interval changes considerably from the previous 5.3 < μ < Yes, apparently confidence interval limits can be very sensitive to outliers. When apparent outliers are discovered in data sets they should be carefully examined to determine if an error has been made. If an error has been made that cannot be corrected, the value should be discarded. If the value appears to be valid, it may be informative to construct confidence intervals with and without the outlier. 34. preliminary values: n = 43, Σx = 358, Σx = 130,930 x = (Σx)/n = (358)/43 = s = [n(σx ) (Σx) ]/[n(n-1)] = [43(130930) (358) ]/[43(4)] = s = 6.18 σ unknown and n>30, alternate method says use s for σ and use z α = 0.01, z = z =.575 x ± z σ/ n ±.575(6.18)/ ± < μ < 57.3 (years) For any α, the z value is smaller than the corresponding t value although the difference decreases as n increases. This creates a smaller E and a narrower confidence interval than one is entitled to i.e., it does not take into consideration the extra uncertainty created by using the sample s instead of the population σ. And so the confidence interval found by the alternative method will always be narrower, but usually by a very small amount. In some situations, however, the unjustified narrowness of the interval could lead to incorrect conclusions.

21 Estimating a Population Mean: σ Not Known SECTION assuming a large population using the finite population N = 465 α = 0.05 & df=99 [100], t df, = t 99,0.0 5 = α = 0.05 & df=99 [100], t df, = t 99,0.05 = E = t s/ n E = [t s/ n ] (N-n)/(N-1) = 1.984(0.0518)/ 100 = [1.984(0.0518)/ 100 ] 365/464 = g = g x ± E x ± E ± ± < μ < (grams) < μ < (grams) The second confidence interval is narrower, reflecting the fact that there are more restrictions and less variability (and more certainty) in the finite population situation when n>.05n. 36. a. In general, one sample value gives no information about the variation of the variable. It is possible, however, that one value plus other considerations can give some insight. If one knows that 0 is a possible value, for example, then one large sample value would indicate a large variance. [For example: If you take a sample of n=1 of the daily snowfall in a US city and find that 10.0 feet of snow fell that day, you would assume that there are days with no snow and that there must be a large variability in the amounts of daily snowfall.] b. The formula for E requires a value for s and a t score with n-1 degrees of freedom. When n=1, the formula for s fails to produce a value [because there is an (n-1) in the denominator] and there is no df=0 row for the t statistic. No confidence interval can be constructed. c. x ± 9.68 x 1.0 ± ± < μ < 18. [which should be adjusted, since negative heights are not possible] 0 < μ < 18. (feet) Is it likely that some other randomly selected Martian may be 50 feet tall? No, if likely is understood to be highly probable. The range for individual heights would be even larger than the 0 18 given for the mean. With so many possibilities over such a wide range, 50 (or any other individual value) is not highly probable. Yes, if likely is understood to be reasonable. Since the confidence interval includes the value 50, it is a reasonable possibility for the mean height of all Martians and any possible mean height would be a possible individual height. 7-5 Estimating a Population Variance 1. We can be 95% confident that the interval from grams to grams includes the true value of the standard deviation in the weights for the population of all M&M s.. Yes; ( g, g) is another format for indicating the confidence interval given in Exercise 1 although the format in Exercise 1 has the advantage if indicating that the parameter of interest is σ, the population standard deviation. No; while the given expression yields the same endpoints given in Exercise 1, it falsely implies that the point estimate for the parameter in question is grams. 3. No; the population of last two digits from 00 to 99 follows a uniform distribution and not a normal distribution. One of the requirements for using the methods of this section is that the population values have a distribution that is approximately normal even if the sample size is large.

22 18 CHAPTE 7 Estimates and Sample Sizes 4. An unbiased estimator is one whose long-run average value is equal to the true value of the population parameter it estimates. The sample variance is an unbiased estimator of the population variance, but the sample standard deviation is not an unbiased estimator of the population standard deviation as illustrated by exercises 10 and 11 in section 6-4 of the previous chapter. 5. α = 0.05 and df = 8 6. α = 0.05 and df = α = 0.01 and df = α = 0.10 and df = 50 = =.180 8,0.975 = = ,0.975 = = ,0.995 = = ,0.95 = = ,0.05 = = ,0.05 = = ,0.005 = = , α = 0.05 and df = 9; (n-1)s / < σ < (n-1)s / = df,1- and = df, (9)(333) /45.7 < σ < (9)(333) / < σ < < σ < α = 0.05 and df = 4; (n-1)s / < σ < (n-1)s / = df,1- and = df, (4)(.3) / < σ < (4)(.3) / < σ < < σ < 3. (mph) 11. α = 0.01 and df = 6; (n-1)s / < σ < (n-1)s / = df,1- and = df, (6)(.019) / < σ < (6)(.019) / < σ < < σ < (cells/microliter) 1. α = 0.01 and df = 7; (n-1)s / < σ < (n-1)s / = df,1- and = df, (7)(0.1) /0.78 < σ < (7)(0.1) / < σ < < σ < 0.3 (seconds) 13. From the upper right section of Table 7-, n = 19,05. No. This sample size is too large to be practical for most applications. 14. From the upper right section of Table 7-, n = 1. Yes. This sample size is practical for most applications. 15. From the lower left section of Table 7-, n = 101. Yes. This sample size is practical for most applications. 16. From the upper left section of Table 7-, n = 11.

