Levine Smume6 Im 08[1]

Solutions to End-of-Section and Chapter Review Problems 403

CHAPTER 8

8.1 X Z n = 85 1.96

864 83.04 86.96

8.2 X Z n = 125 2.58

2436 114.68 135.32

8.3 Since the results of only one sample are used to indicate whether something has gone wrong in the production process, the manufacturer can never know with 100% certainty that the specific interval obtained from the sample includes the true population mean. In order to have 100% confidence, the entire population (sample size N) would have to be selected.

8.4 Yes, it is true since 5% of intervals will not include the population mean.

8.5 If all possible samples of the same size n=100 are taken, 95% of them will include the true population mean annual income within the interval developed. Thus you are 95 percent confident that this sample is one that does correctly estimate the true mean annual income.

8.6 (a) You would compute the mean first because you need the mean to compute the standard deviation. If you had a sample, you would compute the sample mean. If you had the population mean, you would compute the population standard deviation.

(b) If you have a sample, you are computing the sample standard deviation not the population standard deviation needed in Equation 8.1. If you have a population, and have computed the population mean and population standard deviation, you don't need a confidence interval estimate of the population mean since you already have computed it.

8.7 If the population mean annual income is $71,000, the confidence interval estimate stated in Problem 8.5 is correct because it contains $71,000.

8.8 Equation (8.1) assumes that you know the population standard deviation. Because you are selecting a sample of 100 from the population, you are computing a sample standard deviation, not the population standard deviation.

8.9 (a) X Z n

0.995 2.580.02

50 0.9877 1.0023

(b) Since the value of 1.0 is included in the interval, there is no reason to believe that the mean is different from 1.0 gallon.

(c) No. Since is known and n = 50, from the Central Limit Theorem, we may assume that the sampling distribution of X is approximately normal.

Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall

404 Chapter 8: Confidence Interval Estimation

8.9 (d) The reduced confidence level narrows the width of the confidence interval.

cont. X Z n

0.995 1.96 0.02

50 0.9895 1.0005

(b) Since the value of 1.0 is still included in the interval, there is no reason to believe that the mean is different from 1.0 gallon.

8.10 (a)

325.5 374.50

(b) No. The manufacturer cannot support a claim that the bulbs last an average 400 hours. Based on the data from the sample, a mean of 400 hours would represent a distance of 4 standard deviations above the sample mean of 350 hours.

(c) No. Since is known and n = 64, from the Central Limit Theorem, we may assume that the sampling distribution of X is approximately normal.

(d) The confidence interval is narrower based on a process standard deviation of 80hours rather than the original assumption of 100 hours.

(a) X Z n

350 1.96 8064 330.4 369.6

(b) Based on the smaller standard deviation, a mean of 400 hours would represent a distance of 5 standard deviations above the sample mean of 350 hours. No, the manufacturer cannot support a claim that the bulbs have a mean life of 400 hours.

8.11 66.8796 83.1204

8.12 (a) df = 9, = 0.05, = 2.2622

(b) df = 9, = 0.01, = 3.2498

(c) df = 33, = 0.05, = 2.0395

(d) df = 64, = 0.05, = 1.9977

(e) df = 15, = 0.1, = 1.7531



8.13 Set 1: 4.5 2.36463.7417

8 1.3719 7.6281

Set 2: 4.5 2.36462.4495

8 2.4522 6.5478

The data sets have different confidence interval widths because they have different values for the standard deviation.

8.14 Original data: 5.8571 2.4469 6.4660

7 – 0.1229 11.8371

Altered data: 4.00 2.4469 2.1602

7 2.0022 5.9978

The presence of an outlier in the original data increases the value of the sample mean and greatly inflates the sample standard deviation.

8.15 (a) $ 2.46 $ 2.64

(b) To estimate the total value of her inventory, she can multiply the 95% confidence interval for the mean by the total number of greeting cards she has in the inventory.

8.16 (a) 29.44 34.56

(b) The quality improvement team can be 95% confident that the population mean turnaround time is now somewhere in between 29.44 hours and 34.56 hours.

(c) The project was a success because the initial turnaround time of 68 hours does not fall inside the 95% confidence interval.

8.17 (a) 184.6581 205.9419

(b) No, a grade of 200 is in the interval.(c) It is not unusual. A tread-wear index of 210 for a particular tire is only 0.69 standard

deviation above the sample mean of 195.3.



