Question

One group of 5 sophomores were given and English achievement test before and after receiving instruction in basic grammar. Is it reasonable to conclude that future students would show higher scores after instruction? Test with α = .05, one tail. Copy the following data to a sheet of paper and calculate the SS and other quantities needed.

1.Calculate the variance of the difference column

2. calculate the standard error

3. enter the degrees of freedom

4. Enter the tabled critical value of t (t_crit )

5. Calculate the t statistic

6. reject or do not reject

7.

Complete the APA formatted statistic: (round t to 2 decimal places)

After	Before	D	(D - MD)	(D - MD)2
18	20
21	18
17	17
18	16
14	12

Answer 1

Suppose random variables X and Y denote scores of English achievement test after and before receiving instruction in basic grammar. Also random variable D (=X-Y) denotes difference in scores after and before receiving instruction.

We are given with paired scores corresponding to 5 (=n) sophomores. We do not know population standard deviation (or variance). So, we have to perform paired t-test.

We have to test foe null hypothesis

against the alternative hypothesis

After (xi)	Before (yi)	di=xi-yi
18	20	-2
21	18	3
17	17	0
18	16	2
14	12	2

1.

Variance of the different column is given by

2.

Standard error is given by

3.

Degrees of freedom

4.

Our alternative hypothesis is of greater than type. So we performing right tailed test.

Level of significance

[Using R-code 'qt(1-0.05,4)']

5.

Our t-statistic is given by

6.

In case of right tailed test we reject our null hypothesis if .

Here we observe that .

So, we can not reject our null hypothesis.

7.

M=1, SD=2, SE=0.89, v=4, t(4)=1.12, p=.16