Chapter 18
18.1 a
b
18.2 a
b
18.3 a Sales = + Space + Space+
b
= 41.15 and = .4068. The model's fit is relatively poor.
F = 7.54, p-value = .0032. However, there is enough evidence to support the validity of the model.
18.4aFirst–order model: a Demand = + Price+
Second–order model: a Demand = + Price + Price+
First–order model:
Second–order model:
c The second order model fits better because its standard error of estimate is 5.96, whereas that of the first–order models is 13.29
d .= 766.9 –359.1(2.95) + 64.55(2.95)= 269.3
18.5aFirst–order model: a Time = + Day+
Second–order model: a Time = + Day + Day+
b First–order model
F = 45.48, p-value = 0. The model is valid.
Second–order model
F = 26.98, p-value = .0005. The model is valid.
c The second–order model is only slightly better because its standard error of estimate is smaller.
18.6a MBA GPA= + UnderGPA + GMAT + Work + UnderGPAGMAT +
b
F = 18.43, p-value = 0; = .790 and = .4674. The model is valid, but the fit is relatively poor.
c MBA example = .788 and = .4635. There is little difference between the fits of the two models.
18.7 a (Excel output shown below)
b
At least on is not equal to 0
F = 80.65, p-value = 0. There is enough evidence to infer that the model is valid.
18.8a
b
c Both models fit equally well. The standard errors of estimate and coefficients of determination are quite similar.
18.9a
b
At least on is not equal to 0
F = 5.36, p-value = .0019. There is enough evidence to infer that the model is valid.
c 0
0
t = –.12, p-value = .9086. There is not enough evidence to infer that there is an interaction effect between face–offs won and penalty minutes differential.
18.10a Yield = + Pressure + Temperature + Pressure
+Temperature+ Pressure Temperature +
b
c = 512 and = .6872. The model's fit is good.
18.11 The number of indicator variables is m – 1 = 5 – 1 = 4.
18.12 a= 1 if Catholic
= 0 otherwise
= 1 if Protestant
= 0 otherwise
b= 1 if 8:00 A.M. to 4:00 P.M.
= 0 otherwise
= 1 if 4:00 P.M. to midnight
= 0 otherwise
c= 1 if Jack Jones
= 0 otherwise
= 1 if Mary Brown
= 0 otherwise
= 1 if George Fosse
= 0 otherwise
18.13 a Macintoshb IBMc other
18.14a
Prediction: MBA GPA will lie between 8.55 and 11.67
b
Prediction: MBA GPA will lie between 8.15 and 11.31
18.15a
b Exercise 18.10: = 3.24 + .451Mother + .411Father + .0166Gmothers + .0869Gfathers
There are large differences to all the coefficients.
c 0
0
t = –5.56, p-value = 0. There is enough evidence to infer that smoking affects longevity.
18.16a
b
At least on is not equal to 0
F = 20.43, p-value = 0. There is enough evidence to infer that the model is valid.
c 0
0
: t = 1.86, p-value = .0713
: t = –1.58, p-value = .1232
Weather is not a factor in attendance.
d0
> 0
t = 3.30, p-value = .0023/2 = .0012. There is sufficient evidence to infer that weekend attendance is larger than weekday attendance.
18.17a
b = 20,487; in this sample on average two–story houses sell for $20,487 more than other houses with the same number of bedrooms, square footage, and lot size.
= 12,795; in this sample on average side–split houses sell for $12,795 more than other houses with the same number of bedrooms, square footage, and lot size.
= 19,512; in this sample on average back–split sell for $19,512 more than other houses with the same number of bedrooms, square footage, and lot size.
Note that "other" refers to houses that are not two–story, side–split, or back–split.
0
0
: t = 2.93, p-value = .0043
: t = 1.76, p-value = .0812
: t = 2.45, p-value = .0162
We can infer that two–story and back–split houses sell for more than other.
18.18a
b0
0
t = –1.43, p-value = .1589. There is not enough evidence to infer that the type of commercial affects memory test scores.
c Let
= 1 if humorous
= 0 otherwise
= 1 if musical
= 0 otherwise
See Excel output below.
d 0
0
I1: t = 1.61, p-value = .1130
I2: t = 3.01, p-value = .0039
There is enough evidence to infer that there is a difference in memory test scores between watchers of humorous and serious commercials.
e The variable type of commercial in parts (a) and (b) is nominal. It is usually meaningless to conduct a regression analysis with such variables without converting them to indicator variables.
18.19 a
b Let= 1 if morning
= 0 otherwise
= 1 if early afternoon
= 0 otherwise
c Model 1:= 6.25 and = .8525.
Model 2:= 3.82 and = .9461.
The second model fits better.
d0
0
I1: t = –4.51, p-value = 0. There is enough evidence to infer that the average time to unload in the morning is different from that in the late afternoon.
I2: t = 4.47, p-value = .0001. There is enough evidence to infer that the average time to unload in the early afternoon is different from that in the late afternoon.
18.20a Let
= 1 if no scorecard
= 0 otherwise
= 1 if scorecard overturned more than 10% of the time
= 0 otherwise
b
c = 4.20 and = .5327. The model's fit is mediocre.
d
There is a high correlation betweenand that may distort the t–tests.
e =.00012; in this sample for each additional dollar lent the default rate increases by .00012 provided the other variables remain the same.
= 4.08; In this sample banks that don't use scorecards on average have default rates 4.08 percentage points higher than banks that overturn their scorecards less than 10% of the time.
