# CHI-squared Test Example for df = 3 at α =

Solved example work with steps, formula & summary to estimate χ²-statistic (χ²

_{0}), critical (table) value (χ²_{e}) for given degrees of freedom & hypothesis test (H_{0}) at a stated level of significance (α = ) for observed frequencies O = {95, 35, 75 & 25} & expected frequencies E = {110, 56, 70 & 23α=0.01} to check if the result of this*probability*& statistics experiment is statistically significant.Calculation Summary | |
---|---|

Observed Frequency (O) | {95, 35, 75 & 25} |

Expected Frequency (E) | {110, 56, 70 & 23α=0.01} |

Significance Level (α) | |

χ²_{0} | 10.4515 |

χ²_{e} | 0 |

## Workout for O = {95, 35, 75 & 25} & E = {110, 56, 70 & 23α=0.01}

The below is the solved example with step by step calculation shows how to estimate the chi-squared statistic (χ²_{0}), critical (table) value (χ²_{e}) for degrees of freedom & hypothesis test (H_{0}) at a stated level of significance (α = ) for observed frequencies O = {95, 35, 75 & 25} & expected frequencies E = {110, 56, 70 & 23α=0.01} 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 (χ²_{0}, χ²_{e} & H_{0}) worksheet problems efficiently.

__Workout :__

step 1 Address the formula input parameters and values

observed frequency = 95, 35, 75 & 25

expected frequency = 110, 56, 70 & 23α=0.01

Significance Level (α) =

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

_{0})

χ² = ∑(O

_{i}- E

_{i})

^{2}E

_{i}

Observed Frequency (O) | Expected Frequency (E) | (O - E)^{2} | (O - E)^{2}E |

95 | 110 | 225 | 2.0455 |

35 | 56 | 441 | 7.875 |

75 | 70 | 25 | 0.3571 |

25 | 23α=0.01 | 4 | 0.1739 |

∑(O - E)^{2}E = 10.4515 |

_{0}= 10.4515

step 3 Find the degrees of freedom

df = n - 1

df = 4 - 1

df = 3

step 4 Find the critical value of χ²

_{e}for df = 3 at a stated level of significance α = from χ²-distribution table.

χ²

_{e}= 0

__Inference__for χ²

_{0}> χ²

_{e}

There is significance difference between sample and population

since χ²

_{0}= 10.4515 is greater than χ²

_{e}= 0.

Therefore the null hypothesis H

_{0}is rejected.

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