Question

Five sophomores were given an English achievement test before and after receiving instruction in basic grammar. The mean difference score is -0.40, and the estimated variance for the difference scores is 8.3. Using the .05 significance level (and five steps of hypothesis testing) is it reasonable to conclude that future students would show higher scores after instruction?
Step I:
-Population 1: -Population 2: -

Research hypothesis:

Null hypothesis:

Step II: Give the characteristics of the comparison distribution -The shape is:

The mean is:
-
Compute the standard deviation using the following

1- Estimated variance of difference scores for sample S2 = 8.3

2-Compute the estimated variance of the distribution of means:
  
3-Compute the estimated standard deviation of the distribution of means:

Step III: What is (are) the cut-off(s)?

Step IV: Determine the sample score position on the comparison distribution

Step V: What is the decision? Write your answer in the APA format.

Answer 1

Step I:

Population 1: Score in English achievement test before receiving instructions in basic grammar

Population 2: Score in English achievement test after receiving instructions in basic grammar

Research Hypothesis:

Null Hypothesis:

Step II:

The shape is:

t distribution

The mean is

0

(1) Estimated variance of different scores of sample = s² = 8.3

(2) Estimated variance of distribution of means = s²/n = 8.3/5 = 1.66

(3) Estimated standard deviation of distribution of means =

Step III:

= 0.05

ndf = 5 - 1 = 4

One Tail- Left side Test

From Table,

cut - off = - 2.1318

Step IV:

Sample score position:

t = - 0.40/1.2884 = - 0.3105

Step 5:
Since the calculated value of t = - 0.3105 is greater than critical value of t = - 2.1318, the difference is not significant. Fail to reject null hypothesis.

Conclusion:

The data do not support the claim that future students would show higher scores after instruction.