Question

Given the following pairs of scores from dependent samples: Pair: 1 2 3 4 5 X: 4 4 6 5 9 Y: 5 2 3 1 6 (a) State formally the hypotheses necessary to conduct a non-directional test of no difference between the two population means. (b) Calculate t using Formula 15.5. t=Dbar-ud/square root sum of D squared - (sum of d)2/n / n(n-1) (c) Calculate t using formula 15.3. t= (xbar -ybar) - (ux-uy)hyp/ square root s2x + s2y - 2rsbarsybar (d) Complete the test at the .05 and .01 levels of significance and state your conclusion.

Answer 1

Let be the true mean of X and be the true mean of Y

Let be the difference in the means of X and Y

a) The  hypotheses necessary to conduct a non-directional test of no difference between the two population means is

Since this is a paired sample, we can also express the above in terms of

The hypothesized value of mean difference is

b) We calculate the following

The difference between X and Ys

Pair	X	Y	D=X-Y
1	4	5	-1
2	4	2	2
3	6	3	3
4	5	1	4
5	9	6	3

the sample mean difference is

The standard error of mean is

The value of test statistics t is

c) First we find the sample means

The sample standard deviations are

the standard errors of means is

The sample correlation coefficient between X,Y is

The sample correlation coefficient between X,Y is

The standard error of the difference in means is (The denominator of the required t statistics)

Now we can get the test statistics t

d) This is a 2 tailed test. The critical value for alpha=0.05 using t distribution and degrees of freedom df=n-1=5-1=4 is

Using the t tables for df=4 and are under the right tail = 0.025 (or area under both the tails=0.05) we get the critical value of t=2.776

We will reject the null hypothesis if the test statistics is not in the acceptance region -2.776 to +2.776.

Here the test statistics is 2.557 (for both b and c) and it lies in the region -2.776 to +2.776.

Hence we do not reject the null hypothesis.

We conclude that at 0.05 level of significance, there is no sufficient evidence to reject the claim that  there is no difference between the two population means.

This is a 2 tailed test. The critical value for alpha=0.01 using t distribution and degrees of freedom df=n-1=5-1=4 is

Using the t tables for df=4 and are under the right tail = 0.005 (or area under both the tails=0.01) we get the critical value of t=4.604

We will reject the null hypothesis if the test statistics is not in the acceptance region -4.604 to +4.604.

Here the test statistics is 2.557 (for both b and c) and it lies in the region -4.604 to +4.604.

Hence we do not reject the null hypothesis.

We conclude that at 0.01 level of significance, there is no sufficient evidence to reject the claim that  there is no difference between the two population means.