Answer to Question #241114 in Statistics and Probability for darv

Question

Below are the bowling scores of 4 groups of 4 players each. At 5% significance level, find out if there is unusual variation among the four groups. Group A: Player A 98, Player B 78, Player C 95, Player D 110. Group B: Player A 100, Player B 95. Player C 90, Player D 102 Player D 85. Group C: Player A 87, Player B 95, Player C 105, Player D 88 and Group D: Player A 90, Player B 93, Player C 95, Player D 97.

 

Answer 1

SOLUTION

Set the hypothesis

Null Hypothesis: H₀: The mean of four groups have a variation.

$H_0:\mu=variation$

Alternative Hypothesis: H_a: The mean of four groups have no variation

$H_a:\mu\not =variation$

I used the excel function data analysis to do the ANOVA-Single Factor analysis to test if there is variation between the mean of the four groups A, B, C, and D.

The summary displays Group C and D having the lowest mean at 93.75 proceeding to the highest mean that is Group D with a mean of 96.75.

Decision making: If the p-value is less or greater the alpha value 0.05, then the reject the null hypothesis, otherwise do not reject the null hypothesis.

Conclusion: Checking the ANOVA table, the p-value is 0.948137307347537. We reject the null hypothesis because the p-value is greater than the alpha value. There is enough evidence that proves a conclusion that there is variance in the mean of the four groups.