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
27.5   13.6   22.6   17.2
23.7   9.6   20.5   16.5
14.6   11.4   7.3   14.2
8.1   8.2   12.9   15.9
11.6   6.8   7.7   10.5
19.9       9.1
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: Not all the means are equal.
Ho: μ1 = μ2 = μ3 = μ4; H1: Exactly three means are equal.
Ho: μ1 = μ2 = μ3 = μ4; H1: All four means are different.
Ho: μ1 = μ2 = μ3 = μ4; H1: Exactly 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 > 0.100
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

(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  
  
Within groups  
Total  

Answer 1

treatment	A	B	C	D
count, ni =	6	5	6	5
mean , x̅ i =	17.567	9.92	13.350	14.86
std. dev., si =	7.426	2.671	6.685	2.678
sample variances, si^2 =	55.151	7.132	44.695	7.173
total sum	105.4	49.6	80.1	74.3	309.4	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	14.06
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	12.271	17.170	0.509	0.634
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	73.627	85.849	3.056	3.171	165.7025758
SS(within ) = SSW = Σ(n-1)s² =	275.753	28.528	223.475	28.692	556.4483

no. of treatment , k =   4
df between = k-1 =    3
N = Σn =   22
df within = N-k =   18
  
mean square between groups , MSB = SSB/k-1 =    55.2342
  
mean square within groups , MSW = SSW/N-k =    30.9138
  
F-stat = MSB/MSW =    1.7867

Ho: μ1 = μ2 = μ3 = μ4; H1: Not all the means are equal.

SS tot=   722.151
SS bet=   165.703
SSW=   556.448
dfBET   =   3
dfW   =   18
MSBET   =   55.234
MSW   =   30.914
      
F=   1.787  

dfn = 3

df d = 18

P-value > 0.100

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

e)

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

f)

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	165.703	3	55.234	1.787	0.1857
Within Groups	556.448	18	30.914
Total	722.151	21