Question

3.       For a research project, a SACAP student is interested in determining the difference of three types of drinks on the concentration span of medical professionals working longs shifts. The following table indicates scores from three groups of medical professionals who were measured on the same night. Using manual computation, perform the appropriate hypothesis test at α = 0.05. Show all calculations and decision making steps. Also calculate size of the treatment effect using Tukey’s Honestly Significance.   

Please make sure you show calculations of the treatment effect using Tukey’s Honestly Significance.   

High caffeine content soda	Espresso shot	Water
3	7	4
2	9	3
3	6	2
5	4	4
2	5	4
6	4	2
7	2	9
4	3	3

                                                                               

Answer 1

Tukey's HSD Procedure compares the observed differences between each and every pair of group means to calculated critical difference.

Here the null hypothesis is there is no significance difference in the means of 3 samples.

If our null hypothesis accepted so we don't required to furthere analysis. if our null hypothesis is reject then we go to furthere analysis.

here the level of significance is 0.05.

We are calculated ANOVA in excel. (copy the data to excelsheet then go to data then data analysis and use ANOVA -Single Factor from this we get output)

then the output is ,

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	6.083333	2	3.041667	0.673254	0.520716	3.4668
Within Groups	94.875	21	4.517857
Total	100.9583	23

this is the ANOVA result from this we can say that the p-value=0.52 which is greater then the level of singificance=0.05.Then we accept the null hypothesis that  there is no significance difference in the means of 3 sample. we need not go for further analysis.

Then we find the mean of each column is,

We use Tukey method for pairwise comparison, we compare pairs of the sample means, using there absolute difference:

Then we calculate Tukey Criterion is defined as,

• where, is Studentized range distribution ,based on c and n-c d.f.

c= no. of treatment (no. of columns) =3

n=Total sample size =8*3=24

• MSE (Mean Square error) from anova table =
• is sample size of the treatment group with the smallest number of observation.

In this example sample size of all group is same which is 8 in each group.

Let's we calculate the Tukey Critirion,

from the table (http:/davidmlane.com/hyperstats/sr_table.com)

substituing and solving,

• Absolues values of the paired means as follows:

then we can say that there is no significance difference in the means of 3 samples.