Does it appear that the average percentage salary increase is the same among the

Question

Does it appear that the average percentage salary increase is the same among the five industries?

A survey was conducted to compare next year's projected salary increases for five industries: Computer services and software, utilities, banking, manufacturing, and retail. Five companies within each industry were contacted and asked to report the average anticipated raise for its employees. The data are as follows Computer Utilities Banking Manufacturing Retail 4.9% 8.0% 75% 7.2% 73% 7.0% 6.2% 6.1% 5.7% 6.0% 6.7 6.6% 6.8% 0 53% 5.9% 62% 6.4% | 6.4% 6.8% Several companies refused to divulge their raise information and were deleted from the study. At = 0.01, find the critical point, p-value and fill in the following ANOVA table Sum of Sqs d.f Mean Square F Source Treatment Error Total

Explanation / Answer

The table below gives the data which has been calculated from the given figures.

The Hypothesis:

H0: There is no difference in the average percentage salary increase among the five industries.

Ha: There is a difference in the average percentage salary increase among at least 2 of the 5 industries.

The ANOVA table is as below. the p value is calculated for F = 8.15 for df1 = 4 and df2 = 16, p-value = 0.0009

The Fcritical is calculated at = 0.01 for df1 = 4 and df2 = 16, Fcritical = 4.77

The Calcutions of the above values have been done after the conclusion for your reference.

The Decision Rule:

If Ftest is > F critical, Then Reject H0.

Also if p-value is < , Then reject H0.

The Decision:

Since Ftest (8.15) is > F critical (4.77), We Reject H0.

Also since p-value (0.0009) is < (0.01), We Reject H0.

The Conclusion: There is sufficient evidence at the 99% level of significance to conclude that there is a difference in the average percentage salary increase among at least 2 of the 5 industries.

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The overall mean = 6.3619

SS treatment = SUM n* ( xi - overall mean)2 = 5 * (7.4 - 6.3619)2 + 4 * (6 - 6.3619)2 + 3 * (6.7 - 6.3619)2 + 5 * (5.9 - 6.3619)2 + 4 * ( 5.75 - 6.3619)2 = 8.82

df1 = k - 1 = 5 - 1 = 4

MSTR = SS treatment/df1 = 8.82 / 4 = 2.205

SSerror = SUM [(n - 1) * Variance] = 4 * 0.1 + 3 * 0.0467 + 2 * 0.01 + 4 * 0.2 + 3 * 0.99 = 4.33

df2 = N - k = 21 - 5 = 16

Therefore MS error = SSerror/df2 = 4.33/16 = 0.2706

F = MSTR/MSE = 2.205/0.2706 = 8.15

Computer Utilities Banking Manufacturing Retail Total Mean(x bar) 7.4 6 6.7 5.9 5.75 SD 0.3162278 0.21610183 0.1 0.447213595 0.995 0 Variance 0.1000 0.0467 0.0100 0.2000 0.9900 0 n 5 4 3 5 4 21 columns 2

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