A start-up cell phone applications company is interested in determining whether house-hold incomes are different for subscribers to three different service providers. A random sample of 25 subscribers to each of the three service providers was taken, and the annual household income for each subscriber was recorded. The partially completed ANOVA table for the analysis is shown here: Based on the sample results, can the start-up firm conclude that there is a difference in household incomes for subscribers to the three service providers? You may assume normal distributions and equal variances. Conduct your test at the alpha= 0.10 level of significance. Be sure to state a critical F-statistic, a decision rule, and a conclusion. A) H0 : µ1 = µ2 = µ3 HA: Not all populations have the same mean F = MSB/MSW = 1,474,542,579/87,813,791 = 16.79 Because the F test statistic = 16.79 Fα = 2.3778, we do reject the null hypothesis based on these sample Data. B) H0 : µ1 = µ2 = µ3 HA : Not all populations have the same mean F = MSB/MSW = 87,813,791 /1,474,542,579= 0.060 Because the F test statistic = 0.060 Fα = 2.3778, we do reject the null hypothesis based on these sample data. D) H0 : µ1 = µ2 = µ3 HA : Not all populations have the same mean F = SSW/MSW = 6,322,592,933/1,474,542,579= 4.28 Because the F test statistic = 4.28 Fα = 2.3778, we do reject the null hypothesis based on these sample Data

Question

A start-up cell phone applications company is interested in determining whether house-hold incomes are different for subscribers to three different service providers. A random sample of 25 subscribers to each of the three service providers was taken, and the annual household income for each subscriber was recorded. The partially completed ANOVA table for the analysis is shown here:  Based on the sample results, can the start-up firm conclude that there is a difference in household incomes for subscribers to the three service providers? You may assume normal distributions and equal variances. Conduct your test at the alpha= 0.10 level of significance. Be sure to state a critical F-statistic, a decision rule, and a conclusion.
A) H0 : µ1 = µ2 = µ3 HA: Not all populations have the same mean F = MSB/MSW = 1,474,542,579/87,813,791 = 16.79 Because the F test statistic = 16.79 > Fα = 2.3778, we do reject the null hypothesis based on these sample Data. 
B) H0 : µ1 = µ2 = µ3 HA : Not all populations have the same mean F = MSB/MSW = 87,813,791 /1,474,542,579= 0.060 Because the F test statistic = 0.060 < Fα = 2.3778, we do not reject the null hypothesis based on these Sample data. 
C) H0 : µ1 = µ2 = µ3 HA : Not all populations have the same mean F = SSW/MSW = 6,322,592,933/87,813,791 = 72 Because the F test statistic = 72 > Fα = 2.3778, we do reject the null hypothesis based on these sample data. 
D) H0 : µ1 = µ2 = µ3 HA : Not all populations have the same mean F = SSW/MSW = 6,322,592,933/1,474,542,579= 4.28 Because the F test statistic = 4.28 > Fα = 2.3778, we do reject the null hypothesis based on these sample Data

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The Answer of A start-up cell phone applications company is interested in determining whether house-hold incomes are different for subscribers to three different service providers. A random sample of 25 subscribers to each of the three service providers was taken, and the annual household income for each subscriber was recorded. The partially completed ANOVA table for the analysis is shown here:  Based on the sample results, can the start-up firm conclude that there is a difference in household incomes for subscribers to the three service providers? You may assume normal distributions and equal variances. Conduct your test at the alpha= 0.10 level of significance. Be sure to state a critical F-statistic, a decision rule, and a conclusion.
A) H0 : µ1 = µ2 = µ3 HA: Not all populations have the same mean F = MSB/MSW = 1,474,542,579/87,813,791 = 16.79 Because the F test statistic = 16.79 > Fα = 2.3778, we do reject the null hypothesis based on these sample Data. 
B) H0 : µ1 = µ2 = µ3 HA : Not all populations have the same mean F = MSB/MSW = 87,813,791 /1,474,542,579= 0.060 Because the F test statistic = 0.060 < Fα = 2.3778, we do not reject the null hypothesis based on these Sample data. 
C) H0 : µ1 = µ2 = µ3 HA : Not all populations have the same mean F = SSW/MSW = 6,322,592,933/87,813,791 = 72 Because the F test statistic = 72 > Fα = 2.3778, we do reject the null hypothesis based on these sample data. 
D) H0 : µ1 = µ2 = µ3 HA : Not all populations have the same mean F = SSW/MSW = 6,322,592,933/1,474,542,579= 4.28 Because the F test statistic = 4.28 > Fα = 2.3778, we do reject the null hypothesis based on these sample Data

A Start-Up Cell Phone Applications Company Is Interested in Determining