Question

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	13.9	22.7	17.8
23.9	9.4	20.5	16.6
14.4	11.5	7.1	14.2
8.1	8.7	12.9	15.4
11.2	6.8	7.7	10.2
	19.2		9.5

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.

H_o: μ₁ = μ₂ = μ₃ = μ₄; H₁: Exactly three means are equal.H_o: μ₁ = μ₂ = μ₃ = μ₄; H₁: All four means are different.     H_o: μ₁ = μ₂ = μ₃ = μ₄; H₁: Not all the means are equal.H_o: μ₁ = μ₂ = μ₃ = μ₄; H₁: Exactly two means are equal.

(b) Find SS_TOT, SS_BET, and SS_W and check that SS_TOT = SS_BET + SS_W. (Use 3 decimal places.)

SSTOT	=
SSBET	=
SSW	=

Find d.f._BET, d.f._W, MS_BET, and MS_W. (Use 3 decimal places for MS_BET, and MS_W.)

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

Answer 1

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
I	5	84.9	16.98	68.327
II	6	69.5	11.58333	19.86967
III	5	70.9	14.18	51.572
IV	6	83.7	13.95	11.575
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	79.575	3	26.525	0.750	0.537	3.160
Within Groups	636.819	18	35.379
Total	716.395	21

Note - I have used Excel in Excel the command is as follows

Data > Data Analysis > Single factor > Select data > ok

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Ho: μ1 = μ2 = μ3 = μ4

H1: Not all the means are equal.

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b) SSTOT = 716.395

SSBET = 79.575

SSW = 636.819

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dfBET = 3

dfW = 18

MSBET = 26.525

MSW = 35.379

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c) P-value = 0.537

so, P-value > 0.100

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d) Since the P-value is greater than the level of significance at α = 0.05, we do not reject H0

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e) At the 5% level of significance there is sufficient evidence to conclude that the means are not all equal.

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f)

ANOVA
Source of Variation	SS	df	MS	F	P-value	Test
Between Groups	79.575	3	26.525	0.750	0.537	Do not reject H
Within Groups	636.819	18	35.379		p-value > 0.100
Total	716.395	21