GM 533 Applied Managerial Statistics Final Exam Answers
(TCO D) PuttingPeople2Work has a growing business placing outofwork MBAs. They claim they can place a client in a job in their field in less than 36 weeks. You are given the following data from a sample.
Sample size: 100
Population standard deviation: 5
Sample mean: 34.2
Formulate a hypothesis test to evaluate the claim.
(TCO B) Assume that customers patronizing the newly opened Nordstrom’s at River City Mall are binomially distributed by wealth (rich = income greater than $250,000/year; poor = the rest of us). If 10 people visit the Nordstrom’s on a typical weekend shopping day, determine the following:
Binomial distribution
10  n 
0.5  p 
X  P(X)  cumulative probability 
0  0.00098  0.00098 
1  0.00977  0.01074 
2  0.04395  0.05469 
3  0.11719  0.17188 
4  0.20508  0.37695 
5  0.24609  0.62305 
6  0.20508  0.82813 
7  0.11719  0.94531 
8  0.04395  0.98926 
9  0.00977  0.99902 
10  0.00098  1.00000 
What is the probability that at least three will be rich?
3.  Question :  (TCO A) Company ABC had sales per month as listed below. Using the Minitab output given, determine:(A) Range (5 points); (B) Median (5 points); and (C) The range of the data that would contain 68% of the results. (5 points).Raw data: sales/month (Millions of $) 23 45 34 34 56 67 54 34 45 56 23 19

4.  Question :  (TCO C, D) Tesla Motors needs to buy axles for their new car. They are considering using Chris Cross Manufacturing as a vendor. Tesla’s requirement is that 95% of the axles are 100 cm ± 2 cm. The following data is from a test run from Chris Cross Manufacturing. Should Tesla select them as a vendor? Explain your answer.Descriptive statistics

5.  Question :  (TCO D) A PC manufacturer claims that no more than 5% of their machines are defective. In a random sample of 100 machines, it is found that 8.5% are defective. The manufacturer claims this is a fluke of the sample. At a .02 level of significance, test the manufacturer’s claim, and explain your answer.

(TCO B) The following table gives the median incomes of families by level of income and geographical region.
East  South  Midwest  West  Totals  
<$80,000/year  102  98  39  62  301 
$80,000 to less than $250,000/year  263  514  120  351  1248 
$250,000/year or more  100  226  65  99  490 
Totals  465  838  224  512  2039 
(A) Referring to the above table, if a family is chosen at random, what is the probability that it is either from the South or from the West? (15 points)
(B) Referring to the above table, given that the family is from the Midwest, what is the probability that it has an annual income of at least $250,000? (15 points)
(TCO B, F) The length of time Americans exercise each week is normally distributed with a mean of 15.8 minutes and a standard deviation of 2.2 minutes
X  P(X≤x)  P(X≥x)  Mean  Stddev 
11  .0146  .9854  15.8  2.2 
15  .3581  .6419  15.8  2.2 
21  .9910  .0090  15.8  2.2 
24  .9999  .0001  15.8  2.2 
p(lower)  p(upper) 
(A) Analyze the output above to determine what percentage of Americans will exercise between 11 and 21 minutes per week. (15 points)
(B) What percentage of Americans will exercise less than 15 minutes? If 1000 Americans were evaluated, how many would you expect to have exercised less than 15 minutes? (15 points)
3.  Question :  (TCO C) A random sample of 16 Google managers yields the following information on annual salaries. The sample mean is $89,000, with a sample standard deviation of $8,000. What is the mean salary of all Google managers? What is the 95% confidence interval for the population mean?OneSample T

