# CHI-squared Test Example for df = 4 at α = Solved example work with steps, formula & summary to estimate χ²-statistic (χ²0), critical (table) value (χ²e) for given degrees of freedom & hypothesis test (H0) at a stated level of significance (α = ) for observed frequencies O = {15, 45, 105, . . . . , 9} & expected frequencies E = {68, 90, 206, . . . . , 100α=0.1} to check if the result of this probability & statistics experiment is statistically significant.
Calculation Summary
Observed Frequency (O){15, 45, 105, . . . . , 9}
Expected Frequency (E){68, 90, 206, . . . . , 100α=0.1}
Significance Level (α)
χ²0376.8441
χ²e0

## Workout for O = {15, 45, 105, . . . . , 9} & E = {68, 90, 206, . . . . , 100α=0.1}

The below is the solved example with step by step calculation shows how to estimate the chi-squared statistic (χ²0), critical (table) value (χ²e) for degrees of freedom & hypothesis test (H0) at a stated level of significance (α = ) for observed frequencies O = {15, 45, 105, . . . . , 9} & expected frequencies E = {68, 90, 206, . . . . , 100α=0.1} to check the quality of variances between two or more samples which are not uniformly distributed, may help learners or grade school students to solve the similar chi-squared test (χ²0, χ²e & H0) worksheet problems efficiently.

Workout :
step 1 Address the formula input parameters and values
observed frequency = 15, 45, 105, . . . . , 9
expected frequency = 68, 90, 206, . . . . , 100α=0.1
Significance Level (α) =

step 2 Refer the below χ² table to find χ²-statistic (χ²0)

χ² = ∑(Oi - Ei)2Ei

 Observed Frequency (O) Expected Frequency (E) (O - E)2 (O - E)2E 15 68 2809 41.3088 45 90 2025 22.5 105 206 10201 49.5194 12 204 36864 180.7059 9 100α=0.1 8281 82.81 ∑(O - E)2E = 376.8441
χ²0 = 376.8441

step 3 Find the degrees of freedom
df = n - 1
df = 5 - 1
df = 4

step 4 Find the critical value of χ²e for df = 4 at a stated level of significance α = from χ²-distribution table.
χ²e = 0

Inference
for χ²0 > χ²e
There is significance difference between sample and population
since χ²0 = 376.8441 is greater than χ²e = 0.
Therefore the null hypothesis H0 is rejected. 