23 17. α = 0.05 and df = 189; (n-1)s / < σ < (n-1)s / = df,1- and = df, Estimating a Population Variance SECTION (189)(645) / < σ < (189)(645) / < σ < < σ < 717 (grams) No. Since the confidence interval includes 696, it is a reasonable possibility for σ. 18. a. = = grams By the range rule of thumb, σ /4 = 0.319/4 = grams. b. α = 0.05 and df = 99 [100]; = df,1- and = df, (n-1)s / < σ < (n-1)s / (99)(0.0518) / < σ < (99)(0.0518) / < σ < < σ < (grams) c. No; the confidence interval does not contain the estimate from part (a). This suggests that the range rule of thumb is not accurate in this case. emember, however, that the range rule of thumb applies to all distributions and that normal distributions (like the weights of the M&M s) have smaller standard deviations than other distributions with the same range because they bunch up near the middle. 19. a. α = 0.05 and df = ; (n-1)s / < σ < (n-1)s / = df,1- and = df, ()(.9) / < σ < ()(.9) / < σ < < σ < 3.4 (minutes) b. α = 0.05 and df = 11; = and = (n-1)s / < σ < (n-1)s / df,1- df, (11)(0.8) /1.90 < σ < (11)(0.8) / < σ < < σ < 35.3 (minutes) c. The two intervals are similar. No, there does not appear to be a difference in the variation of lengths of PG/PGF-13 movies and movies. 0. a. α = 0.01 and df = 39 [40]; (n-1)s / < σ < (n-1)s / = df,1- and = df, (39)(11.3) / < σ < (39)(11.3) / < σ < < σ < 15.5 (beats/minute) b. α = 0.01 and df = 39 [40]; = and = (n-1)s / < σ < (n-1)s / df,1- df, (39)(1.5) / < σ < (39)(1.5) / < σ < < σ < 17. (beats/minute) c. The two intervals are similar. No, there does not appear to be a difference in the variation of pulse rates of men and women.

24 0 CHAPTE 7 Estimates and Sample Sizes 1. preliminary values: n = 1, Σx = 5118, Σx = 8,07,688 x = (Σx)/n = (5118)/1 = s = [n(σx ) (Σx) ]/[n(n-1)] = [1(807688) (5118) ]/[1(11)] = 155, s = α = 0.01 and df = 11; = df,1- and = df, (n-1)s / < σ < (n-1)s / (11)(394.91) /6.757 < σ < (11)(394.91) / < σ < < σ < (seconds). preliminary values: n = 1, Σx = 10008, Σx = 8,360,13 x = (Σx)/n = (1008)/1 = 84.0 s = [n(σx ) (Σx) ]/[n(n-1)] = [1(836013) (10008) ]/[1(11)] = s = α = 0.05 and df = 7; = df,1- and = df, (n-1)s / < σ < (n-1)s / (11)(34.98) /1.90 < σ < (11)(34.98) / < σ < < σ < 59.4 (mm) Yes, the interval contains the traditionally believed value of 35 mm. 3. preliminary values: n = 6, Σx = 9.3, Σx = x = (Σx)/n = (9.3)/6 = s = [n(σx ) (Σx) ]/[n(n-1)] = [6(3.5197) (9.13) ]/[6(5)] = s = α = 0.05 and df = 5; = df,1- and = df, (n-1)s / < σ < (n-1)s / (5)(3.664)/1.833 < σ < (5)(3.664)/ < σ < < σ < (micrograms per cubic meter) Yes. One of the requirements to use the methods of this section is that the original distribution be approximately normal, and the fact that 5 of the 6 sample values are less than the mean suggests that the original distribution is not normal. 4. a. preliminary values: n = 10, Σx = 71.5, Σx = x = (Σx)/n = (71.5)/10 = 7.15 s = [n(σx ) (Σx) ]/[n(n-1)] = [10(513.7) (71.5) ]/[10(9)] = 0.7 s = 0.48 α = 0.05 and df = 9; = df,1- and = df, (n-1)s / < σ < (n-1)s / (9)(0.7)/19.03 < σ < (9)(0.7)/ < σ < < σ < 0.87 (minutes) b. preliminary values: n = 10, Σx = 71.5, Σx = x = (Σx)/n = (71.5)/10 = 7.15 s = [n(σx ) (Σx) ]/[n(n-1)] = [10(541/09) (71.5) ]/[10(9)] = s = 1.8