In: Statistics and Probability
A steel company is considering the relocation of one of its manufacturing plants. The company’s executives have selected four areas that they believe are suitable locations. However, they want to determine if the average wages are significantly different in any of the locations, since this could have a major impact on the cost of production. A survey of hourly wages of similar workers in each of the four areas is performed with the following results. Do the data indicate a significant difference among the average hourly wages in the three areas? Hourly Wages ($) Area 1 Area 2 Area 3 22 16 17 15 8 23 12 22 24 21 17 25 15 11 9 17 9 19 14 11 12 21 10 23 Step 1 of 2 : Find the value of the test statistic to test for a difference in the areas. Round your answer to two decimal places, if necessary.
Area 1 | Area 2 | Area 3 |
22 | 24 | 9 |
16 | 21 | 19 |
17 | 17 | 14 |
15 | 25 | 11 |
8 | 15 | 12 |
23 | 11 | 21 |
12 | 9 | 10 |
22 | 17 | 23 |
Method: One way Anova
Null hypothesis | All means are equal |
Alternative hypothesis | Not all means are equal |
Significance level | α = 0.05 |
Equal variances were assumed for the analysis.
Means
Factor | N | Mean | StDev | 95% CI |
Area 1 | 8 | 16.88 | 5.30 | (12.84, 20.91) |
Area 2 | 8 | 17.38 | 5.76 | (13.34, 21.41) |
Area 3 | 8 | 14.88 | 5.38 | (10.84, 18.91) |
Pooled StDev = 5.48428
Analysis of Variance
Source | DF | Adj SS | Adj MS | F-Value | P-Value |
Factor | 2 | 28.00 | 14.00 | 0.47 | 0.634 |
Error | 21 | 631.63 | 30.08 | ||
Total | 23 | 659.63 |
Therefore the value of test statistics = 0.47
and p-value = 0.63
Decision: since the p-value is 0.63 which is quite large value. so we don't have sufficient evidence to reject the null hypothesis. So, Ho can't be rejected.
Conclusion: There is no significant difference among the average hourly wages in the three area.