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CHI-squared Test Example for df = 4 at α = 0.1

Chi-squared Test CalculatorSolved example work with steps, formula and summary to estimate χ²-statistic (χ²0), critical (table) value (χ²e) for given degrees of freedom and hypothesis test (H0) at a stated level of significance (α = 0.1) for observed frequencies O = {15, 45, 105, . . . . , 9} and expected frequencies E = {68, 90, 206, . . . . , 100} to check if the result of this probability and statistics experiment is statistically significant.
Calculation Summary
Observed Frequency (O){15, 45, 105, . . . . , 9}
Expected Frequency (E){68, 90, 206, . . . . , 100}
Significance Level (α)0.1
χ²0376.8441
χ²e7.779

Workout for O = {15, 45, 105, . . . . , 9} and E = {68, 90, 206, . . . . , 100}

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 and hypothesis test (H0) at a stated level of significance (α = 0.1) for observed frequencies O = {15, 45, 105, . . . . , 9} and expected frequencies E = {68, 90, 206, . . . . , 100} to check the quality of variances between two or more samples which are not uniformly distributed, may help students and learners solve similar chi-squared test (χ²0, χ²e and 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, . . . . , 100
Significance Level (α) = 0.1

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
1568280941.3088
4590202522.5
1052061020149.5194
1220436864180.7059
9100828182.81
∑(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 = 0.1 from χ²-distribution table.
χ²e = 7.779

Inference
for χ²0 > χ²e
There is significance difference between sample and population
since χ²0 = 376.8441 is greater than χ²e = 7.779.
Therefore the null hypothesis H0 is rejected.

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