8.18 PHStat output:Confidence Interval Estimate for the Mean

DataSample Standard Deviation 4.389608942Sample Mean 36.53Sample Size 6Confidence Level 95%

Intermediate CalculationsStandard Error of the Mean 1.792050347Degrees of Freedom 5t Value 2.570581835Interval Half Width 4.606612068

Confidence IntervalInterval Lower Limit 31.93Interval Upper Limit 41.14

(a) = 36.53 31.9267 41.1399

(b) You can be 95% confident that the population mean price for two tickets with onlineservice charges, large popcorn, and two medium soft drinks is somewhere between $31.93 and $41.14.

8.19 PHStat output:Data

Sample Standard Deviation 2.657536453Sample Mean 27.4375Sample Size 16Confidence Level 95%



(a) = 27.4375 26.0214 28.8536

(b) You can be 95% confident that the population mean miles per gallon of 2009 family sedans priced under $20,000 is somewhere between 26.0214 and 28.8536.

(c) Since the 95% confident interval for population mean miles per gallon of 2007 small SUVs does not overlap with that for the population mean miles per gallon of 2007 family sedans, we are 95% confident that the population mean miles per gallon of 2007 small SUVs is lower than that of 2007 family sedans.




Sample Standard Deviation 2.245768969Sample Mean 19.95652174Sample Size 23Confidence Level 95%



(a) = 19.9565 18.9854 20.9277

(b) You can be 95% confident that the population mean miles per gallon of 2009 small SUVs priced under $30,000 is somewhere between 18.9854 and 20.9277.

(c) Since the 95% confident interval for population mean miles per gallon of 2009 small SUVs priced under $30,000 does not overlap with that for the population mean miles per gallon of 2009 family sedans priced under $20,000, we are 95% confident that the population mean miles per gallon of 2009 small SUVs priced under $30,000 is lower than that of 2009 family sedans priced under $20,000.

8.21 (a) PHStat output:Data

Sample Standard Deviation 3485663.093Sample Mean 7,765,479Sample Size 10Confidence Level 95%



5,271,985.94 10,258,972.26



8.21 (b) PHStat output:cont.

DataSample Standard Deviation 3485663.093Sample Mean 7,765,479Sample Size 10Confidence Level 99%



4,183,304.15 11,347,654.05(c) The higher the level of confidence, the wider is the confidence interval.

8.22 (a) 31.12 54.96

(b) The population distribution needs to be normally distribution.(c)

Normal Probability Plot

0

20

40

60

80

100

120

140

160

180

-3 -2 -1 0 1 2 3

Z Value

Day

s



8.22 (c)cont.

Box-and-whisker Plot

Days

0 50 100 150

Both the normal probability plot and the boxplot suggest that the distribution is skewed to the right.

(d) Even though the population distribution is not normally distributed, with a sample of 50, the t distribution can still be used due to the Central Limit Theorem.

8.23 (a) 33.89 53.89

(b) The population distribution needs to be normally distributed.(c)


Time

10 30 50 70 90


0102030405060708090

100

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

Z Value

Tim

e

Both the normal probability plot and the boxplot show that the population distribution is not normally distributed and is skewed to the right.



8.24 (a) PHStat output:Confidence Interval Estimate for the Mean

DataSample Standard Deviation 0.327289904Sample Mean 0.93Sample Size 14Confidence Level 95%



0.7367 1.1147

(b) The population distribution needs to be normally distributed.(c)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

Cost

($)

Z Value


(c)

Cost($)

0 0.5 1 1.5 2

Boxplot

Both the normal probability plot and the boxplot show that the distribution for the cost per ounce is right-skewed.



8.25 (a) –0.000566 0.000106

(b) The population distribution needs to be normally distributed. However, with a sampleof 100, the t distribution can still be used as a result of the Central Limit Theorem even if the population distribution is not normal.


-0.004-0.003-0.002-0.001

00.0010.0020.0030.0040.0050.006

-3 -2 -1 0 1 2 3

Z Value

Erro

r

(c)


Error

-0.006 -0.004 -0.002 0 0.002 0.004 0.006

Both the normal probability plot and the boxplot suggest that the distribution is skewed to the right.

