Question

Anela is a computer scientist who is formulating a large and complicated program for a type of data processing. She has three ways of storing and retrieving data: cloud storage, disk, or hard drive. As an experiment, she sets up her program in three different ways: one using cloud storage, one using disks, and the other using a hard drive. Then she makes four test runs of this type of data processing on each program. The time required to execute each program is shown in the following table (in minutes). Use a 0.01 level of significance to test the hypothesis that the mean processing time is the same for each method.

Hard Drive	Cloud	Disks
8.5	7.1	6.5
9.4	9.6	6.6
7.5	7.6	10.2
7.8	7.8	8.4

In the case of one-way ANOVA, make a summary table. (Round your answers to three decimal places.)

Source of Variation	Sum of Squares	Degrees of Freedom	MS	F Ratio	P-Value	Test Decision
Between groups
Within groups
Total

Answer 1

Using Minitab software, (Stat -> ANOVA -> One way) , we get the following output :

Minitab-Untitled- [Session] File Edit Data Calc Stat Graph Editor Tools Window Help Assistant Welcome to Minitab, press Fl for help. One-way ANOVA: Response versus Storage Factor Method Null hypothesis Alternative hypothesis At least one mean is different Significance level All means are equal α 0.05 Equal variances were assumed for the analysis. Factor Information Factor Levels Values Storage Factor 3 Cloud, Disks, Hard Drive Analysis of Variancee DF Adj SS Adj MS F-Value P-Value 0.914 Source Storage Factor 2 0.3017 0.1508 Error Total 0.09 9 14. 8950 1.6550 11 15.1967 Model Summary S R-sq R-sq (adj) R-sq (pred) 0. 00% 1.28647 1.99% 0. 00% Means 95% CI Storage Factor N Mean StDev Cloud Disks Hard Drive 4 8.025 1.090 (6.570, 9.480) 4 7.925 1.750(6.470, 9.380) 4 8.300 0.845 (6.845, 9.755) Pooled StDev = 1 . 2 8 64 7

Since p-value = 0.914 > 0.01, so we fail to reject H₀ at 1% level of significance and we can conclude that that the mean processing time for each storage type is not significantly different.