6. a) ABC ABD ABE ABO ACD ACE ACO ADE ADO AEO BCD BCE BCO BDE BDO BEO CDE CDO CEO DEO

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Dortha Harvey
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1 CHAPTER 7. a) ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE b) ABCD ABCE ABDE ACDE BCDE 4. a) WX WY WZ XY XZ YZ b) With a sample size of and a population size of 4, we ll divide 4 by to determine the sampling interval. In this case 4/ = tells us to select one of the first names on the list and then every nd name after that. The possible samples are thus WY and XZ. 6. a) ABC ABD ABE ABO ACD ACE ACO ADE ADO AEO BCD BCE BCO BDE BDO BEO CDE CDO CEO DEO b) With a sample size of 3 and a population size of 6, divide 6 by 3 to determine the sampling interval. In this case 6/3 = tells us to select one of the first countries on the list and then every nd country after that. The possible samples are thus ACE and BDO. c) Select two Latin American countries: possible samples are AC AE CE. Select two Middle East countries: possible samples are BD BO DO. So the combined sample possibilities are: ACBD ACBO ACDO AEBD AEBO AEDO CEBD CEBO CEDO 8. a) and b) The 10 samples, together with the sample means, are: Sample Sample VWX VYZ 110 VWY WXY 110 VWZ 110 WXZ VXY WYZ VXZ 110 XYZ c) The sampling distribution of the sample mean is P( ) /10 = /10 = /10 =.30 Total = 1.00

2 The bar chart is P( ) d) From part c), P( = 110) = Sample Average Sales 10. a) and b) The 6 samples, together with the sample means, are: Sample ABCDE 11 ABCDF 1 ABCEF 13 ABDEF 1 ACDEF 13 BCDEF 14 c) The sampling distribution of the sample mean is The bar chart is P( ) 11 1/6 = /6 = /6 = /6 =.167 Total = 1.00

3 P( ) Sample Average Number of Employees d) From part c), P( = 11) = a) The sampling distribution is The bar chart is P( ).40 P( ) Sample Average Sales There s a hint of a bell-shape but the population and sample sizes are too small to show a true normal distribution shape. b) and c) The population mean µ is ( )/5 = 110 The mean of the sampling distribution, (E( ), as shown below, is also 110. P( ) P( ) ( - E( )) P( )

4 E( ) = 110 The population standard deviation is = 6.66 = =.58 = ( ) + ( ) + ( ) 5 + ( ) (10 110) = 40 = 6.3 The standard deviation of the sampling distribution, as shown in the table above, is =.58. The fpc here is 5 3 =.707, so we can show = a) The sampling distribution of the sample mean is 6.3 fpc = (.707) =.58 n 3 The bar chart is P( ) Total = 1.00 P( ) Sample Average Number of Employees There s a hint of a bell-shape but the population and sample sizes are too small to show a true normal distribution shape. b) and c) The population mean µ is ( )/6 = 1.5 The mean of the sampling distribution, E( ), as shown below, is also 1.5.

5 P( ) P( ) ( - E( )) P( ) E( ) = 1.5 The population standard deviation is =.916 =. 916 =.957 = (5-1.5) =. 9 = (10-1.5) + (15-1.5) + (10-1.5) 6 (15 1.5) + (0 1.5) The standard deviation of the sampling distribution, as shown in the table above, is =.957 The fpc here is 6 5 =.447, so we can show = fpc = (.447) =.957 n Population: All young people in the US between the ages of 1 and 17. Characteristic of Interest: µ, the average time spent online per week by this population. a) 8. ± 1.( ) which gives 8. ±.19 or 8.01 hours to 8.39 hours b) We can be 95% confident that the interval 8.01 hours to 8.39 hours contains the average time spent on line for the population of American teenagers between the ages of 1 and 17. Our level of confidence is based on the fact that if we were to repeat this procedure a large number of times, approimately 95% of the intervals we construct would contain the population of American teenagers between the ages of 1 and 17 average time spent on line, µ. That is, in 95% of the cases, the population mean would be within.19 hours of the sample mean. 18. Population: All teachers in California public schools. Characteristic of Interest: µ, the mean salary for the population of California public school teachers a) 55,693 ± 1.( ), 00 which gives 55,693 ± 94 or $54,751 to $56, b) 55,693 ±.58( ), 00 which gives 55,693 ± 140 or $54,453 to $56,933

