How productive are U.S. workers? One way to answer this question is to study annual profits per employee. A random sample of companies in computers (I), aerospace (II), heavy equipment (III), and broadcasting (IV) gave the following data regarding annual profits per employee (units in thousands of dollars).
I | II | III | IV |
27.3 | 13.5 | 22.2 | 17.5 |
23.8 | 9.8 | 20.3 | 16.9 |
14.9 | 11.9 | 7.9 | 14.7 |
8.4 | 8.8 | 12.4 | 15.3 |
11.4 | 6.1 | 7.4 | 10.5 |
19.8 | 9.2 |
Shall we reject or not reject the claim that there is no difference in population mean annual profits per employee in each of the four types of companies? Use a 5% level of significance.
(a) What is the level of significance?
(b) Find SSTOT, SSBET, and SSW and check that SSTOT = SSBET + SSW. (Use 3 decimal places.)
SSTOT | = | |
SSBET | = | |
SSW | = |
Find d.f.BET, d.f.W,
MSBET, and MSW. (Use 3 decimal
places for MSBET, and
MSW.)
dfBET | = | |
dfW | = | |
MSBET | = | |
MSW | = |
Find the value of the sample F statistic. (Use 3 decimal
places.)
What are the degrees of freedom?
(numerator)
(denominator)
(f) Make a summary table for your ANOVA test.
Source of Variation |
Sum of Squares |
Degrees of Freedom |
MS | F Ratio |
P Value | Test Decision |
Between groups | ||||||
Within groups | ||||||
Total |
applying one way ANOVA on above data:
a) level of significance=0.05
b)
SSTOT=705.458
SSBET=82.891
SSW=622.567
dfBET=3
dfW=18
MSBET=82.891/3=27.630
MSW=622.567/18=34.587
value of the sample F statistic =MSBET/MSE=27.630/34.587=0.799
degrees of freedom (numerator)=3
degrees of freedom (denominator)=18
Source of variation | SS | df | MS | F | p vlaue | Decision | |
treatments | 82.891 | 3.00 | 27.630 | 0.799 | 0.5105 | Do not reject Ho | |
error | 622.567 | 18.00 | 34.587 | ||||
total | 705.458 | 21.00 |
Get Answers For Free
Most questions answered within 1 hours.