Stats final HW[1]

Stats final HW[1] - Stats 215 1 Anova Single Factor SUMMARY...

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Stats 215 1. Anova: Single Factor SUMMARY Groups Count Sum Average Variance Credit Union 15 63.91 4.260667 0.434378 Other Fin Inst. 15 69.69 4.646 0.572383 Bank 20 92.38 4.619 0.27342 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 1.442959 2 0.721479 1.757915 0.18355 3.195056 Within Groups 19.28963 47 0.410418 Total 20.73259 49 H 0 =U 1 =U 2 =U 3 H 1 = U 1 ≠U 2 ≠U 3 Alpha Level= .05 which means that we have 5 chance in 100 of making a type 1 error. We can see that mean CD rate by the Credit union (4.26) is lower than that of either the Bank (4.619) or other financial institution (4.646). But the question I pose is these differences statistically significant? According to the test result F=1.75. With a critical value of .05, the critical F = 3.19. Therefore, since the F statistic is smaller than the critical value, we fail to reject the null hypothesis. The null hypothesis was that all 3 of these groups' means were equal. So, we accept that the three financial institutions have the same rate offered. Accept H 0 because the F statistic of 1.75 is less that the Critical Value of 3.195….1.75< 3.195 Also Accept H 0 because the P value of .18355 is greater than the Alpha level of .05 .18355 > .05
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2. Anova: Single Factor SUMMARY Groups Count Sum Average Variance Downtown 6 1443 240.5 14543.1 University 6 1380 230 2070 River 6 1810 301.6667 696.6667 Interstate 6 1215 202.5 777.5 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 31533 3 10511 2.324508 0.105672 3.0 98 39 1 Within Groups 90436.33 20 4521.817 Total 121969.3 23 H 0 =U 1 =U 2 =U 3 =U 4 H 1 = U 1 ≠U 2 ≠U 3 ≠U 4 Alpha Level= .05 which means that we have 5 chance in 100 of making a type 1 error. We can see that the mean Sales for burgers on the River (202.5) are much higher than that of either downtown (240.5), at the University (230) and on the Interstate (202.5). But the question I pose is these differences statistically significant? According to the test result F=2.32. With a critical value of .05, the critical F = 3.09. Therefore, since the F statistic is smaller than the critical value, we fail to reject the null hypothesis. The null hypothesis was that all 4 of these groups' means were equal. So, we accept that the four burger locations over the past six weeks have been statistically the. Accept H
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Stats final HW[1] - Stats 215 1 Anova Single Factor SUMMARY...

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