CHAPTER 4: Solutions to Selected Exercises

4.1 a.The plot of price versus size has a straight-line appearance and thus the model is appropriate. The plot of price versus rating has a straight-line appearance. The model is appropriate. Combining these two models, we obtain the model .

b.is the mean (or average) of the sales prices of all houses having 20 hundred (that is, 2000) square feet and a niceness rating of 9.

c.= mean sales price of all houses having 0 square feet and 0 niceness rating – meaningless.

= change in mean sales price associated with each increase in house size of 100 square feet, when niceness rating stays constant.

= change in mean sales price associated with each increase in niceness rating of 1, when house size remains constant.

The error term represents all factors other than the square footage and the niceness rating.

One such factor is the ability and effort of the real estate agent listing the house.

4.2The plots suggest a linear relationship between each of the independent variables and hours.

= meaningless

= The change in mean monthly labor hours associated with an increase of one x-ray exposure, when the other independent variables are held constant.

- The change in mean monthly labor hours associated with an increase of one in the number of monthly bed days, when the other independent variables are held constant.

= The change in mean monthly labor hours associated with a one day increase in the average length of stay, when the other independent variables are held constant.

The error term represents all factors other than . Such factors could involve patient load, size of hospital, etc.

4.3a.

= meaningless

= 5.6128 implies that we estimate that mean sales price increases by $5,612.80 for each increase of 100 square feet in house size, when the niceness rating stays constant.

= 3.8344 implies that we estimate that mean sales price increases by $3,834.40 for each increase in niceness rating of 1, when the square footage remains constant.

172.28. From

4.4a.=1523.38924 =.05299 =.97848 = -320.95083

b.16,065 is computed by

c.hours

Actual hours exceeds predicted hours by 1,142.31

4.5a.

b. Total variation = 7447.5

Unexplained variation = 73.6

Explained variation = 7374

= .987

d. F(model) =

e. Based on 2 and 7 degrees of freedom, = 4.74. Since F(model) = 350.87 > 4.74, we

reject by setting =.05.

f. Based on 2 and 7 degrees of freedom, = 9.55. Since F (model) = 350.87 > 9.55, we

reject by setting = .01.

g. p-value = 0.000 (which means less than.001). Since this p-value is less than = .10,

.05, .01, and .001, we have extremely strong evidence that is false. That is, we have extremely strong evidence that at least one of is significantly related to y.

4.6a.SSE = 4,913,399

Total Variation = 494,712,540

Unexplained Variation = 4,913,399

Explained Variation = 489,799,142

d. F (model) =

Based on 3 and 13 degrees of freedom, = 3.41

F (model) = 431.97 > =3.41. Reject at =.05

Based on 3 and 13 degrees of freedom, =4.35.

F (model) = 431.97 > =4.35. Reject at =.01

p-value < .0001. Reject . We have extremely strong evidence that at

least one of is significantly related to y.

4.7We first consider the intercept

a.= 29.347, =4.891, t = 6.00

where t =

b. We reject (and conclude that the intercept is significant) with if

Since = 2.365 (with n – (k + 1) = 10 – (2 + 1) = 7 degrees of freedom), we have

t = 6.00 > = 2.365.

We reject with and conclude that the intercept is significant at the .05 level.

We reject with if |t| >

Since = 3.499, we have t = 6.00 > t= 3.499.

We reject with and conclude that the intercept is significant at the .01 level.

The Minitab output tells us that the p-value for testing is 0.000. Since this p-value is less than each given value of , we reject at each of these values of . We can conclude that the intercept is significant at the .10, .05, .01, and the .001 levels of significance.

A 95% confidence interval for is

= [29.347 2.365 (4.891)]

= [17.780, 40.914]

This interval has no practical interpretation since is meaningless.

A 99% confidence interval for is

= [12.233, 46.461]

We next consider .

where= 5.6128 / .2285 = 24.56

b., c., and d.:

We reject (and conclude that the independent variable is significant) at level of significance if (based on n – (k + 1) = 10 – 3 = 7 d.f.)

For = .05, , and for = .01,

Since t = 24.56 > , we reject with = .05.

Since t = 24.56 > , we reject with = .01.

Further, the Minitab output tells us that the p-value related to testing is 0.000. Since this p-value is less than each given value of , we reject at each of these values of (.10, .05, .01, and .001).

The rejection points and p-value tell us to reject with = .10, = .05, = .01, and = .001. We conclude that the independent variable (home size) is significant at the .10, .05, .01, and .001 levels of significance.

e. and f.:

95% interval for :

99% interval for :

For instance, we are 95% confident that the mean sales price increases by between $5072 and $6153 for each increase of 100 square feet in home size, when the rating stays constant.

Last, we consider .

where= 3.8344 / .4332 = 8.85

b., c. and d.:

We reject (and conclude that the independent variable is significant) at level of significance if .