6 0. Population: All Canadian furniture stores. Characteristic of Interest: µ, the average advertising budget for the population of furniture stores.. a) 7.1 ± 1.( ), which gives 7.1 ±.68 or 6.4 to b) 7.1 ± 1. ( ), which gives 7.1 ±.43 or 6.67 to c) here the margin of error is +.1. This means 1.( ) =.1. Solving for n, n 1.(.) n = = a) standard error = = 5 49 b) margin of error = 1.( 35 ) = Population: All U.S. retailers. Characteristic of Interest: µ, the average e-commerce sales increase. a) 4.7 ± 3.1 or 1.6 to 7.8 b) from a), (1.) (standard error) = 3.1. This means standard error = 3.1/1. = 1.58 Therefore, for a 90% interval, the margin of error must 1.65(1.58) = a) look up a right tail area of.10, t = 1.35 b) look up a right tail area of =.10, giving t =.60 c) look up a right tail area of (.01)/ =.005, giving t = ± a) look up a right tail area of.05, giving t = b) look up a right (left) tail area of =.01, giving t = c) look up a right tail area of (.10)/ =.05, giving t = ± (10 130) + ( ) + ( ) 30. a) = = 130 s = = b) use the t table with df = and a right tail area of.05, giving a t of The interval is or or 105. to 154.8

7 3. Population: All gas stations nationwide. Characteristic of Interest: µ, the average gas price for the population of stations. a) b) or or $.45 to $.395 or or $.06 to $ a) For a.10 right tail and df = 40, the t value is 1.303; z = 1.8 b) For a.005 right tail and df = 9, the t value is 3.50; z =.58 (rounded) 36. a) For a.05 right tail and df =, the t value is 1.0; z = 1. b) For a.005 right tail and df =, the t value is.576; z =.58 (rounded) 38. Population: All Milestone subscribers. Characteristic of Interest: µ, the average number of advertisements recalled by the population of Milestone subscribers a) b) c) or or 6.1 to or or 6.71 to 9.69 or or 7.6 to Population: All professional baseball players. Characteristic of Interest: µ, the average daily consumption of nutritional supplements taken by the population of professional baseball players. 60 a) or or to b) or or to Note: Using a z of 1.65 instead of the t value of would change things very little here and would be an acceptable switch since sample size is more than c) or or to Note: Using a z of 1.65 instead of the t value of would change things very little here and would be an acceptable switch since sample size is more than 30.

8 1.65(54) 4. n = 6 = 1 1.(0) 44. n = 5 = 98 1.(100) 46. Step 1: n = = Step : n = = (446) 48. Step 1: n = = 1911 (same as above) 0 Step : n = = a) AB AC AD AE BC BD BE CD CE DE b) 1/10 =.10 or 10% c) 1/10 =.10 or 10% 5. For the list of possible sample means that you produced in Eercise 43, a) The bar chart is P( ) b) The population mean µ is ( )/4 = 8

9 The mean of the sampling distribution, E( ), as shown below, is also 8. c) P( ) P( ) E( ) = 8 P( ) ( - E( )) P( ) The population standard deviation is =. =. =.471 = (8-8) + (6-8) + (10-8) 4 + (8-8) = = The standard deviation of the sampling distribution, as shown in the table above, is = The fpc here is =.577, so we can show = fpc = (.577) = n a) 4879 ± 1.( ), which gives 4789 ± (1.)(189.74) 50 or 4789 ± or $ to $ b) We can be 95% confident that the interval $ to $ will contain the average signing bonus for the population of newly hired pharmacists. This level of confidence is based on the fact that if we were to repeat this procedure a large number of times, approimately 95% of the intervals we construct would contain the population average signing bonus, µ a) $6,540 ±.58( ), which gives $6,540 ± (.58)(73.34) 1500 or $6, 540 ± or $6, to $6,79.3 b) We can be 99% confident that the interval $6, to $6,79.3 will contain the average salary for the population of IT professionals. If we were to repeat this procedure a large

