# F-Test Solved Example for x̄ = 78.8, ȳ = 76 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 = {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}
F01.2867
Fe4.1072

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

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

S12 = ∑(x - x̄)2n1-1

S12 = 1358.85-1 = 1358.84

S12 = 339.7

S22 = ∑(y - ȳ)2n2-1

S12 = 10565-1 = 10564

S22 = 264

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 = 339.7/264

F0 = 1.2867

step 5 Find critical value of F (Fe) using stated level of significance (α = 0.1) & degrees of freedom (ν1, ν2) from F-distribution table
ν1 = n1 - 1
ν2 = n2 - 1
ν1 = 5 - 1 = 4
ν2 = 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 F0 = 1.2867 is smaller than the table value Fe = 4.1072
Therefore the null hypothesis H0 is accepted. 