Answer:
There is no significant difference in the average daily sales of this dog food based on shelf height
Step-by-step explanation:
Null hypothesis: There is no significant difference in the average sales
Alternate hypothesis: There is a significant difference in the average daily sales
Using the F-table, at 0.05 significance level and (2, 21) degrees of freedom (3-1, 24-3), the critical value is 3.47
Conclusion: Reject the null hypothesis if the computed F exceeds 3.47
Data values
Knee level: 77,88,85,82,94,85,86,93 (Mean is 86.25)
Waist level: 87,78,90,81,81,91,80,86 (Mean is 84.25)
Eye level: 94,79,77,90,87,81,87,93 (Mean is 86)
Grand mean = (86.25+84.25+86)/3 = 85.5
Sum of Squares (SS) for knee level= summation(sales - grand mean)^2 = 220
SS for waist level = 180
SS for eye level = 288
SStotal = 220+180+288 = 688
Sum of Squares due to Treatment (SST) for knee level = n(mean - grand mean)^2 = 8(86.25 - 85.5)^2 = 4.5
SST for waist level = 12.5
SST for eye level = 0.25
Total SST = 4.5+12.5+0.25 =
17.25
Mean Sum of Treatment (MST)= SST/(number of food types - 1) = 17.25/(3-1) = 17.25/2 = 8.625
Sum of Squares due to Error (SSE) = SStotal - SST = 688 - 17.25 = 670.75
Mean Sum of Error (MSE) = SSE/(24-3) = 670.75/21 = 31.940
F = MST/MSE = 8.625/31.940 = 0.27
The computed F 0.27 is less than the critical value 3.47, so we fail to reject the null hypothesis
Conclusion: There is no significant difference in the average daily sales of this dog food based on shelf height
