Question

The manager of a computer software company wishes to study the number of hours per week senior executives by type of industry spend at their desktop computers. The manager selected a sample of five executives from each of three industries. At the 0.05 significance level, can she conclude there is a difference in the mean number of hours spent per week by industry?

Banking	Retail	Insurance
32	28	30
30	28	28
30	26	26
32	28	28
30	30	30

1. State the null hypothesis.

2. What is the decision rule? (Round your answer to 2 decimal places.)

3. Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places.)

Analysis of Variance

Source	SS	df	MS	F
Treatments
Error

Answer 1

Here the test is to check whether the mean number of hours spent per week by the 3 industries viz. Banking, Retail and Insurance are different or not.

The null hypothesis is H0: µ1=µ2=µ3

vs the alternative H1:µi's are not equal for atleast one i=1,2,3 i.e to see whether there is difference in the true mean of hours spent per week by the 3 industries.

where, µ1,µ2 and µ3 are the true means or population means of Banking, Retail and Insurance respectively.

Let yij be the observations. Here the no of treatments or categories is 3 , i=1,2,..3

j=1,...,ni , ni is the no of observations in each class. here there are 5 observations in each class i.e n1=n2=n3=5

Assuming that the basic assumptions are met we will carry out the one-way ANOVA test.

Decision Rule:

The Decision rule is if the the observed F is > the critical value of F at level of significance alpha=0.05, then we will reject the null hypothesis , otherwise we wil fail to reject the null hypothesis.

The F critical value for alpha=0.05, and degrees of freedom (k-1) and (n-k) , where k is the number of treatments and n is the total number of observations. i.e degrees of freedom = k-1 =3-1=2 and n-k=15-3=12

F critical value =F(0.05,2,12) =3.89

So, Decision rule is if observed F is > 3.89, then we will reject the null hypothesis, otherwise we will fail to reject the null hypothesis.

Calculation for ANOVA table:

Now, Sum of Squares between the treatments (SSB) =∑i ni (yi0 - y00)^2   

with degrees of freedom = (k-1), where k is the no of classes.

Total Sum of Squares (TSS) =∑i ∑j (yij - y00)^2

with degrees of freedom = (n-1), where n is the total no of observations

Sum of Squares of Error= TSS - SSB with degrees of freedom = (n-k)

For simplicity of calculation we will use:

(SSB) =∑i ni (yi0 - y00)^2 = ∑i (Ti0^2)/ni - T00^2/n

where Ti0 are the total of each class and T00 is the grand total

So, T10=total for Banking industry=32+30+30+32+30=154

T20 = total for Retail Industry=28+28+26+28+30=140

T30=total for Insurance Industry= 30+28+26+28+30=142

and T00= grand total= T10+T20+T30=154+140+142=436

n =total no of observations=15, and n1=n2=n3=5

So, SSB=  ∑i (Ti0^2)/ni - T00^2/n

= ( T10^2/n1+T20^2/n2+T30^2/n3) - (436^2)/15

=(154^2/5+140^2/5+142^2/5) - (436^2)/15

=12696-12673.0667

=22.9333=22.93 with degrees of freedom = (3-1)=2

TSS=∑i ∑j (yij - y00)^2 = ∑i ∑j yij^2 - T00^2/n

∑i ∑j yij^2 is the sum of squares of all the observations =32^2+30^2+30^2+.....+26^2+28^2+30^2=12720

T00^2/n =12673.0667 [from above]

So, TSS = 12720 -12673.0667=46.9333=46.93 with degrees of freedom =n-1=15-1=14

SSE=TSS -SSA =46.9333-22.93333= 24 with degrees of freedom =n-k=15-3=12

So The ANOVA table is as follows:

Sources	SS	DF	MS	F
Treatments	22.93	2	11.47	5.74
Error	24	12	2
Total	46.93	14

[*MS= SS/DF and F =MS( Between Treatments)/ MS(Error) is the F statistic and P is the p-value corresponding to the observed F statistic at degrees of freedom 2 and 12.

MS(Between Treatments)=22.9333/2=11.47

MS(Error) =24/12=2

F=11.47/2 =5.74

So, according to the decision rule, since the observed F=5.74 is > critical value=3.89, we will reject the null hypothesis at α=0.05

Conclusion: We have enough evidence to conclude that there are significant differences between the mean hours spent per week by the 3 industries.