For = .05, , and for = .01,

Since t = 8.85 > , we reject with = .05.

Since t = 8.85 > , we reject with = .01.

The rejection points and p-value tell us to reject with = .10, = .05, = .01, and = .001.

We conclude that the independent variable (niceness rating) is significant at the .10, .05, .01, and .001 levels of significance.

e. and f.:

95% interval for:

99% interval for :

For instance, we are 95% confident that the mean sales price increases by between $2810 and $4860 for each increase of one rating point, when the home size stays constant.

4.8Same process as Exercise 4.7

. Do not reject at

. Reject at , not .01

. Reject at

. Do not reject at , .01

p-value for testing is .0205. Reject at .

is <.0001. Reject at =.001.

is .0583. Do not reject at

but can reject a .

95% C.I.

99% C.I.

4.9a.Point estimate is

95% confidence interval is [168.56, 175.99]

b.Point prediction is

95% prediction interval is [163.76, 180.80]

Stdev Fit =

This implies that Distance value = (1.57 /s)

= (1.57 / 3.242)

= 0.2345

The 99% confidence interval for mean sales price is

= [172.28 3.499 (1.57)]

= [172.28 5.49]

= [166.79, 177.77]

The 99% prediction interval for an individual sales price is

= [172.28 3.499 (3.242)]

= [172.28 12.60]

= [159.68, 184.88]

4.10y = 17,207.31 is within the P.I. interval [14,511, 17,618]. There is no statistical evidence to say the labor hours are unusually high or low for this hospital.

4.11=30,626 + 3.893 (28,000) – 29,607 (1.56) + 86.52 (1821.7) = 251,056.564

4.12For a house of a given square footage and age, each additional room adds $6321.78 to the price of the home when the number of bedrooms stays constant, while adding a bedroom deducts $11,103.16 from the price when the number of rooms stays constant.

4.13a.The straight line appearance of the plot of y versus suggests that the model

might appropriately relate y to . The possibly quadratic appearance of the plot of y versus suggests that the model

might appropriately relate y to . Combining these models, we obtain the model

which might appropriately relate y to and .

The p-value related to F(model) is 0.000. Since this p-value is less than .001, we have extremely strong evidence that the model is significant.

The p-values related to and are, respectively, 0.000, 0.000, and 0.006. Since each of these p-values is less than .01, we have very strong evidence that each of and is significant (important).

The point prediction is = 171.222 ($171,222)

The 95% prediction interval is [166.365, 176.079]

We are 95% confident that the sales price for an individual house with 2,000 square feet and a “niceness rating” of 8 will be between $166,365 and $176,079.

4.14a.The plots have a quadratic appearance.

b.(1)= 35.0261

95% C.I. = [34.4997, 35.5525]

(2)= 35.0261

95% P.I. = [35.5954, 36.4568]

4.15 a.The p-value for is .014. Since this p-value is less than .05, we have strong evidence that is important.

b.171.751, [168.835, 174.666]

The length of this interval is 174.666 – 168.835 = 5.83

The length of the interval for the model in Figure 4.27 is 176.079 – 166.36 = 9.72

Hence, the interaction model is giving us a more accurate estimate for the sales price of a house with 2,000 square feet and a rating of 8.

4.16

when = 13when = 22

= 39.8934 + 5.31076 (13)= 39.8934 + 5.31076 (22)

= 39.8934 + 69.03988 = 39.8934 + 116.83672

= 108.93328 = 156.73012

when = 13when = 22

= 51.7668 + 5.99914 (13)= 51.7668 + 5.99914 (22)

= 51.7668 + 77.98882 = 51.7668 + 131.98108

= 129.75562 = 183.74788

c.One can see from the slopes in the two equations that the slope of 5.99914 when is somewhat larger than the slope of 5.31076 when . The graph also shows that the line for rises a little faster. Thus we estimate that as square feet (home size) increases, the mean sales price increases faster when the rating is 8 than when the rating is 2.

4.17 = 6.0599; 95% P.I. = [3.7578, 8.3620]; 95% confident the actual profit for a future construction project with a contract size of $480,000 and a supervisor with 6 years experience will be between $375,780 and $836,200.

4.18 a.

When ,

When .

When

Plot of against (for = 3, 4, and 5) when

When =3,

When =4,

When =5,

Plot of against (for = 3, 4, and 5) when

When =3,

When =4,

When =5,

Plot of against (for = 3, 4, and 5) when

Plot of against (for = 3, 4, and 5)

These plots suggest that:

For a low level of supervisor experience (), profit decreases substantially as the contract size () increases.

For a moderate level of supervisor experience (), profit decreases when the contract size () becomes large.

For a high level of supervisor experience (), profit increases as the contract size () increases.

These results suggest that inexperienced supervisors should be assigned to smaller contracts, while the most experienced supervisors should be assigned to the largest contracts. Medium sized contracts can be assigned to supervisors with medium experience levels with slightly better results than would be obtained when an inexperienced supervisor is assigned, and the slightly worse results than would be obtained when a more experienced supervisor is assigned.