(d) We are 95% confident that the mean difference between the actual length of the steelpart and the specified length of the steel part is between -0.000566 and 0.000106 inches, which is narrower than the plus or minus 0.005 inches requirement. The steel mill is doing a good job at meeting the requirement. This is consistent with the finding in Problem 2.33.

8.26 = 0.25

0.19 0.31

8.27 = 0.0625

0.0313 0.0937



8.28 (a)

= 0.27

0.22 0.32(b) The manager in charge of promotional programs concerning residential customers

can infer that the proportion of households that would purchase an additional telephone line if it were made available at a substantially reduced installation cost is between 0.22 and 0.32 with a 99% level of confidence.

8.29 (a) = 0.25

0.22 0.28

(b) 0.4404

0.41 0.47(c) You can be 95% confident that the population proportion of moms employed

full-time that are dissatisfied with their work-life balance is somewhere between 0.22 and 0.28 while the population proportion of moms employed full-time that would take a pay cut to spend more time with their kids is somewhere between 0.41 and 0.47.



8.30 (a) PHStat output:Data

Sample Size 500Number of Successes 260Confidence Level 95%

Intermediate CalculationsSample Proportion 0.52Z Value -1.95996398Standard Error of the Proportion 0.022342784Interval Half Width 0.043791052


= 0.52

0.4762 0.5638(b) Since the 95% confidence interval contains 0.50, you cannot claim that more than

half of all U.S. workers have negotiated a pay raise.

(c) (a) = 0.52

0.5062 0.5338(b) Since the lower limit of the 95% confidence interval is greater than 0.50, you

can claim that more than half of all U.S. workers have negotiated a pay raise.(d) The larger the sample size, the narrow is the confidence interval holding everything

else constant.


Sample Size 1000Number of Successes 760Confidence Level 95%



= 0.76

0.7335 0.7865



8.32 (a) PHStat output:

DataSample Size 2395Number of Successes 1916Confidence Level 95%



= 0.8

0.7840 0.8160(b) PHStat output:

DataSample Size 2395Number of Successes 1269Confidence Level 95%



= 0.5299

0.5099 0.5498(c) Since the two confidence intervals do not overlap, you can conclude that with 95%

confidence that the population proportion of adults who report that emails are easy to misinterpret is higher than the population proportion of adults who report that telephone conversations are easy to misinterpret.



8.33 (a) = 0.3340

0.29 0.38

(b) = 0.1797

0.15 0.21(c) You can be 95% confident that the population proportion of employees who

typically took work with them on vacation is somewhere between 0.29 and 0.38 and the population proportion of employees who said that there are unwritten and unspoken expectations that they stay connected is somewhere between 0.15 and 0.21.

(d) The two confidence interval estimates do not overlap because the two separate point estimates are far enough apart.

8.34 n Z 2 2

e2 1.962 152

52 = 34.57 Use n = 35

8.35 n Z 2 2

e2 2.582 1002

202 = 166.41 Use n = 167

8.36 = 1,040.06 Use n = 1,041

8.37 = 2,304.96 Use n = 2,305

8.38 (a) n Z 2 2

e2 1.962 4002

502 = 245.86 Use n = 246

(b) = 983.41 Use n = 984

8.39 n Z 2 2

e2 1.962 (0.02)2

(0.004)2 = 96.04 Use n = 97

8.40 n Z 2 2

e2 1.962 (100)2

(20)2 = 96.04 Use n = 97

8.41 n Z 2 2

e2 1.962 (0.05)2

(0.01)2 = 96.04 Use n = 97



8.42 (a) n Z 2 2

e2 2.582 252

52 = 166.41 Use n = 167

(b) n Z 2 2

e2 1.962 252

52 = 96.04 Use n = 97

8.43 (a) n Z 2 2

e2 1.6452 452

52 = 219.19 Use n = 220

(b) n Z 2 2

e2 2.582 452

52 = 539.17 Use n = 540

8.44 (a) = 245.85 Use n = 246

(b) = 384.15 Use n = 385

(c) = 553.17 Use n = 554

(d) When there is more variability in the population, a larger sample is needed to accurately estimate the mean.

8.45 (a) = 576.2188 Use n = 577

(b) = 995.2345 Use n = 996

(c) = 2,304.8753 Use n = 2,305

(d) = 3,980.9380 Use n = 3,981

(e) The higher the level of confidence desired, the larger is the sample size required. The smaller the sampling error desired, the larger is the sample size required.

