# t-Test for x̄_{1} = 25, x̄_{2} = 18, n_{1} = 16, n_{2} = 20 & α = 0.2

Solved example work with steps, formula & calculation summary to estimate the t-statistic (t_{0}), critical (table) value (t_{e}) for degrees of freedom ν & hypothesis test (H_{0}) at a stated level of significance α = 0.2 for difference between two small sample *means* x̄_{1} = 25 & x̄_{2} = 18, s_{1} = 7 & s_{2} = 11 for *sample size* n_{1} = 16 & n_{2}= 20.
The below is the calculation summary for the test of significance or hypothesis using student's t-test for difference between two small sample means x̄_{1} = 25 & x̄_{2} = 18.

Calculation Summary | |
---|---|

sample mean (x̄_{1}) | 25 |

sample mean (x̄_{2}) | 18 |

variance (s_{1}²) | 7 |

variance (s_{2}²) | 11 |

sample size (n_{1}) | 16 |

sample size (n_{2}) | 20 |

Significance Level (α) | 0.2 |

t_{0} | 6.6787 |

t_{e} | 1.31 |

## Work with Steps for x̄_{1} = 25, x̄_{2} = 18 at α = 20%

Solved example work with steps to find t-statistic (t_{0}), critical (table) value (t_{e}) for the test of significance at α = 0.2 using student's two tailed t-test for degrees of freedom ν = 34, sample means x̄_{1} = 25 & x̄_{2} = 18, s_{1} = 7 & s_{2} = 11, sample size n_{1} = 16 & n_{2}= 20 in statistical experiment may helpful for beginners or grade school students to learn how to solve t-test worksheet or homework problems efficiently by using t-statistic & critical value of t from t-distribution table.

__Workout :__

step 1 Address the formula, input parameters and values

Sample mean (x̄

_{1})= 25

Sample mean (x̄

_{2})= 18

Sample variance (s

_{1}²) = 7

Sample variance (s

_{2}²) = 11

Sample size (n

_{1}) = 16

Sample size (n

_{2}) = 20

Level of significance (α) = 0.2

t

_{0}=

x̄

_{1}- x̄

_{2}

√

^{2}x {(1 / n

_{1}) + (1 / n

_{2})}

^{2}

S

^{2}=n

_{1}s

_{1}² + n

_{2}s

_{2}²n

_{1}+ n

_{2}- 2

S

^{2}=(16 x 7) + (20 x 11)16 + 20 - 2

S

^{2}=33234

= 9.7647

step 3 Substitute x̄

_{1}, x̄

_{2}, variance 1,variance 2, n

_{1}, n

_{2}value in the below formula for test of significance for difference between two means.

=25 - 18

√

step 4 To find t

_{0}, simplify the above expression

=7

√

=7

√

=7

√

=71.0481

t

_{0}= 6.6787

step 4 Find expected t

_{e}using α , degrees of freedom from t-distribution table

d.f = n

_{1}+ n

_{2}- 2

d.f = 16 + 20 - 2

d.f = 34

The critical value of t

_{e}from the t-distribution table for the degrees of freedom (n

_{1}+ n

_{2}- 2) at 0.2 significance level.

The critical value t

_{e}(34) from t-distribution table at 0.2 of significance level is 1.31

t

_{e}= 1.31

__Inference__t

_{0}> t

_{e}

The null hypothesis H

_{0}is rejected since t

_{0}= 6.6787 is greater than the critical value for degrees of freedom t

_{e}(33) = 1.31. Therefore, there is significant difference between the sample & population means.