Question

3. Five university students were given an English achievement test before and after receiving instruction in basic grammar. Their scores are shown below:
Student Before After
A 20 18
B 18 22
C 17 15
D 16 17
E 12 9
Should we conclude that future students would show higher scores after instruction? Use the .05 significance level. Use hypothesis testing.

Answer 1

Answer on Question #49819 – Math – Statistics and Probability

Five university students were given an English achievement test before and after receiving instruction in basic grammar. Their scores are shown below:

Should we conclude that future students would show higher scores after instruction? Use the .05 significance level. Use hypothesis testing.

Solution

$$n = n_1 = n_2 = 5.$$

The mean difference is

$$\bar{d} = \overline{(x_a - x_b)} = \frac{\sum(x_a - x_b)}{n} = \frac{(18 - 20) + (22 - 18) + (15 - 17) + (17 - 16) + (9 - 12)}{5} = -0.4.$$

Sample standard deviation of difference is

$$\sum (x_a - x_b)^2 = (18 - 20)^2 + (22 - 18)^2 + (15 - 17)^2 + (17 - 16)^2 + (9 - 12)^2 = 34$$ $$s_d = \sqrt{\frac{\sum (x_a - x_b)^2 - n\bar{d}^2}{n - 1}} = \sqrt{\frac{34 - 5 \cdot (-0.4)^2}{5 - 1}} = 2.881.$$

Hypotheses:

$$H_0: \bar{d} \leq 0; H_a: \bar{d} > 0.$$

Decision Rule:

$$\alpha = 0.05$$

Degrees of freedom $n - 1 = 5 - 1 = 4$

Critical t-score from t-table $t^* = 2.132$.

Reject $H_0$ if $t > 2.132$.

Test Statistic:

$$t = \frac{\bar{d}}{\frac{s}{\sqrt{n}}} = \frac{-0.4}{\frac{2.881}{\sqrt{5}}} = -0.310.$$

Decision (in terms of the hypotheses):

Since $t = -0.310 < t^* = 2.132$ we fail to reject $H_0$.

Conclusion (in terms of the problem):

There is no sufficient evidence at 0.05 significance level that future students would show higher scores after instruction.

www.AssignmentExpert.com