Report ene Mean 191.5 319.7 342.3 2845 Std. Deviation treated control1 Control2

ID: 3326602 • Letter: R

Question

Report ene Mean 191.5 319.7 342.3 2845 Std. Deviation treated control1 Control2 Total 17 17 17 51 223.6 181.8 188.3 ANOVA time Sum of Squares 224891 xXX 773061 Mean Square xXX 32253.54 df xxX 3.486 0.0386 Between Groups Within Groups Total 50 a. (8 points) Fill out the ANOVA table for the data above (there are 4 "xxx" items to fill in). b. (8 points) Perform an appropriate hypothesis test to determine whether there is a difference in mean time to completion among the 3 groups. Be sure to include your hypotheses, test statistic, degrees of freedom if appropriate, either the p-value or critical value, and your conclusion in terms of the problem

Explanation / Answer

Solution:

Part a

Completed ANOVA table is given as below:

Within groups SS = Total SS – Between groups SS

Within groups SS = 1773061 – 224891 = 1548170

Total number of groups = k = 3

Degrees of freedom for between groups = k – 1 = 3 – 1 = 2

Degrees of freedom for within groups = total df – between df = 50 – 2 = 48

Mean square between groups = SS for between groups / df for between groups

Mean square between groups = 224891/2 = 112445.5

ANOVA table is given as below:

Source of variation

Sum of Squares

Mean Square

P-value

Between Groups

224891

112445.5

3.486299

0.038573

Within Groups

1548170

32253.54167

Total

1773061

Part b

Here, we have to perform ANOVA F test for checking whether there is any significant difference exists between the mean time to completion among the three groups or not. The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: There is no any statistically significant difference exists between population averages time of completion for three groups.

Alternative hypothesis: Ha: There is a statistically significant difference exists between population averages time of completion for three groups.

H0: µ1 = µ2 = µ3 versus Ha: At least two µi’s are not same.

We assume level of significance as = 0.05.

For this ANOVA test, the test statistic value is given as below:

F = MSB/MSW

F = 112445.5/32253.54167 = 3.486299

DF1 = 2

DF2 = 48

Critical value = 3.190727336

P-value = 0.038573

= 0.05

P-value < = 0.05

So, we reject the null hypothesis at 5% level of significance.

We reject the null hypothesis that there is no any statistically significant difference exists between population averages time of completion for three groups.

There is sufficient evidence to conclude that there is a statistically significant difference exists between population averages time of completion for three groups.