# CHI-squared Test Example for df = 4 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 = {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) | |

χ²_{0} | 376.8441 |

χ²_{e} | 0 |

## 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 (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 = {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 (χ²_{0}, χ²_{e} & H_{0}) 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 (χ²

_{0})

χ² = ∑(O

_{i}- E

_{i})

^{2}E

_{i}

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

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)^{2}E = 376.8441 |

_{0}= 376.8441

step 3 Find the degrees of freedom

df = n - 1

df = 5 - 1

df = 4

step 4 Find the critical value of χ²

_{e}for df = 4 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}= 376.8441 is greater than χ²

_{e}= 0.

Therefore the null hypothesis H

_{0}is rejected.

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