Business, 24.05.2021 19:00, ezequielromero41

The manager of a computer software company wishes to study the number of hours 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 .05 significance level, can she conclude there is a difference in the mean number of hours spent per week by industry? Banking Retail Insurance
12 8 10
10 8 8
10 6 6
12 8 8
10 10 10

Answers: 1

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Since the calculated value of F= 5.733 falls in the critical region we reject the null hypothesis and conclude all three means are not equal.

Explanation:

The given data is

Banking Retail Insurance

12 8 10

10 8 8

10 6 6

12 8 8

10 10 10

The results of excel are:

Anova: Single Factor

SUMMARY

Groups CountSumAverage Variance

Column 1554 10.8 1.2

Column 2540 8 2

Column 3542 8.4 2.8

1) Let the null and alternate hypotheses be

H0: u1=u2=u3 i.e all the three means are equal and

Ha: Not all three means are equal

2) The significance level is set at ∝ =0.05

3)The test statistic to use is

F= sb²/ sw²

which has F distribution with v1= k-1 →3-1=2 and v2= n-k →15-3=12 degrees of freedom

After calculations the following table is obtained.

ANOVA

Source SS df MS F P-value F crit

of Variation

B/w Groups 22.93 2 11.467 5.733 0.01788 3.885

Within Groups 24 12 2

Total 46.93 14

4) The critical region is F ≥ F(0.05, 2,12) = 3.885

5) Since the calculated value of F= 5.733 falls in the critical region we reject the null hypothesis and conclude all three means are not equal.

The manager of a computer software company wishes to study the number of hours senior executives by

answered: Guest

the answer is 10c + 6b = 32

step-by-step explanation:

answered: Guest

Ibelieve it's the left middle finger

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