Hypothesis Testing
The given data for this problem is referred to as a two-sample test because we are provided with data from two different populations. There is no unique relationship between any data point in set one of the BMI index of male and female and a data point in set two. Such like data is referred to as independent samples and the two-sample test as a test for independent data. This is the kind of t test you are most likely to run in two-sample t tests.
If you collect a data point from a number of cultures, specimens, cells, mice, people, or any other group of individuals and then do not collect data from those same individuals again, it is concluded that the collected data samples are independent of each other. In this case, we are assuming that the two data seta have different standard deviations. If you have equal numbers of data points, or the numbers are nearly the same, then e can safely apply the two-sample test for equal variances. Otherwise, one should use Levene’s test, Bartlett’s test, or the Brown-Forsythe test, or test for equal variances graphically using a normal quantile plot. However, in this case, I am going to use excel to conduct my analysis.
| BMI | ||
| MALE | FEMALE | |
| 23.8 | 19.6 | |
| 23.2 | 23.8 | |
| 24.6 | 19.6 | |
| 26.2 | 29.1 | |
| 23.5 | 25.2 | |
| 24.5 | 21.4 | |
| 21.5 | 22.0 | |
| 31.4 | 27.5 | |
| 26.4 | 33.5 | |
| 22.7 | 20.6 | |
| 27.8 | 29.9 | |
| 28.1 | 17.7 | |
| 25.2 | 24.0 | |
| 23.3 | 28.9 | |
| 31.9 | 37.7 | |
| 33.1 | 18.3 | |
| 33.2 | 19.8 | |
| 26.7 | 29.8 | |
| 26.6 | 29.7 | |
| 19.9 | 31.7 | |
| 27.1 | 23.8 | |
| 23.4 | 44.9 | |
| 27.0 | 19.2 | |
| 21.6 | 28.7 | |
| 30.9 | 28.5 | |
| 28.3 | 19.3 | |
| 25.5 | 31.0 | |
| 24.6 | 25.1 | |
| 23.8 | 22.8 | |
| 27.4 | 30.9 | |
| 28.7 | 26.5 | |
| 26.2 | 21.2 | |
| 26.4 | 40.6 | |
| 32.1 | 21.9 | |
| 19.6 | 26.0 | |
| 20.7 | 23.5 | |
| 26.3 | 22.8 | |
| 26.9 | 20.7 | |
| 25.6 | 20.5 | |
| 24.2 | 21.9 | |
| MEAN | 25.9975 | 25.74 |
| STD DEV | 3.430742 | 6.16557 |
The t-Test is used to test the null hypothesis that the means of two populations are equal.Above are the set of data that represent the BMI index of male and females to be tested. The hypotheses being tested are as below:
H0: μ1 – μ2 = 0
H1: μ1 – μ2 ≠ 0
Real Statistics Data Analysis Tool: The Real Statistics Resource Pack provides a data analysis tool called T Tests and Non-parametric Equivalents, which combines the analyses for equal and unequal variances, as well as providing confidence intervals and the Cohen effect size. A second measure of effect size is also provided, which we will study in Dichotomous Variables and the t-test (Wilcox, 2005).
For this case, I used excel to test the null hypothesis and found the results as given below:
| t-Test: Two-Sample Assuming Unequal Variances | ||
| 23.8 | 19.6 | |
| Mean | 26.05385 | 25.89744 |
| Variance | 11.94939 | 37.9971 |
| Observations | 39 | 39 |
| Hypothesized Mean Difference | 0 | |
| df | 60 | |
| t Stat | 0.138212 | |
| P(T<=t) one-tail | 0.445268 | |
| t Critical one-tail | 1.670649 | |
| P(T<=t) two-tail | 0.890536 | |
| t Critical two-tail | 2.000298 | |
We do a two-tail test (inequality). lf t Stat < -t Critical two-tail or t Stat > t Critical two-tail, we reject the null hypothesis(Wilcox, 2005). This is not the case, -2.000< 1.382<2.000. Therefore, we do not reject the null hypothesis. The observed difference between the sample means (23.8–19.6) is not convincing enough to say that the average body mass index (BMI Index) of female and male students differ significantly. The two taided diagram is shown as below.
References
Wilcox, R. R. (2005). Introduction to robust estimation and hypothesis testing. Amsterdam: Elsevier/Academic Press.
