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Student[Statistics][TwoSampleTTest] Overview

overview of the Two Sample T-Test

Description

 • The Two Sample T Test is used to test if the means of two samples, each assumed to follow a normal distribution, are equal.
 • To apply this test, the standard deviations of the two populations do not need to be known. If the standard deviations of the two populations are provided, then the Two Sample Z Test can be used.
 • If the two samples are paired, then the Two Sample Paired T Test can be used.
 • Requirements for using Two Sample T Test:
 1 The two populations studied are assumed to be normally distributed.
 2 The two standard deviations of the two populations are unknown.
 • The formula is:

$T=\frac{\mathrm{Mean}\left(X\right)-\mathrm{Mean}\left(Y\right)-\mathrm{\beta }}{\sqrt{\frac{{\mathrm{s1}}^{2}}{\mathrm{N1}}+\frac{{\mathrm{s2}}^{2}}{\mathrm{N2}}}}$

 where $X$ is the sample drawn from the first population, $Y$ is the sample drawn from the second population, $\mathrm{\beta }$ is the test value of the difference, $\mathrm{s1}$: The sample standard deviation of $X$, $\mathrm{s2}$ is the sample standard deviation of $Y$, $\mathrm{N1}$ is the sample size of $X$, and $\mathrm{N2}$ is the sample size of $Y$.
 • $T$ follows Student's T distribution with degree of freedom computed as follows: $\mathrm{DF}=\frac{{\left(\frac{{\mathrm{s1}}^{2}}{\mathrm{N1}}+\frac{{\mathrm{s2}}^{2}}{\mathrm{N2}}\right)}^{2}}{\frac{{\mathrm{s1}}^{4}}{{\mathrm{N1}}^{2}\left(\mathrm{N1}-1\right)}+\frac{{\mathrm{s2}}^{4}}{{\mathrm{N2}}^{2}\left(\mathrm{N2}-1\right)}}$

Example

A research team conducted a survey to know the difference between the weights of male and female subjects of ages from 20 to 30 years. They knew that the weights should be normally distributed, but they did not know the standard deviations. The researchers randomly selected 10 men and 8 women for the survey, and their weights were recorded as follows:

 $X$ Sample of the weights of male subjects $Y$ Sample of the weights of female subjects Male Weight(kg) Female Weight(kg) 1 75 1 61 2 80 2 55 3 73 3 65 4 88 4 53 5 84 5 48 6 67 6 66 7 77 7 50 8 66 8 60 9 77 10 69

The researchers hypothesized that due to differences in the body structure, males should on average be 20 kg heavier than females. Now they want to test their hypothesis with the data that was collected.

 1 Determine the null hypothesis:
 Null Hypothesis: On average, men are 20 kg heavier than women ($\mathrm{\beta }$ is 20 for this case)
 2 Substitute the information into the formula:
 $\mathrm{Mean}\left(X\right)=75.6,\mathrm{Mean}\left(Y\right)=57.25,\mathrm{\beta }=20,\mathrm{s1}=7.18331,\mathrm{s2}=6.75595,\mathrm{N1}=10,\mathrm{N2}=8$
 $t=\frac{\left(\left(75.6-57.25\right)-20\right)}{\sqrt{\left(\frac{{7.18331}^{2}}{10}+\frac{{6.75595}^{2}}{8}\right)}}=-0.500567$
 $\mathrm{DF}=\frac{{\left(\frac{{7.18331}^{2}}{10}+\frac{{6.75595}^{2}}{8}\right)}^{2}}{\frac{{\left(\frac{{7.18331}^{2}}{10}\right)}^{2}}{9}+\frac{{\left(\frac{{6.75595}^{2}}{8}\right)}^{2}}{7}}=15.51621$
 3 Compute the p-value:
 $p-\mathrm{value}=\mathrm{Probability}\left(|T|>|-0.500567|\right)=\mathrm{Probability}\left(T<-0.500567\right)+\mathrm{Probability}\left(T>0.500567\right)=0.623702$, $T˜\mathrm{StudentT}\left(15.51621\right)$
 4 Draw the conclusion:
 This statistical test does not provide enough evidence to conclude that the null hypothesis is false, so we fail to reject the null hypothesis.