Question

In: Statistics and Probability

The manager of a computer software company wishes to study the number of hours per week...

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

  1. State your decision regarding the null hypothesis.

Solutions

Expert Solution

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

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