10 number of times, approimately 99% of the intervals we construct would contain the population average salary, µ c) margin of error:.58( 800 ) = $189.3 standard error: = $ a) Here we ll replace in the calculation of the standard error with s and replace z with t. For 95% confidence, to find t we ll use the t table for a.05 right tail and df = 0-1 = 19. The required t value is.093: 13.6 ± (.093)(1.16) or 13.6 ±.4 or years to 16.0 years For 99% confidence, to find t we ll use the t table for a.005 right tail and df = 0-1 = 19. The required t value is.861: 13.6 ± (.861)(1.16) or 13.6 ± 3.3 or 10.8 years to 16.9 years For 80% confidence, to find t we ll use the t table for a.10 right tail and df = 0-1 = 19. The required t value is 1.38: 13.6 ± (1.38)(1.16) or 13.6 ± 1.54 or 1.06 years to years b) Because the sample is less than thirty, we must assume that the normal population condition is met. That is, we have to assume the years of seniority for the population of all employees working for our company follows a normal distribution. 60. a) Population: all assemblies of the new product. b) 95% confidence: t =.145 (df = 14) ± (.145), which gives ± (.145)(1.45) or 38.4 ± 3.10 or 35.3 minutes to 41.5 minutes c) margin of error error: 3.10 standard error: First calculate and s: = = 6 6 s = (6 6) + (10 6) + (4 6) + (5 6) (7 6) + (4 6) =.8 90% confidence: t =.015 (df = 6-1 = 5)

11 .8 6 ±.015( ), which gives 6 6 ±.015 (.93) or 6 ± 1.87 or 4.13 visits to 7.87 visits 64. a) Estimating the population average: 1560 ±.58( 30 ), which gives ±.58(41.3) or 1560 ± or units to units b) Estimate total daily output for the industry as a whole: (1560)(300) ± (106.6)(300) or 3,588,000 ± 45,180 or 3,34,80 units to 3,833,180 units 66. a) 8 ± 1.44( ) which gives 8 ± (1.44)(.35) or 8 ±.50 or 7.5 employees to 8.5 employees b) Estimate of total number of people in China employed by individual hotels: (8)(7,500) ± (.5)(7,500) or 770,000 ± 13,750 or 756,50 people to 783, 750 people 68. The interval is: ± t ( s 5.3 ) or 16. ±.13( ) n 16 a) the t distribution: Yes, since the population standard deviation will be estimated by the sample standard deviation and the sample size is less than 30. b) an assumption that the population distribution here is approimately normal: Yes, this is a small sample situation and requires the normal population assumption in order to justify the sort of interval-building procedure we ve discussed. c) division by n-1 rather than n: we ll assume that the division by n-1 was done in the calculation of the sample standard deviation, s. 70. a) False. For large enough samples (n 30), the shape of the sampling distribution of sample means will be approimately normal, regardless of the shape of the parent population. b) False. Just check the normal table. The z for 90% confidence is 1.65 and for 45% the z is.6. c) False. The best estimate of the population standard deviation would be the sample standard deviation, 1.6 interviews. (The standard error would be equal to 0. interviews.) d) False. The confidence interval does not show where individual values in the population are

12 likely to fall. It shows where the MEAN of the population is likely to fall. e) True. This is precisely the nature of the interval estimation procedure. 1.(30) 7. n = = 865 commuters = = 58 5 (55 58) + (68 58) + (77 58) + (49 58) s = (14.5) n = 1 = 573 new accountants + (41 58) = (10) 76. n = 0 = 40 nurses 78. a) N = 100 retail outlets 1.($.50) Step 1: n = = retail outlets $.50 Step : n = = 49 retail outlets b) N = 500 retail outlets n = = 81 retail outlets c) N = 5,000 retail outlets n = = 94 retail outlets As you can see, using the adjustment has very little impact on the sample size. The sample size here is a small % of the population size (n/n is less than 5%) d) N = 0,000 retail outlets n = = retail outlets 1+ 0,000 As you can see, using the adjustment has virtually no impact on the sample size. The sample size here is a small % of the population size (n/n is less than 5%).

Education Assistance In your opinion is the cost to a student of a university education a good long-term investment or not?

Education Assistance In your opinion is the cost to a student of a university education a good long-term investment or not? In your opinion is the cost to a student of a university education a good long-term investment or not? Proportions/Means: Columns Tested (5% risk level) - A/B/C/D/E/F - G/H - I/J - K/L/M/O - N/O REGION