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} |
| F0 | 1.2867 |
| Fe | 4.1072 |
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
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.
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.
