Test 4

Posted: August 26th, 2021

Student’s Name

Instructor’s Name

Course

Date

Test 4

Question 1: Independent t-test

Given the data below;

Table 1: Sample Data

Group 1	Group 2
32	34
43	57
35	54
36	43
48	45
39	56
33

Analysis

To solve the above problem, the following is the hypothesis test for the case;

Null Hypothesis, H0: (μ₁ – μ₂ = 0)

Alternative Hypothesis, H1: (μ₁ – μ₂ ≠ 0)

The excel analysis is used to obtain the results of the analysis. The following are the results;

Table 2: t-Test: Two samples assuming unequal variances

Details	Group 1	Group 2
Mean	38	48.166
Variance	33.333	82.166
Observations	7	6
Hypothesized Mean Difference	0
df	8
t Stat	-2.3665
P(T<=t) one-tail	0.02274
t Critical one-tail	1.8595
P(T<=t) two-tail	0.04549
t Critical two-tail	2.3060

From Table 1 above, the mean value for Group 1 is 38.333, while the mean value for Group 2 is 48.166. In this case, we consider the results of the two-tail because the aim is to assess inequality. Therefore, if t Stat < -t Critical two tail or the t Stat > t Critical two-tail, reject the null hypothesis. In the case shown under Table 1, t Stat is -2.3665 while t Critical two-tail is 2.306. Since t Stat (-2.3665) < -t Critical two-tail (-2.3060). Therefore, we reject the null hypothesis. Hence, the evidence of the difference in the means between Group 1 and Group 2 (38-48.166) is not statistically significant at the 0.05 significance level to claim that there is a difference between the two groups.

Question 2: Dependent t-Test

Given the data below;

Table 3: Sample Data

Pre Post 32 25 28 15 41 19 36 26 34 41

Analysis

Hypothesis Test;

Null Hypothesis, H0: μ₁ = μ₂

Alternative Hypothesis, H1: μ₁> μ2

Table 4 below shows the results of the analysis;

	Pre	Post
Mean	34.75	25.25
Variance	28.9166	130.916
Observations	4	4
Pearson Correlation	0.10428
Hypothesized Mean Difference	0
df	3
t Stat	1.56709
P(T<=t) one-tail	0.10754
t Critical one-tail	2.3534
P(T<=t) two-tail	0.21508
t Critical two-tail	4.1824

The mean value for pre results is 34.75, and the post results are 25.25. To fail to reject the null hypothesis, the p-value should be less than 0.05 (P-(T=t) two-tail < 0.05. However, the p-value, in this case, is 0.21508 hence; reject the null hypothesis. Thus, it implies that the pre-test results are significantly different from the posttest results.