In: Statistics and Probability
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 |
State the null hypothesis.
What is the decision rule? (Round your answer to 2 decimal places.)
Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places.)
Analysis of Variance
State your decision regarding the null hypothesis.
Let be mean number of hours spent per week by Banking industry,
be mean number of hours spent per week by Retail industry,
be mean number of hours spent per week by Insurance industry
Here we have to test that
Null hypothesis :
Alternative hypothesis : At least one is different.
32 | 1.44 | 28 | 0 | 30 | 2.56 | |
30 | 0.64 | 28 | 0 | 28 | 0.16 | |
30 | 0.64 | 26 | 4 | 26 | 5.76 | |
32 | 1.44 | 28 | 0 | 28 | 0.16 | |
30 | 0.64 | 30 | 4 | 30 | 2.56 | |
Sum | 154 | 4.8 | 140 | 8 | 142 | 11.2 |
Mean | 154/5 = 30.8 | 140/5 = 28 | 142/5 =28.4 |
Grand mean :
(Round to 2 decimal)
k = number of groups = 3
n = n1+n2+n3 = 5 + 5 + 5 = 15
Degrees of freedom for between groups = k - 1 = 3 - 1 = 2
Degrees of freedom for within groups = n - k = 15 - 3 = 12
Total degrees of freedom = n - 1 = 15 - 1 = 14
Between group Sum of square(SSG) :
SSG = 22.93 (Round to 2 decimal)
Within group Sum of square(SSE) :
SSE = 24
Total sum of squares(SST) = SSG + SSE = 22.93 + 24 = 46.93
SST = 46.93
Mean square for between groups (MSG) :
where is degrees of freedom for between groups
Mean square for with groups (MSE) :
where is degrees of freedom for within groups
Test statistic :
F = 5.74 (Round to 2 decimal)
Test statistic = F = 5.74
Significance level = = 0.05
Numerator degrees of freedom = Between degrees of freedom =
Denominator degrees of freedom = Within degrees of freedom =
F critical value for = 0.05, , from excel using function:
=F.INV.RT(0.05,2,12)
=3.89 (Round to 2 decimal)
F critical value = 3.89
Here test statistic > critical value
So we reject the null hypothesis.
Conclusion : The manager can conclude that there is a difference in the mean number of hours spent per week by industry.
ANOVA Table :
Source of variation | df | SS | MS | F |
Between group | 2 | 22.93 | 11.47 | 5.74 |
Within group | 12 | 24 | 2 | |
Total | 14 | 46.93 |