Solutions to Chapter 13 homework problems asking for pairwise comparisons

MGT 249-1 Winter 1998


This page contains answers to the following homework problems from Chapter 13: 14, 15, 18, 32 and 59b. The answers differ from the answers in the book because in all cases I have solved the problems using the Bonferroni adjustment. I have also used a confidence interval in every case, not using the LSD test procedures described in the text.


14. Here we will use the output from number 6. I will first use the book method (Fisher's LSD), but then use the correction for doing multiple confidence intervals. I will do this for both 95% and 94% CI's.

Here are the numbers from the output for number 6.

xbar1 xbar2 MSW n1 n2 df
23 28 4.888889 4 4 9

Using the book's method:

a t   L R
0.05 2.262   -8.537 -1.463

So we are 95% confident that the difference between (population) mean 1 and mean 2 is between -8.537 and -1.463. Since 0 is not included in the interval, we can conclude that the difference is significant.


Using the Bonferroni adjustment to get a 95% confidence interval:

Here we need the number of possible pairwise comparisons. Since there are 3 groups, there are 3 comparisons.

a m t   L R
0.05 3 2.933   -9.586 -0.414

(NOTE: I found the value of t using =tinv(.05/3,9).)

With this method the interval is wider, but it still does not include 0.

Using the Bonferroni adjustment to get a 94% confidence interval:

Here we need the number of possible pairwise comparisons. Since there are 3 groups, there are 3 comparisons.

a m t   L R
0.06 3 2.821   -9.411 -0.589


15. This time I will only use Bonferroni adjustment method, and a 95% confidence interval. We again have m=3, since there are 3 groups.

xbar1 xbar2 xbar3 MSW n1 n2 n3 df
5 4.5 6 0.5 6 6 6 15

Comparison a m t L R
1 and 2 0.05 3 2.694 -0.600 1.600
1 and 3 0.05 3 2.694 -2.100 0.100
2 and 3 0.05 3 2.694 -2.600 -0.400

The only interval that does not include 0 is the third one, indicating that the means of groups 2 and 3 are significantly different.


18. I will again use Bonferroni adjustment method, and a 95% confidence interval. We again have m=3, since there are 3 groups.

xbar1 xbar2 xbar3 MSW n1 n2 n3 df
9.95 14.75 13.5 13.01963 12 8 10 27

Comparison a m t L R
1 and 2 0.05 3 2.552 -9.003 -0.597
1 and 3 0.05 3 2.552 -7.493 0.393
2 and 3 0.05 3 2.552 -3.118 5.618

Here only the first interval contains 0, indicating that industries 1 and 2 (banking and financial services) have different mean P/E ratios.


32. I will use the Bonferroni adjustment to build 95% confidence intervals.

xbar1 xbar2 xbar3 MSW n1 n2 n3 df
17 20.4 25 5.092222 7 7 7 18

Comparison a m t L R
1 and 2 0.05 3 2.639 -6.583 -0.217
1 and 3 0.05 3 2.639 -11.183 -4.817
2 and 3 0.05 3 2.639 -7.783 -1.417

None of the intervals contain 0, indicating that all differences are significant.


59. b. I will use the Bonferroni adjustment to build 95% confidence intervals.

xbar1 xbar2 xbar3 MSW n1 n2 n3 df
4.25 5.25 5.75 1.166667 8 8 8 21

Comparison a m t L R
1 and 2 0.15 3 2.08 -2.123 0.123
1 and 3 0.15 3 2.08 -2.623 -0.377
2 and 3 0.15 3 2.08 -1.623 0.623

The second interval does not contain 0, indicating that there is a significant difference between nonbrowsers and heavy browsers. There is not a significant difference between light and nonbrowsers and between light and heavy browsers.


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Last updated Feb. 11, 1998