A. Do a Multiple Regression of Passengers Against Advertising and National Income. (12)

6/10/99 252zz9943

5. Data from problem 4 is repeated below. (Use )

a. Do a multiple regression of passengers against advertising and National Income. (12)

b. Compute and adjusted for degrees of freedom for both this and the previous problem. Compare the values of adjusted between this and the previous problem. Use an F test to compare here with the from the previous problem.(5)

c. Compute the regression sum of squares and use it in an F test to test the usefulness of this regression. (5)

d. Use your regression to predict the number of passengers when we spend $13 (thousand) on advertising and National Income is $3.5 (trillion).(2)

e. The regression on the previous page was run with the command

MTW > regress C1 on 1 C3;

SUBC > dw.

As a result, the last line of the regression read

Durbin-Watson statistic = 0.71

Solution: a) First, we compute . Second, we compute , , , , and . Third, we compute our spare parts , , , , and . (Note that some of these were computed for the last problem.) Fourth, we substitute these numbers into the Simplified Normal Equations:

which are

and solve them as two equations in two unknowns for . We do this by multiplying the second equation by 4.5115, which is 23.0979 divided by 5.11975 so that the two equations become , we then subtract the second equation from the first to get , so that . The first of the two normal equations can now be rearranged to get , which gives us . Finally we get by solving . Thus our equation is

b) The coefficient of determination is . (The standard error is

6/10/99 252zz9943

, but we don’t need it yet.) Our results can be summarized below as:

.8104 / 15 / 1 / .7958
.9430 / 15 / 2 / .9335

, which is adjusted for degrees of freedom, has the formula , where is the number of independent variables. adjusted for degrees of freedom seems to show that our second regression is better.

The easiest way to do the F test and have it look right is to note that . For the regression with one independent variable the regression sum of squares is . For the regression with two independent variables the regression sum of squares is . The difference between these is 25.954. the remaining unexplained variation is 195.733 –184.576 = 11.157. the ANOVA table is

Source / SS / DF / MS / /
/ 158.622 / 1 / 158.622
/ 25.954 / 1 / 25.954 / 27.9105 /
Error / 11.157 / 12 / 0.9299
Total / 195.733 / 14

Since our computed is larger than the table , we reject our null hypothesis that has no effect.

c) We computed the regression sum of squares in the previous section.

Source / SS / DF / MS / /
, / 184.576 / 2 / 92.288 / 99.245 /
Error / 11.157 / 12 / 0.9299
Total / 195.733 / 14

Since our computed is larger than the table , we reject our null hypothesis that and do not explain .

d) =11.103.

e) A Durbin-Watson Test is a test for autocorrelation. For , and , the test table gives and .According to the text, the null hypothesis is ‘No Autocorrelation’ and our rejection region is or . We really should use the value for , but a check of the table leaves us sure that it is below .70. thus the D-W statistic of 0.71 is not in the rejection region. Check the examples to see that it could be in the “possibly significant” region.

6/14/99 252zz9943

6.(Watch it!) Three methods are used to train candidates for the FAA pilots exam. Scores for trainees are shown below classified by method.

Methoda. Assume that the data is normal and compare the means

Video Audio for the first two methods (Assume unequal variances) (5)

CassetteCassette Classroomb. Do the same for all three methods (You may assume

72 73 68 equal variances now) (7)

86 75 83

80 60 50c. Test column 1 to see if it has the normal distribution (5)

91 52 91

46 84 84

68 76 77

75 94

Note: For the first column:

Note:

Note: In spite of the words “Watch it!, ” many people assumed that this was identical to a problem with similar data on an earlier exam. You have to read the question before answering it!

Solution: Note:

a) Assume unequal variances. From Table 3 of the Syllabus Supplement:

Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical Value
Difference
between Two
Means(
unknown,
variances
assumed
unequal) /
/
Same as
/ /

6/14/99 252zz9943

, so use 10 degrees of freedom.

, so, using a test ratio . Since this is between , do not reject or, using a critical value, . Since is between these values, do not reject .

b) 1-way ANOVA

Method
72 / 73 / 68
86 / 75 / 83
80 / 60 / 50
91 / 52 / 91
46 / 84 / 84
68 / 76 / 77
75 / 94
81
92
. / . / 90
Sum / 518 / + 420 / +810 / = 1748
/ 7 / + 6 / + 10 / = 23
/ 74 / 70 / 81 / 76
SS / 39626 / + 30090 / + 67240 / = 136956
/ 5476 / 4900 / 6561

Note that is not a sum, but is . .

. ()

Source

SS

DF

MS

/ / /
Between (Methods) / 494 / 2 / 247 / 1.365 / ns / Column means equal
Within (Error) / 3614 / 20 / 181
Total / 4108 / 22

Because our computed is smaller than , we cannot reject

6/14/99 252zz9943

c) We use the Lilliefors method because we are testing for the Normal distribution, we have a small sample and the population mean and variance are unknown. The column is the cumulative distribution computed from the Normal table. is , which was computed for you.

Cumulative

46 1 1 .14286 -1.91 .0281 .1148

68 1 2 .28571 -0.41 .3409 .0552

72 1 3 .42857 -0.14 .4443 .0157

75 1 4 .57142 0.07 .5279 .0435

80 1 5 .71428 0.41 .6591 .0552

86 1 6 .85714 0.82 .7939 .0632

91 1 7 1.00000 1.16 .8770 .1230

From the Lilliefors Table, the critical value for a 95% confidence level is .300. Since the largest number in is not above this value, we do not reject .

6/14/99 252zz9943

7. (Watch it!) Three methods are used to train candidates for the FAA pilots exam. Scores for trainees are shown below classified by method.

Methoda. Using a sign test, check to see if it has a median of 85. (4)

Video Audio

CassetteCassetteClassroomb. Repeat the test on using a more powerful method. (5)

72 73 68 c. Apply the Runs Test as follows:

86 75 83Write down the numbers in and together in order.

80 60 50Underneath the numbers write down A if the number comes from

91 52 91 and C if it comes from. You will have a sequence like AACACC

46 84 84….. . In case of a tie remove both tying numbers from your test.

68 76 77Do a runs test on the resulting sequence to see if the A’s and C’s

75 94appear randomly.

81Congratulations! You have just done a Wald-Wolfowitz Test for the

92equality of means in two (nonnormal) samples. If the sequence is

90random, the means are equal. (6)

Solution:

(corrected)

68 -17 9 9-

83 -2 2 2-

50 -35 10 10-

91 6 5 5

84 -1 1 1-

77 -8 7 7-

94 9 8 8

81 -4 3 3-

92 7 6 6

90 5 4 4

a) . To do a sign test, note that there are 6 numbers below 85 and 4 above. Using the binomial table with

If this p-value is above the significance level and we do not reject .

b) To do a Wilcoxon Signed Rank Sum Test, rank the differences from 85 and put the sign of the difference next to these ranks. To check , note that . From the table the critical value is 8, since both Ts are above this value, do not reject .

c) The numbers written out in order are:

If we eliminate ties we get:

The number of is and the number of is and there are runs. If we look this up in the table entitled “Critical Values of r for the Runs Test, ” we fine that the upper critical value is 11 and the lower critical value is 3. Since 8 lies between these values we do not reject the null hypothesis of randomness. Out final conclusion is that the means of the populations from which the two samples come are equal.