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} |
| F0 | 1.4368 |
| Fe | 20.0891 |
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
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.
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.
