In: Statistics and Probability
A random sample of companies in electric utilities (I), financial services (II), and food processing (III) gave the following information regarding annual profits per employee (units in thousands of dollars).
| I | II | III |
| 49.9 | 55.9 | 39.4 |
| 43.4 | 24.5 | 37.7 |
| 32.7 | 41.5 | 10.3 |
| 27.1 | 29.1 | 32.2 |
| 38.9 | 39.5 | 15.7 |
| 36.8 | 42.4 | |
| 20.6 |
Shall we reject or not reject the claim that there is no difference in population mean annual profits per employee in each of the three types of companies? Use a 1% level of significance.
(a) What is the level of significance?
State the null and alternate hypotheses.
Ho: μ1 = μ2 = μ3; H1: Not all the means are equal.Ho: μ1 = μ2 = μ3; H1: Exactly two means are equal. Ho: μ1 = μ2 = μ3; H1: All three means are different.Ho: μ1 = μ2 = μ3; H1: At least two 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 = 0.01, we do not reject H0.Since the P-value is less than or equal to the level of significance at α = 0.01, we reject H0. Since the P-value is greater than the level of significance at α = 0.01, we reject H0.Since the P-value is less than or equal to the level of significance at α = 0.01, we do not reject H0.
(e) Interpret your conclusion in the context of the
application.
At the 1% level of significance there is insufficient evidence to conclude that the means are not all equal.At the 1% level of significance there is sufficient evidence to conclude that the means are all equal. At the 1% level of significance there is insufficient evidence to conclude that the means are all equal.At the 1% 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)
level of significance =0.01
| Ho:μ1=μ2=μ3 | ||
| Ha:Not all the means are equal. | ||
b)
| SSTOT= | 2293.738 |
| SSBET= | 216.155 |
| SSW = | 2077.583 |
| dfBET= | 2 |
| dfW | 15 |
| MSBET | 108.078 |
| MSW= | 138.506 |
| value of the sample F statistic= | 0.780 | ||
c)P-value > 0.10
d)
| Since the P-value is greater than the level of significance at α = 0.01, we do not reject H0 |
e)
| At the 1% level of significance there is insufficient evidence to conclude that the means are not all equal. |
f)
| Source of variation | SS | df | MS | F | p vlaue | ||
| between | 216.155 | 2 | 108.0776 | 0.780 | p value>0.10 | fail to reject Ho | |
| within | 2077.583 | 15 | 138.5055 | ||||
| total | 2293.738 | 17 |