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

Solved example with steps, formula & summary to estimate the F-statistic (F0), F-critical (table) value (Fe) for given degrees of freedom & hypothesis test (H0) 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}
F01.4368
Fe20.0891

## F0, Fe & H0 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 (F0), critical (table) value (Fe) for degrees of freedom & hypothesis test (H0) 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 (F0, Fe & H0) 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 = ∑xn1 & ȳ = ∑yn2

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

S12 = ∑(x - x̄)2n1-1

S12 = 71.428571437-1 = 71.428571436

S12 = 11.904761905

S22 = ∑(y - ȳ)2n2-1

S12 = 49.714285727-1 = 49.714285726

S22 = 8.2857142866667

step 4 Find F-statistic (F0) using the below formula

F0= S12S22

It should be noted that the numerator is always greater than the denominator in F0
F0 = 11.904761905/8.2857142866667

F0 = 1.4368

step 5 Find critical value of F (Fe) using stated level of significance (α = 0.001) & degrees of freedom (ν1, ν2) from F-distribution table
ν1 = n1 - 1
ν2 = n2 - 1
ν1 = 7 - 1 = 6
ν2 = 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 F0 = 1.4368 is smaller than the table value Fe = 20.0891
Therefore the null hypothesis H0 is accepted.