If we set 6.562 + 1.9476 we find 12.743 = 3.4342. (See part a when . See part b when .) If we set 6.562 + 1.9476- .7522() = -18.924 + 8.816- .7522 () we find 25.486 = 6.8684or = 3.71. (See part a when . See part c when .) Therefore, all three curves meet at a point corresponding to a contract size of = 3.71.

(i)The curve when . Supervisors with 2 years of experience should be assigned to contracts of size less than = 3.71.

(ii)The curve when . Supervisors with 6 years of experience should be assigned to contracts of size greater than = 3.71.

(iii)Contract sizes “near” = 3.71 should be assigned to supervisors with 4 years of experience.

4.19The effect on attendance from a promotion for a day game if all other variables remain the same is (Promotion = 1, Daygame = 1)

4,745 (1) + 5,059 (1) (1) – 4,690 (1) (Weekend) + 696.5 (1) (Rival)

The effect on attendance from a promotion for a night game if all other variables remain the same is (Promotion = 1, Daygame = 0)

4,745 (1) + 5,059 (1) (0) – 4,690 (1) (Weekend) + 696.5 (1) (Rival)

Thus, the impact of promotion on attendance is greater by 5,059 for a day game. For example, a promotion for a day game on a weekday against a rival is expected to increase average attendance by 4,745 + 5,059 – 0 + 696.5 = 10,500.5. For a night game on a weekday against a rival the increase would be 4745 + 0 – 0 + 696.5 = 5,441.5 (i.e. 5,059 less). By a similar argument, promotion on a weekend would decrease the increase in attendance by 4,690 compared to promotion on a weekday.

4.20a.The lines relating the size of the firm and average months to adoptions are parallel, indicating no interaction of size and type of firm. Also, the two lines are different.

b.equals the difference between the mean innovation adoption times of stock companies and mutual companies.

c.p-value is less than .001; Reject at both levels of .

is significant at both levels of .

95% C.I. for : [4.9770, 11.1339]; 95% confident that for any given size of insurance firm, the mean adoption time for an insurance innovation is between 4.9770 and 11.1339 months longer if the firm is a stock company rather than a mutual company.

No interaction.

4.21a.

F = 184.57, p-value < .001:

Reject ; conclude the means are not equal, that is, at least one is different.

95% C.I. for

d.77.20, [75.040, 79.360], [71.486, 82.914]

e.[25.700 2.131 (1.433)] = [22.646, 28.754]

For Or since 0 is not in 95% C.I. for

4.22a.

95% C.I. for

Both p-values < .01, are significant.

95% C.I. for mean demand: [8.4037, 8.5977]

We are 95% confident that the mean demand for all sales periods when the price difference is .20, the advertising expenditure is 6.50, and campaign C is used will be between 840,370 and 859,770.

95% P.I. for individual demand: [8.2132, 8.7881]

We are 95% confident that the actual demand in a particular sales period when the price difference is .20, the advertising expenditure is 6.50, and campaign C is used will be between 821,322 bottles and 878,813 bottles.

c.[.0363, .2999]; p-value = .0147 significant at = .05.

95% C.I. for: [.16809 2.069 (.06371)] = [.0363, .2999]

4.23a.

Both differences increased with the larger value of a.

95% P. I. [8.2249, 8.7988] length = 8.7988 – 8.2249 = .5739

For P.I. in Exercise 4.22 (Figure 4.36),

Length = 8.7881 - 8.2132 = .5739

The intervals are essentially the same length. This is not surprising since the p-values for the two additional interaction terms both exceed = .10.

4.24To test in Model 2, we note that Model 2 is the complete model, and therefore k = 6. If is true, Model 2 reduces to Model 1, which is the reduced model, and therefore k – g = 2. It follows that

Based on 2 and 23 degrees of freedom = 3.42 and = 5.66. Since F = 19.599 is greater than and, we reject at = .05 and = .01. We have seen in Exercise 4.22 that Therefore, if , it follows that That is, is equivalent to If is true, the mean demand does not differ with type of advertising. We have shown that at = .01, the mean demand for at least one type of advertising is different from the other two means when price difference and advertising expenditures remain the same.

4.25Model 3 – complete

Model 1 – reduced

= 2.84 based on 4 and 21 degrees of freedom.

= 4.37 based on 4 and 21 degrees of freedom.

Since 10.634 > 4.37, reject at = .05 and .01.

The type of advertising has an impact on mean demand.

4.26Model 3 – complete

Model 2 – reduced

= 3.47 based on 2 and 21 degrees of freedom.

Since 1.248 < 3.47, do not reject at = .05 and .01.

The effect of the type of advertising on demand is not altered by amount spent on advertising.

4.27

4.28a.group 0: