In: Statistics and Probability
Atlanta | New York | Los Angeles | Chicago |
650 | 250 | 850 | 540 |
480 | 525 | 700 | 450 |
550 | 300 | 950 | 675 |
600 | 175 | 780 | 550 |
675 | 500 | 600 | 600 |
minitab<stat<anova<one way anova
Results for: Worksheet 2
One-way ANOVA: Atlanta, New York, Los Angeles, Chicago
Method
Null hypothesis All means are equal
Alternative hypothesis At least one mean is different
Significance level α = 0.05
Equal variances were assumed for the analysis.
Factor Information
Factor Levels Values
Factor 4 Atlanta, New York, Los Angeles, Chicago
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Factor 3 456630 152210 11.03 0.000
Error 16 220770 13798
Total 19 677400
Model Summary
S R-sq R-sq(adj) R-sq(pred)
117.465 67.41% 61.30% 49.08%
Means
Factor N Mean StDev 95% CI
Atlanta 5 591.0 78.5 (479.6, 702.4)
New York 5 350.0 155.1 (238.6, 461.4)
Los Angeles 5 776.0 134.6 (664.6, 887.4)
Chicago 5 563.0 82.7 (451.6, 674.4)
Pooled StDev = 117.465
Tukey Pairwise Comparisons
Grouping Information Using the Tukey Method and 95% Confidence
Factor N Mean Grouping
Los Angeles 5 776.0 A
Atlanta 5 591.0 A B
Chicago 5 563.0 B
New York 5 350.0 C
Means that do not share a letter are significantly different.
a) since p-value =0.000
since p-value is less than level of significance so we reject null hypothesis.
so there is significant difference between means.
b) there significant difference between pair of means
los Angeles and chigago and los Angeles and newyork
Altanta and newyork
chicago and newyork