Question

You (Dr. Gore) are a researcher examining the effects of a new teaching approach

on student learning of mathematics concepts. The data below reflect the scores on a math

concept exam. The data represent the number correct before the teaching approach was

used (Column A) and after the teaching approach was used (Column B) for the same

subjects. Conduct a dependent samples t-test on the data and decide whether you can

reject the null hypothesis that the approach had no effect. Be sure to list all the steps

needed in conducting a hypothesis test.

A                             B

3                              5

4                             6

3                             7

6                              6

3                             5

2                              4

1                              3

6                              9

8                              9

Part 2) Imagine that the data are collected using a between subjects methodology in

which two groups of subjects are compared. Analyze the data using an independent

samples t-test and test the hypothesis described above. Be sure to list all the steps needed

in conducting a hypothesis test.

Answer 1

Solution:

Part1:

The null and alternative hypotheses are as follows:

i.e. The population means of the scores is same in the both the situations.

i.e. The population means of the scores is not same in the both the situations.

To test the hypothesis we shall use dependent samples t-test.The test statistic is given as follows:

Where, is the mean of the difference of scores, S is sample standard deviation of the difference of the scores and n is sample size.

and   

A	B	d​​​​​​i = A - B
3	5	3 - 5 = -2
4	6	4 - 6 = -2
3	7	3 - 7 = -4
6	6	6 - 6 = 0
3	5	3 - 5 = -2
2	4	2 - 4 = -2
1	3	1 - 3 = -2
6	9	6 - 9 = -3
8	9	8 - 9 = -1

The value of the test statistic is -5.3666.

Our test is two-tailed test, therefore we shall obtain two-tailed p-value for the test statistic, which is given as follows:

p-value = 2P(T > |t|)

Since, |t| = 5.3666

p-value = 2P(T > 5.3666)

p-value = 0.00067

We make decision rule as follows:

If p-value is greater than the significance level, then we fail to reject H_{​​​​​​0} at given significance level.

If p-value is less than the significance level, then we reject the H_{​​​​​​0} at given significance level.

In our question significance level is not given. Generally significance level of 0.05 or 0.01 is used.

We shall use significance level of 0.01.

p-value = 0.00067 and significance level = 0.01

(0.00067 < 0.01)

Since, p-value is less than the significance level of 0.01, therefore we shall reject H_{​​​​​​0} at 0.01 significance level.

Conclusion: At significance level of 0.01, there is enough evidence to conclude that new teaching approach has significant effect on learning.

Part2:

The null and alternative hypotheses are as follows:

i.e. The population means of the scores is same in the both the situations.

i.e. The population means of the scores is not same in the both the situations.

To test the hypothesis we shall use independent samples t-test. The test statistic is given as follows:

Where, and are the sample means, S_{​​​​​​pooled} is pooled standard deviation, n1 and n2 are the sample sizes.

, and

The value of the test statistic is -1.9728.

Degrees of freedom = n1 + n2 - 2 = (9+9-2) = 16

Our test is two-tailed test, therefore we shall obtain two-tailed p-value for the test statistic, which is given as follows:

p-value = 2P(T > |t|)

Since, |t| = 1.9728

p-value = 2P(T > 1.9728)

p-value = 0.06606

We make decision rule as follows:

If p-value is greater than the significance level, then we fail to reject H_{​​​​​​0} at given significance level.

If p-value is less than the significance level, then we reject the H_{​​​​​​0} at given significance level.

In our question significance level is not given. Generally significance level of 0.05 or 0.01 is used.

We shall use significance level of 0.01.

p-value = 0.06606 and significance level = 0.01

(0.06606 > 0.01)

Since, p-value is greater than the significance level of 0.01, therefore we shall be fail to reject H_{​​​​​​0} at 0.01 significance level.

Conclusion: At significance level of 0.01, there is not enough evidence to conclude that new teaching approach has significant effect on learning.