In: Statistics and Probability
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 |
27.2 | 13.2 | 22.8 | 17.8 |
23.4 | 9.6 | 20.1 | 16.2 |
14.6 | 11.7 | 7.4 | 14.1 |
8.9 | 8.1 | 12.3 | 15.4 |
11.9 | 6.9 | 7.0 | 10.8 |
19.6 | 9.6 |
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?
State the null and alternate hypotheses.
Ho: μ1 = μ2 = μ3 = μ4; H1: All four means are different.Ho: μ1 = μ2 = μ3 = μ4; H1: Exactly two means are equal. Ho: μ1 = μ2 = μ3 = μ4; H1: Not all the means are equal.Ho: μ1 = μ2 = μ3 = μ4; H1: Exactly three means are equal.
(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)
(c) Find the P-value of the sample test statistic.
P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.001 < P-value < 0.010P-value < 0.001
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis?
Since the P-value is greater than the level of significance at α = 0.05, we do not reject H0.Since the P-value is less than or equal to the level of significance at α = 0.05, we reject H0. Since the P-value is greater than the level of significance at α = 0.05, we reject H0.Since the P-value is less than or equal to the level of significance at α = 0.05, we do not reject H0.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance there is insufficient evidence to conclude that the means are not all equal.At the 5% level of significance there is sufficient evidence to conclude that the means are all equal. At the 5% level of significance there is insufficient evidence to conclude that the means are all equal.At the 5% level of significance there is sufficient evidence to conclude that the means are not all equal.
(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 | ---Select--- p-value > 0.100 0.050 < p-value < 0.100 0.025 < p-value < 0.050 0.010 < p-value < 0.025 0.001 < p-value < 0.010 p-value < 0.001 | ---Select--- Do not reject H0. Reject H0. | ||||
Within groups | ||||||
Total |
(a) What is the level of significance?
Level of significance = 0.05
Group 1 | Group 2 | Group 3 | Group 4 | |
27.2 | 13.2 | 22.8 | 17.8 | |
23.4 | 9.6 | 20.1 | 16.2 | |
14.6 | 11.7 | 7.4 | 14.1 | |
8.9 | 8.1 | 12.3 | 15.4 | |
11.9 | 6.9 | 7.0 | 10.8 | |
19.6 | 9.6 | |||
Sum = | 86 | 69.1 | 69.6 | 83.9 |
Average = | 17.2 | 11.517 | 13.92 | 13.983 |
\sum_i X_{ij}^2 =∑iXij2= | 1721.38 | 900.67 | 1178.9 | 1224.05 |
St. Dev. = | 7.781 | 4.58 | 7.247 | 3.189 |
SS = | 242.18 | 104.86833333333 | 210.068 | 50.848333333334 |
n = | 5 | 6 | 5 | 6 |