8.46 (a) = 0.28

0.2522 0.3078

(b) = 0.19

0.1657 0.2143

(c) = 0.07

0.0542 0.0858



8.46 (d) (a) = 1,936.0952 Use n = 1,937

cont.

(b) = 1,478.0013 Use n = 1,479

(c) = 625.1974 Use n = 626

8.47 (a) = 315/500 = 0.63

0.5877 0.6723(b) You are 95% confident that the proportion of companies who informally monitored

social networking sites to stay on top of information related to their company is somewhere between 0.5877 and 0.6723.

(c) = 8,954.44 Use n = 8,955

8.48 (a) If you conduct a follow-up study to estimate the population proportion of individuals who said that the most important way to judge a company was how a company responded to a crisis, you would use a π of 0.84 in the sample size formula since it provides the most conservative value based on past information on the proportion. If you are really paranoid, you could use 0.5 to ensure that you would never underestimate the sample size needed.

(b) = 573.67 Use n = 574

8.49 (a) = 1,654.30 Use n = 1,655

(b) = 595.55 Use n = 596

(c) The higher the precision you require of your confidence interval, the larger is the sample size required.

8.50

$10,721.53 Population Total $14,978.47



8.51 Using PHStat,Confidence Interval Estimate for the Total Difference

DataPopulation Size 10000Sample Size 200Confidence Level 95%

Intermediate CalculationsSum of Differences 200.63Average Difference in Sample 1.00315Total Difference 10031.5Standard Deviation of Differences 4.998502FPC Factor 0.989999Standard Error of the Total Diff. 3499.126Degrees of Freedom 199t Value 1.971957Interval Half Width 6900.128


3131.37 Total Difference in the Population 16931.63

8.52 (a)

(b)

(c)

8.53 =

$ 2,467.13 Population Total $ 2,632.87

8.54

$543,176.96 Population Total $1,025,223.04



8.55


8.56

$ 5,126.26 Total Difference in the Population $54,546.28Note: The t-value of 2.6092 for 99% confidence and d.f. = 149 was derived on Excel.

8.57

– $4,046.99 Total Difference in the Population

8.58 (a)

(b) Since the upper bound is higher than the tolerable exception rate of 0.04, the auditor should request a larger sample.

8.59 (a)

(b) With 95% level of confidence, the auditor can conclude that the rate of noncompliance is less than 0.0347, which is less than the 0.05 tolerable exception rate for internal control, and, hence, the internal control compliance is adequate.

8.60 The only way to have 100% confidence is to obtain the parameter of interest, rather than a sample statistic. From another perspective, the range of the normal and t distribution is infinite, so a Z or t value that contains 100% of the area cannot be obtained.

8.61 The t distribution is used for obtaining a confidence interval for the mean when is unknown.

8.62 If the confidence level is increased, a greater area under the normal or t distribution needs to be included. This leads to an increased value of Z or t, and thus a wider interval.

8.63 When estimating the rate of noncompliance, it is commonplace to use a one-sided confidence interval instead of a two-sided confidence interval since only the upper bound on the rate of noncompliance is of interest.

8.64 In some applications such as auditing, interest is primarily on the total amount of a variable rather than the mean amount.



8.65 Difference estimation involves determining the difference between two amounts, rather than a single amount.

8.66 940.50 1007.50

Based on the evidence gathered from the sample of 34 stores, the 95% confidence interval for the mean per-store count in all of the franchise’s stores is from 940.50 to 1007.50. With a 95% level of confidence, the franchise can conclude that the mean per-store count in all of its stores is somewhere between 940.50 and 1007.50, which is larger than the original average of 900 mean per-store count before the price reduction. Hence, reducing coffee prices is a good strategy to increase the mean customer count.

8.67 (a) Turn off lights, power strips, unplug things:

= 0.73

0.6911 0.7689Recycle aluminum, plastic, newspapers, cardboard:

= 0.47

0.4263 0.5137Recycle harder-to-recycle products:

= 0.36

0.3179 0.4021Buy products with least packaging:

= 0.34

0.2985 0.3815Ride a bike or walk:

= 0.23

0.1931 0.2669(b) The population proportion of what adults do to conserve energy, in descending order,

is “turn off lights, power strips, unplug things”; “recycle aluminum, plastic, newspapers, cardboard”; “recycle harder-to-recycle products” and “buy products with least packaging” (which are not statistically different because the confidence intervals overlap); and finally “ride a bike or walk”.



