The two sample paired t-test is used to test if two samples come from the same population by comparing paired values between the samples. A paired samples t-test is based on a matched-pairs sample, where each value in one sample has a similar match in terms of other measured variables in the other sample.

The Student[Statistics][TwoSamplePairedTTest] function computes the paired t-test upon two datasets X1 and X2. This means that every entry of the population of X1 is related to an entry of the population of X2; in the samples, X1[i] is related to X2[i]; and the procedure tests whether the mean of the population of differences between related pairs, X1[i] minus X2[i], is equal to a value beta, under the assumption that these differences are normally distributed.

Requirements for using the Two Sample Paired T Test:

In order for the data in the two samples to be paired the sample size of the two samples must be the same.

The paired differences are assumed independent.

The paired differences are all assumed to follow identical normal distribution (same mean and variance).

The formula is:

where $Z$ is the sample whose data is generated as the difference between the data in each pair ($Z_i=X_i-Y_i$ ; $X$ is Sample One and $Y$ is Sample Two), $\mathrm{\beta }$ is the test value of difference between the means, $s$ is the sample standard deviation of $Z$, and $N$ is the sample size of $Z$.

$T$ follows Student's T distribution with $N-1$ degrees of freedom.

Determine the null hypothesis:

Null Hypothesis: Children aged 7 and children aged 12 think WashBill RoundShirt is equally funny.

$\mathrm{\beta }=0$ in this case

Substitute the information into the formula:

$Z=X-Y$, where $X$ is the sample of group of children aged 7, $Y$ is the sample of group of children aged 12.

Thus $Z=\left[1\,0\,1\,\mathrm{-1}\,1\,0\,\mathrm{-1}\,\mathrm{-2}\,3\,0\right]$

$t=\frac{\left(0.2-0\right)}{\left(\frac{1.39841}{\sqrt{10}}\right)}=0.452267$

Compute the p-value by checking the table of Student's T distribution with 9 degrees of freedom:

$p-\mathrm{value}=\mathrm{Probability}\left(\left|T\right|>0.452267\right)=\mathrm{Probability}\left(T<-0.452267\right)+\mathrm{Probability}\left(T>0.452267\right)=0.66178$, $T˜\mathrm{StudentT}\left(9\right)$

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.