(TCO E and F) The U.S Department of Transportation and Safety performed an analysis to determine safe driving speeds. To obtain information about the safe driving speed, it analyzed data from multiple countries comparing the maximum allowed speed limit to the observed death rate. The analysis revealed the following:
Refer to the Minitab output below to answer questions A through G.
Regression Equation  
Death rate (per 100 million vehicles = 0.535979 + 0.0789418 Speed limit (miles per hour)  
Coefficients  
Term  Coef  SE Coef  T  P  95% CI  
Constant  0.535979  2.34352  0.22871  0.825  (5.94014, 4.86818)  
Speed limit (miles per hour)  0.078942  0.03849  2.05106  0.074  (0.00981, 016770) 
Summary of Model  
S = 0.836621  RSq = 34.46%  RSq(adj) = 26.27% 
PRESS = 10.8252  RSq(pred) = 26.70% 
Analysis of Variance  
Source  DF  Seq SS  Adj SS  Adj MS  F 
Regression  1  2.94453  2.94453  2.94453  4.20687 
Speed limit (miles per hour)  1  2.94453  2.94453  2.94453  4.20687 
Error  8  5.59947  5.59947  0.69993  
LackofFit  3  3.37947  3.37947  1.12649  2.53714 
Pure Error  5  2.22000  2.22000  0.44400  
Total  9  8.54400  
Source  P  
Regression  0.074385  
Speed limit (miles per hour)  0.074385  
Error  
LackofFit  0.170419  
Pure Error  
Total 
Predicted Values for New Observations  
New Obs  Fit  SE Fit  95% CI  95% PI 
1  4.20053  0.265262  (3.58883, 4.81222)  (2.17663, 6.22443) 
Values of Predictors for New Observations  
New Obs  Speed limit (miles per hour) 
1  60 
(A) Analyze the above output to determine the regression equation. (10 points)
(B) What conclusions are possible using the meaning of b0 (intercept) and b1 (regression coefficient) in this problem? (That is, explain the meaning of the coefficients.) (10 points)
(C) What conclusions are possible using the coefficient of determination (rsquared)? (6 points)
(D) Calculate the coefficient of correlation. Interpret this value. (6 points)
(E) Does this data provide significant evidence (a=0.05) that the death rate is associated with the speed limit? Find the pvalue and interpret. (6 points)
(F) Determine the average death rate for a speed limit of 60 miles per hour. (6 points)
(G) What is the 95% confidence interval for the death rate for a speed limit of 60 miles per hour? What conclusion is possible using this interval? (6 points)
(TCO E and F) A national trade association is concerned with increasing competition from foreign companies. They decide, in close consultation with their membership, to evaluate the sales performance of 25 randomly selected U.S. companies, so that all companies can benefit from their collective experience.
The association’s research director, with substantial input from member companies’ sales managers, has decided to measure the performance, y, of each company by using the yearly sales of the same product for all of the companies.
The research director and the sales managers believe that sales performance, y (measured in hundreds of units), substantially depends on three independent variables:
x_{1} = sales of the product and all competing products in the company (Market Potential, in hundreds of units)
x_{2} = dollar advertising expenditures in the company (Advertising, in hundreds of dollars)
x_{3} = weighted average of the company’s market share over the previous four years (Market Share)
Refer to the Minitab output below to answer questions A through G.
Multiple Regression and Model Building: Minitab Output
Coefficients
Term  Coef  SE Coef  T  P  95% CI 
Constant  1603.58  505.550  3.17195  0.005  (2654.93, 552.231) 
Market Potential, x1  0.05  0.007  7.26321  0.000  (0.04, 0.070) 
Advertising, x2  0.17  0.044  3.78288  0.001  (0.08, 0.260) 
Market Share, x3  282.75  48.756  5.79927  0.000  (181.35, 384.139) 
Summary of Model
S = 545.515  RSq = 84.90%  RSq(adj) = 82.74% 
RESS = 8616510  P RSq(pred) = 79.18% 
Analysis of Variance
Source  DF  Seq SS  Adj SS  Adj MS  F  P 
Regression  3  35130240  35130240  11710080  39.3502  0.0000000 
Market Potential, x1  1  14788185  15698916  15698916  52.7542  0.0000004 
Advertising, x2  1  10333786  4258521  4258521  14.3102  0.0010905 
Market Share, x3  1  10008270  10008270  10008270  33.6315  0.0000093 
Error  21  6249309  6249309  297586  
Total  24  41379549 
Predicted Values for New Observations
New Obs  Fit  SE Fit  95% CI  95% PI 
1  4251.56  178.023  (3881.34, 4621.78)  (3058.22, 5444.90) 
Values of Predictors for New Observations
New Obs  Marketing Potential, x1  Advertising, x2  Market Share, x3 
1  35182.7  7281.65  9.64 
Find and interpret the 95% Prediction interval for Sales, when Market Potential = 35182.7, Advertising = 7281.65, and Market Share = 9.64. (10 points)