8.68 (a) 14.085 16.515

(b) 0.530 0.820

(c) n Z 2 2

e 2 1.962 52

22 = 24.01 Use n = 25

(d) = 784 Use n = 784

(e) If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 784) should be used.

8.69 (a) 1,638.69 1,879.31

(b) 0.485 0.715

8.70 (a) 8.049 11.351

(b) 0.284 0.676

(c) n Z 2 2

e 2 1.962 4.52

1.52 = 34.57 Use n = 35

(d) = 120.268 Use n = 121


8.71 (a) n Z 2 2

e 2 2.582 182

52 = 86.27 Use n = 87

Note: If the Z-value used is carried out to 2.5758, the value of n is 85.986 and only 86 women would need to be sampled.

(b) = 334.07 Use n = 335

If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 335) should be used.



8.72 (a) $25.80 $31.24

(b) 0.3037 0.4963

(c) n Z 2 2

e 2 1.962 102

22 = 96.04 Use n = 97

(d) = 422.82 Use n = 423


8.73 (a) $19.14 $23.54

(b)

0.2764 0.4664

(c) n Z 2 2

e 2 1.962 102

1.52 = 170.74 Use n = 171

(d) = 334.08 Use n = 335

(e) If a single sample were to be selected for both purposes, the larger of the two samplesizes (n = 335) should be used.

8.74 (a) $36.66 $40.42

(b) 0.2027 0.3973

(c) n Z 2 2

e 2 1.962 82

1.52 = 109.27 Use n = 110

(d) = 422.82 Use n = 423




8.75 (a) = 384.16 Use n = 385

If we assume that the population proportion is only 0.50, then a sample of 385 would be required. If the population proportion is 0.90, the sample size required is cut to 103.

(b)

0.7384 0.9416(c) The representative can be 95% confidence that the actual proportion of bags that will do

the job is between 74.5% and 93.5%. He/she can accordingly perform a cost-benefit analysis to decide if he/she wants to sell the Ice Melt product.

8.76 (a)

0.2013(b) Since the upper bound is higher than the tolerable exception rate of 0.15, the auditor

should request a larger sample.

8.77 (a) = 138.292

= 137.53 Use n = 138

Note: The sample size is computed using the finite population correction factor.

(b)

0.0476 0.1263(c) Using Excel, we find the t-value for 95% confidence and 137 degrees of

freedom is t = 1.9774.

$87.90 $99.50



8.77 (d)

cont.

$2,202,427.61 Population Total $2,493,066.79

(e)

$14,937.30 Total Difference in the Population $72,577.14

8.78 (a) = 26.5396 Use n = 27

(b)


8.79 (a)

(b) Since 5.5 grams is within the 99% confidence interval, the company can claim that the mean weight of tea in a bag is 5.5 grams with a 99% level of confidence.

(c) The assumption is valid as the weight of the tea bags is approximately normally distributed.

8.80 (a)

(b) With 95% confidence, the population mean width of troughs is somewhere between 8.41 and 8.43 inches. Hence, the company's requirement of troughs being between 8.31 and 8.61 is being met with a 95% level of confidence.

(c) The assumption is valid as the width of the troughs is approximately normally distributed.



8.81 (a)

(b)

(c)


3000

3050

3100

3150

3200

3250

3300

-4 -3 -2 -1 0 1 2 3 4

Z Value

Bos

ton


3550

3600

3650

3700

3750

3800

3850

3900

-4 -3 -2 -1 0 1 2 3 4

Z Value

Verm

ont

The weight for Boston shingles is slightly skewed to the right while the weight for Vermont shingles appears to be slightly skewed to the left.

(d) Since the two confidence intervals do not overlap, the mean weight of Vermont shingles is greater than the mean weight of Boston shingles.



8.82 (a)

(b)

(c)


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

-3 -2 -1 0 1 2 3

Z Value

Verm

ont


0

0.2

0.4

0.6

0.8

1

1.2

-3 -2 -1 0 1 2 3

Z Value

Bos

ton

The amount of granule loss for both brands are skewed to the right but the sample sizes are large enough so the violation of the normality assumption is not critical.

