# F-Test Solved Example for x̄ = 14.7143, ȳ = 15.5714

Solved example with steps, formula & summary to estimate the F-statistic (F

_{0}), F-critical (table) value (F_{e}) for given degrees of freedom & hypothesis test (H_{0}) at a stated level of significance for sample x = {11, 13, 14, 16, 18, 11 & 20} & y = {20, 13, 12, 14, 15, 17 & 18} to check if the results of analysis between two*variances*is statistically significant.Calculation Summary | |
---|---|

Data set value x | {11, 13, 14, 16, 18, 11 & 20} |

Data set value y | {20, 13, 12, 14, 15, 17 & 18} |

F_{0} | 1.4368 |

F_{e} | 20.0891 |

## F_{0}, F_{e} & H_{0} Workout for x̄ = 14.7143, ȳ = 15.5714 & α = 0.001

The below is the work with step by step calculation shows how to estimate the F-statistic (F_{0}), critical (table) value (F_{e}) for degrees of freedom & hypothesis test (H_{0}) at a stated level of significance (α = 0.001) for x = 11, 13, 14, 16, 18, 11 & 20 and y = 20, 13, 12, 14, 15, 17 & 18 to check the quality of variances among two or more sample dataset may help learners or grade school students to solve the similar F-test (F_{0}, F_{e} & H_{0}) worksheet problems efficiently.

__Workout :__

step 1 Address the formula, input parameters and values

x = {11, 13, 14, 16, 18, 11 & 20}

y = {20, 13, 12, 14, 15, 17 & 18}

α = 0.001

step 2Refer the below

*F-table*to conduct the test of significance

x̄ = ∑xn

_{1}& ȳ = ∑yn

_{2}

x̄ = 1037& ȳ = 1097

x̄ =14.7143 & ȳ =15.5714

x | x - x̄ | (x - x̄)^{2} | y | y - ȳ | (y - ȳ)^{2} |

11 | -3.7143 | 13.79602449 | 20 | 4.4286 | 19.61249796 |

13 | -1.7143 | 2.93882449 | 13 | -2.5714 | 6.61209796 |

14 | -0.7143 | 0.51022449 | 12 | -3.5714 | 12.75489796 |

16 | 1.2857 | 1.65302449 | 14 | -1.5714 | 2.46929796 |

18 | 3.2857 | 10.79582449 | 15 | -0.5714 | 0.32649796 |

11 | -3.7143 | 13.79602449 | 17 | 1.4286 | 2.04089796 |

20 | 5.2857 | 27.93862449 | 18 | 2.4286 | 5.89809796 |

∑x = 103 | ∑(x-x̄) = -9.9999999997991E-5 | ∑(x-x̄)^{2} = 71.42857143 | ∑y = 109 | ∑(y-ȳ) = 0.00019999999999598 | ∑(y-ȳ)^{2} = 49.71428572 |

step 3 Substitute the above table to below formulas

S

_{1}

^{2}= ∑(x - x̄)

^{2}n

_{1}-1

S

_{1}

^{2}= 71.428571437-1 = 71.428571436

S

_{1}

^{2}= 11.904761905

S

_{2}

^{2}= ∑(y - ȳ)

^{2}n

_{2}-1

S

_{1}

^{2}= 49.714285727-1 = 49.714285726

S

_{2}

^{2}= 8.2857142866667

step 4 Find F-statistic (F

_{0}) using the below formula

F

_{0}= S

_{1}

^{2}S

_{2}

^{2}

It should be noted that the numerator is always greater than the denominator in F

_{0}

F

_{0}= 11.904761905/8.2857142866667

F

_{0}= 1.4368

step 5 Find critical value of F (F

_{e}) using stated level of significance (α = 0.001) & degrees of freedom (ν

_{1}, ν

_{2}) from

*F-distribution table*

ν

_{1}= n

_{1}- 1

ν

_{2}= n

_{2}- 1

ν

_{1}= 7 - 1 = 6

ν

_{2}= 7 - 1 = 6

The critical value F(6, 6) from F-distribution table at 0.001 of significance level is 20.0891

F

_{e}= 20.0891

__Inference__There is no significance difference

since the calculated value of F

_{0}= 1.4368 is smaller than the table value F

_{e}= 20.0891

Therefore the null hypothesis H

_{0}is accepted.