Homework Problem: An experiment was conducted to test the effectof data file siz

Question

Homework Problem:

An experiment was conducted to test the effectof data file size on the ability to access the files (as measuredby read time, in milliseconds). Three different levels of data filesize were considered:small (50,000 characters),medium (75,000characters) and large (100,000 characters). A sample of eightfiles of each size was evaluated. The following Data One and DataTwo (Figure 1 and Figure 2 below) are located in theacess.xls file. This file contains access read times inmilliseconds.

Question# 2: At the 5% level of significance, isthere evidence of a difference among the mean access read times forthe three file sizes?

     DataOne            (AccessRead Times)   Data Two

         Small Med Large                                     Small Med Large

          2.05  2.24  2.08                                 20   2.05  2.24  2.08

          2.04   2.21  2.34                                 20   2.04  2.21  2.34

         2.21  2.23   2.33                                 20   2.21   2.23  2.33

          2.12   2.09  2.24                                 20  2.12   2.09  2.24

          2.32  2.52   2.71                                 40   2.32   2.52  2.71

          2.31   2.62   2.73                                 40   2.31  2.62  2.73

          2.48   2.57  2.9                                   40   2.48  2.57  2.9

          2.42   2.61  2.72                                 40   2.42   2.61 2.72

Do I conductthe one-way ANOVA on Data One or Data Two (5% significance)? Iwould think Data Two because it has the read access times inmilliseconds (20,20,20 etc.). Therefore I could conduct the ANOVAsummary table (Data Two) at the 5% significance level, F= MSA/MSW ,F= 1525.95/28.61133=53.3340. The p-value=1.06E-11. This p-value isless than .5, I would reject the null hypothesis, there is nodifferences in the access read time files.

If I conductthe one-way ANOVA the ANOVA summary table at the 5% significancelevel, F=MSA/MSW, where F =0.13815/0.053198=2.5969. The p-value =0.0982. This p-value is also less than .5 significance level,therefore I would reject the null also.

Which DATA isappropriate to use?

Explanation / Answer

x.j5 the data set 1 ONE - WAY ANOVA is appropriate, since the experimental units are distinguished among themselves only by the size of the data file they access. Applying MINITAB ANOVA on this data we have the following output.

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