# F-Test Solved Example for x̄ = 78.8, ȳ = 76

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 = {50, 72, 85, 96 & 91} & y = {55, 65, 77, 89 & 94} to check if the results of analysis between two*variances*is statistically significant.Calculation Summary | |
---|---|

Data set value x | {50, 72, 85, 96 & 91} |

Data set value y | {55, 65, 77, 89 & 94} |

F_{0} | 1.2867 |

F_{e} | 4.1072 |

## F_{0}, F_{e} & H_{0} Workout for x̄ = 78.8, ȳ = 76 & α = 0.1

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 = 0.1) for x = 50, 72, 85, 96 & 91 and y = 55, 65, 77, 89 & 94 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 = {50, 72, 85, 96 & 91}

y = {55, 65, 77, 89 & 94}

α = 0.1

step 2Refer the below

*F-table*to conduct the test of significance

x̄ = ∑xn

_{1}& ȳ = ∑yn

_{2}

x̄ = 3945& ȳ = 3805

x̄ =78.8 & ȳ =76

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

50 | -28.8 | 829.44 | 55 | -21 | 441 |

72 | -6.8 | 46.24 | 65 | -11 | 121 |

85 | 6.2 | 38.44 | 77 | 1 | 1 |

96 | 17.2 | 295.84 | 89 | 13 | 169 |

91 | 12.2 | 148.84 | 94 | 18 | 324 |

∑x = 394 | ∑(x-x̄) = 1.4210854715202E-14 | ∑(x-x̄)^{2} = 1358.8 | ∑y = 380 | ∑(y-ȳ) = 0 | ∑(y-ȳ)^{2} = 1056 |

step 3 Substitute the above table to below formulas

S

_{1}

^{2}= ∑(x - x̄)

^{2}n

_{1}-1

S

_{1}

^{2}= 1358.85-1 = 1358.84

S

_{1}

^{2}= 339.7

S

_{2}

^{2}= ∑(y - ȳ)

^{2}n

_{2}-1

S

_{1}

^{2}= 10565-1 = 10564

S

_{2}

^{2}= 264

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}= 339.7/264

F

_{0}= 1.2867

step 5 Find critical value of F (F

_{e}) using stated level of significance (α = 0.1) & degrees of freedom (ν

_{1}, ν

_{2}) from

*F-distribution table*

ν

_{1}= n

_{1}- 1

ν

_{2}= n

_{2}- 1

ν

_{1}= 5 - 1 = 4

ν

_{2}= 5 - 1 = 4

The critical value F(4, 4) from F-distribution table at 0.1 of significance level is 4.1072

F

_{e}= 4.1072

__Inference__There is no significance difference

since the calculated value of F

_{0}= 1.2867 is smaller than the table value F

_{e}= 4.1072

Therefore the null hypothesis H

_{0}is accepted.