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 |
49.3 | 55.7 | 38.7 |
43.3 | 25.0 | 37.3 |
32.6 | 41.5 | 10.9 |
27.1 | 29.4 | 32.9 |
38.5 | 39.9 | 15.5 |
36.5 | 42.2 | |
20.4 |
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: At least two means are equal.Ho: μ1 = μ2 = μ3; H1: All three means are different.
(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 |
Applying one way ANOVA: (use excel: data: data analysis: one way ANOVA: select Array): |
a)
level of significance =0.01
Ho: μ1 = μ2 = μ3; H1: Not all the means are equal.
b)
SSTOT | 2221.783 |
SSBET | 221.706 |
SSW | 2000.077 |
dfBET | 2 |
dfW | 15 |
MSBET | 110.853 |
MSW | 133.338 |
value of test statistic for factor A = | 0.831 | |
df(numerator) = | 2 | |
df(Denominator) = | 15 |
c)
P-value > 0.100
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 | SS | df | MS | F | P value |
treatment | 221.706 | 2 | 110.853 | 0.831 | 0.4546 |
error | 2000.077 | 15 | 133.338 | ||
Total | 2221.783 | 17 | |||
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