# t-Test for x̄_{1} = 156.75, x̄_{2} = 167, n_{1} = 4, n_{2} = 5 & α = 10%

Solved example work with steps, 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 = 0.1 for difference between two small samples x = {150, 160, 152 & 165} & y = {173, 154, 164, 179 & 165}.
The below is the calculation summary for the test of significance or hypothesis using student's t-test for difference between two small samples with *means* x̄ = 156.75 & ȳ = 167.

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

Dataset (x) | {150, 160, 152 & 165} |

Dataset (y) | {173, 154, 164, 179 & 165} |

Significance level (α) | 0.1 or 10% |

t-statistic (t_{0}) | 1.7923 |

Critical value (t_{e}) | 1.89 |

## Work with Steps for x̄ = 156.75, ȳ = 167 at α = 0.1

The t-test work with steps to find t-statistic (t_{0}), t-critical (t_{e}) and hypothesis (H_{0}) for dataset x = {150, 160, 152 & 165} & y = {173, 154, 164, 179 & 165} to estimate the test of significance for difference between two small sample means.

__Workout :__

step 1 Address the formula, input parameters and values

Dataset (x) = 150, 160, 152 & 165

Dataset (y) = 173, 154, 164, 179 & 165

Level of significance (α) = 0.1

t

_{0}=

x̄ - ȳ

√

S

^{2}x {(1 / n_{1}) + (1 / n_{2})}step 2 To find x̄, ȳ, & s

^{2}

x | x-x̄ (x-156.75) | (x-x̄)^{2} |

150 | -6.75 | 45.5625 |

160 | 3.25 | 10.5625 |

152 | -4.75 | 22.5625 |

165 | 8.25 | 68.0625 |

∑x = 627 | x-x̄ = 0 | (x-x̄)^{2} = 146.75 |

y | y-ȳ (y-167) | (y-ȳ)^{2} |

173 | 6 | 36 |

154 | -13 | 169 |

164 | -3 | 9 |

179 | 12 | 144 |

165 | -2 | 4 |

∑y = 835 | y-ȳ = 0 | (y-ȳ)^{2} = 362 |

x̄ =∑xn

_{1}

x̄ =627/4 = 156.75

ȳ =∑yn

_{2}

ȳ =835/5 = 167

s

^{2}= 1/n1 + n2 - 2[ ∑(x - x̄)

^{2}+ ∑(y - ȳ)

^{2}]

s

^{2}= 1/4 + 5 - 2[ 146.75 + 362 ]

s

^{2}= 1/7 x 508.75

s

^{2}= 72.6786

step 3 Substitute x̄ , ȳ, s

^{2}, n

_{1}& n

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

=156.75 - 167

√

72.6786 x {(1 / 4) + (1 / 5)}

step 4 To find t

_{0}, simplify the above expression

=10.25

√

72.6786 x {(1 / 4) + (1 / 5)}

=10.25

√

72.6786 x 0.45

=10.25

√

32.7054

=10.255.7189

t

_{0}= 1.7923

step 5 Find expected t

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

d.f = n

_{1}+ n

_{2}- 2

d.f = 4 + 5 - 2

d.f = 7

The critical value t

_{e}from the two tailed t-distribution table for the degrees of freedom ν = 7 at 0.1 significance level.

t

_{e}= 1.89

__Inference__t

_{0}< t

_{e}

The null hypothesis H

_{0}is accepted since t

_{0}= 1.7923 is smaller than the critical value for degrees of freedom t

_{e}(6) = 1.89. Therefore, there is no significance difference between the sample & population means.