(d) Because the two confidence intervals do not overlap, you can conclude that the mean granule loss of Boston shingles is higher than that of Vermont shingles



8.84 The 95% confidence interval for the various parameters:

Category:Average risk:

Intermediate Government: 0.36 0.61 Short Term Corporate: 0.39 0.64

Above average risk:Intermediate Government: 0.44 0.69 Short Term Corporate: 0.31 0.56

Below average risk:Intermediate Government: 0.33 0.58 Short Term Corporate: 0.42 0.67

Asset:Average risk: 289.58 1453.48Above average risk: 324.96 1073.39Below average risk: 326.20 909.38

Fee:Average risk:

Yes: 0.18 0.42No: 0.58 0.82

Above average risk:Yes: 0.24 0.49No: 0.51 0.76

Below average risk:Yes: 0.11 0.31No: 0.69 0.89

Expense ratio:Average risk: 0.66 0.78Above average risk: 0.66 0.83Below average risk: 0.62 0.78

Return 2008:Average risk: 1.04 4.05Above average risk: -3.81 2.09Below average risk: 1.20 3.13

3-year return:Average risk: 3.60 4.71Above average risk: 2.11 4.28Below average risk: 3.54 4.35



8.84 5-year return:cont. Average risk: 3.03 3.77

Above average risk: 1.88 3.33Below average risk: 2.94 3.51

8.85 The 95% confidence interval for the various parameters:Gender:

Male: 0.38 0.66Female: 0.34 0.62

Age: 20.01 21.91Height: 66.03 68.01Class:

Freshman: Not both X and n – X > 5Sophomore: 0.32 0.60Junior: 0.23 0.49Senior: Not both X and n – X > 5

Major:Accounting: 0.11 0.33Economics/Finance: 0.07 0.29International Business: Not both X and n – X > 5Information System: Not both X and n – X > 5Management: 0.07 0.29Marketing/Retailing: 0.09 0.31Other: Not both X and n – X > 5Undecided: Not both X and n – X > 5

Grad School:Undecided: 0.21 0.47Yes: 0.23 0.49No: 0.17 0.43

GPA: 3.01 3.24Expected Salary: $45.19 thousands $49.41 thousandsAnnual Salary in 5 Years: $72.37 thousands $82.91 thousandsEmployment Status:

Full-time: Not both X and n – X > 5Part-time: 0.64 0.88Unemployed: 0.11 0.33



8.85 Number of Affiliations: 0.30 0.78cont. Satisfaction Advisement:

1: Not both X and n – X > 52: Not both X and n – X > 53: 0.12 0.364: 0.14 0.385: 0.14 0.386: Not both X and n – X > 57: Not both X and n – X > 5

Spending: $432.05 $509.35

8.87 The 95% confidence interval for the various parameters:Gender:

Male: 0.45 0.75Female: 0.25 0.55

Age: 28.24 31.21Height: 65.95 68.60Major:

Accounting: 0.12 0.38Economics/Finance: 0.10 0.35International Business: Not both X and n – X > 5Information System: Not both X and n – X > 5Management: 0.06 0.29Marketing/Retailing: 0.06 0.29Undecided: Not both X and n – X > 5

GRADGPA: 3.42 3.61Undergrad Specialization:

Biological Science: Not both X and n – X > 5Business Administration: 0.14 0.41Computers or Math: 0.10 0.35Education: Not both X and n – X > 5Engineering: Not both X and n – X > 5Humanities: Not both X and n – X > 5Performing Arts: Not both X and n – X > 5Physical Sciences: Not both X and n – X > 5Social Sciences: Not both X and n – X > 5Other: Not both X and n – X > 5

UNDERGRADGPA: 3.20 3.39GMAT: 540.61 568.89Employment Status:

Full-time: 0.71 0.94Part-time: Not both X and n – X > 5Unemployed: Not both X and n – X > 5

Expected Salary: $63.82 thousands $76.18 thousands



8.87 Annual Salary in 5 Years: $96.77 thousands $119.48 thousandscont. Number of Jobs: 1.75 2.50

Satisfaction Advisement:1: Not both X and n – X > 52: Not both X and n – X > 53: Not both X and n – X > 54: 0.32 0.635: 0.06 0.296: 0.04 0.267: Not both X and n – X > 5

Spending: $174.35 $292.40


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