Hypothesis testing of paired data may include two populations that have the equal standard deviations. The t-test for differences considered in a previous post used the standard deviation of the differences. In this test, we want to consider if one population is different in some way than the other and we use the samples from each population directly.

## Test Setup

The null hypothesis for a paired t-test is H_{o}: Î¼_{1} = Î¼_{2}.

The three alternate hypothesis become:

Î¼_{1} â‰¥ Î¼_{2}

Î¼_{1} â‰¤ Î¼_{2}

Î¼_{1} â‰ Î¼_{2}

We are considering the two population standard deviations are the same. You can use the F-test hypothesis test to check this assumption.

We calculate the test statistic using

$$ \large\displaystyleÂ t=\frac{{{{\bar{X}}}_{1}}-{{{\bar{X}}}_{2}}}{{{s}_{p}}\sqrt{\frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{2}}}}}$$

where, s_{p} is the pooled standard deviation and calculated from the samples by

$$ \large\displaystyleÂ {{s}_{p}}=\sqrt{\frac{\left( {{n}_{1}}-1 \right)s_{1}^{2}+\left( {{n}_{2}}-1 \right)s_{2}^{2}}{n+n-2}}$$

The degrees of freedom used to determine the critical value is df = n_{1} + n_{2} – 2.

The critical value (or rejection region) for the three tests given a (1-Î±)100% confidence level becomes:

Reject H_{o} if t > t_{Î±,df}

Reject H_{o} if t < t_{Î±,df}

Reject H_{o} if |t| > t_{Î±/2,df}

## Example

Letâ€™s say we have two processes creating the top tubes of bicycle and we want to know if the two processes differ. The weight is important, so we measure five tubes from each process. The data follows:

Sample | Process 1 | Process 2 |

1 | 3.125 | 3.110 |

2 | 3.120 | 3.095 |

3 | 3.135 | 3.115 |

4 | 3.130 | 3.120 |

5 | 3.125 | 3.125 |

The means are X-bar_{1} = 3.127 and X-bar_{2} = 3.113, s_{1} = 0.0057 and s_{2} = 0.0115. The five samples from each process provides degrees-of-freedom of df = 5 + 5 – 2 = 8.

The critical value is t_{0.05, 8 }= 2.306 given an Î± = 0.05 or a 95% confidence level.

The pooled standard deviation is

$$ \large\displaystyleÂ {{s}_{p}}=\sqrt{\frac{\left( 5-1 \right)0.0057_{{}}^{2}+\left( 5-1 \right)0.01115_{{}}^{2}}{5+5-2}}=0.0091$$

test statistic is

$$ \large\displaystyleÂ t=\frac{3.127-3.113}{0.0091\sqrt{\frac{1}{5}+\frac{1}{5}}}=2.43$$

Since 2.43 is larger than 2.306 and in the rejection region, the null hypothesis is rejected. This means there is convincing evidence the two process do not create the same weight fork tubes.

Related:

Hypothesis un-equal varianceÂ (article)

Paired-Comparison Hypothesis TestsÂ (article)

Two Proportions Hypothesis TestingÂ (article)

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