= 10.18; In this sample banks that overturn their scorecards more than 10% of the time on average have default rates 10.18 percentage points higher than banks that overturn their scorecards less than 10% of the time.
f
We predict that the bank's default rate will fall between 1.39 and 18.49%.
18.21 a Let
= 1 if welding machine
= 0 otherwise
= 1 if lathe
= 0 otherwise
b = 2.54; in this sample for each additional month repair costs increase on average by $2.54 provided that the other variable remains constant.
= –11.76; in this sample welding machines cost on average $11.76 less to repair than stamping machines for the same age of machine.
= –199.4; in this sample lathes cost on average $199.40 less to repair than stamping machines for the same age of machine.
c0
< 0
t = –.60, p-value .5531/2 = .2766. There is no evidence to infer that welding machines cost less to repair than stamping machines.
18.22a
b0
0
t = 3.11, p-value = .0025. There is enough evidence to infer that the availability of shiftwork affects absenteeism.
c0
0
t = –5.36, p-value =(5.99E-07) /2 = 3.00×10-7= virtually 0. There is enough evidence to infer that in organizations where the union–management relationship is good absenteeism is lower.
18.23
a
At least on is not equal to 0
F = 344.04, p-value = 0. There is enough evidence to infer that the model is valid.
b 0
> 0
t = 1.72, p-value = .0879/2 = .0440. There is evidence that male professors are better paid than female professors with the same qualifications.
18.24
In this case male–dominated jobs are paid on average $.039 (3.9 cents) less than female–dominated jobs after adjusting for the value of each job.
18.25 All weights = .2
In this case male–dominated jobs are paid on average $.26 (26 cents) more than female–dominated jobs after adjusting for the value of each job.
18.26 The strength of this approach lies in regression analysis. This statistical technique allows us to determine whether gender is a factor in determining salaries. However, the conclusion is very much dependent upon the subjective assignment of weights. Change the value of the weights and a totally different conclusion is achieved.
18.27
Probability of heart attack =
18.28
Probability of heart attack =
18.29
Probability of heart attack =
18.30
Probability of heart attack =
18.31The variable to increase is the one with the largest coefficient. Thus, the fastest way of improving a credit rating is getting 1 year older.
18.32
Applicant 1: = 2.5668
Probability of repaying loan =
Applicant 2: = 3.7040
Probability of repaying loan =
Applicant 3: = 2.6392
Probability of repaying loan =
For each applicant the probability of repaying the loan increased.
18.33 Applicant 1: = 2.5088
Probability of repaying loan =
Applicant 2: = 3.6460
Probability of repaying loan =
Applicant 3: = 2.5812
Probability of repaying loan =
For each applicant the probability of repaying the loan increased.
18.34
Probability of repaying the loan =
18.35 An increase in the value of any of the variables results in a decrease in the probability of repaying the loan.
18.36
= 1.0154
Probability of repaying loan =
18.37
= 1.6197
Probability of repaying loan =
18.38a
b The stepwise regression equation includes the independent variables that are significantly related to the dependent variable. It does not include the variables undergraduate GPA,
18.39a
b In this equation only income has not been included.
c In the regression equation produced in Exercise 17.9 age and income were not statistically significant. By excluding income the independent variable age is now found to be statistically significant using stepwise regression.
18.40a
b In the stepwise regression equation both the number of minor league home runs and the number of years as a profession are significant. In Exercise 17.4 only the number of minor league home runs was significant.
cThe difference is that by eliminating age as an independent variable allowed the stepwise regression to reveal that the number of years as a profession is significant.
18.41a
b In the regression analysis in Exercise 17.13 age and children were not statistically significant. Stepwise regression excludes these variables.
18.42a Mileage = + Speed +Speed+
b
c = 3.86 and = .7102. The model fits moderately well.
18.43a Apply a first–order model with interaction.
b
c:
At least on is not equal to 0
F = 54.14, p-value = 0. There is enough evidence to infer that the model is valid.
18.44a
b F = 32.65, p-value = 0. There is enough evidence to infer that the model is valid.
18.45 a Let
= 1 if ad was in newspaper
= 0 otherwise
= 1 if ad was on radio
= 0 otherwise
b
b
At least on is not equal to 0
F = 14.91, p-value = 0. There is enough evidence to infer that the model is valid.
c 0
0
I–1: t = –1.48, p-value = .1467
I–2: t = –2.83, p-value = .0067
There is enough evidence to infer that the advertising medium makes a difference.
18.46 (See Excel output below)
b 0
< 0
t = –8.61, p-value = 0. There is enough evidence to infer that a team that fires its manager within 12 months wins less frequently than other teams.
18.47a Units = +Years +Years+
b
c = 87.98 and= .1893. The model fits poorly.
18.48a Depletion = +Temperature +PH–level +PH–level+ ++
where
= 1 if mainly cloudy
= 0 otherwise
= 1 if sunny
= 0 otherwise
b
c
At least on is not equal to 0
F = 77.00, p-value = 0. There is enough evidence to infer that the model is valid.
d 0
> 0
t = 6.78, p-value = 0. There is enough evidence to infer that higher temperatures deplete chlorine more quickly.
e 0
> 0
t = 18.07, p-value = 0. There is enough evidence to infer that there is a quadratic relationship between chlorine depletion and PH level.
f 0
0
: t = –1.53, p-value = .1282. There is not enough evidence to infer that chlorine depletion differs between mainly cloudy days and partly sunny days.
: t = 1.65, p-value = .0997. There is not enough evidence to infer that chlorine depletion differs between sunny days and partly sunny days.
Weather is not a factor in chlorine depletion.
1