Grades in a statistics course and an operations research course taken simultaneously were as follows for a group of students.

Statistics Grade/ Operation Research Grade | A | B | C | Other |
---|---|---|---|---|

A | 25 | 6 | 17 | 13 |

B | 17 | 16 | 15 | 6 |

C | 18 | 4 | 18 | 10 |

Other | 10 | 8 | 11 | 20 |

Are the grades in statistics and operations research related? Use $\alpha = 0.01$ in reaching your conclusion. What is the P-value for this test?

#### Solution

The observed data is

OR A | OR B | OR C | OR Other | Sum | |
---|---|---|---|---|---|

ST A | 25 | 6 | 17 | 13 | 61 |

ST B | 17 | 16 | 15 | 6 | 54 |

ST C | 18 | 4 | 18 | 10 | 50 |

ST Other | 10 | 8 | 11 | 20 | 49 |

Sum | 70 | 34 | 61 | 49 | 214 |

Number of rows $r=4$, number of coulmns $c=4$.

**Step 1** The null and alternative hypothesis are as follows:

$H_0:$ The grades in Statistics and Operation Research are not related.

$H_1:$ The grades in Statistics and Operation Research are related.

**Step 2** Test statistic:

The test statistic for testing above hypothesis is

` $$ \begin{equation*} \chi^2= \sum \sum \frac{(O_{ij} -E_{ij})^2}{E_{ij}} \sim \chi^2_{(r-1)(c-1)}\\ \end{equation*} $$ `

**Step 3** Level of Significance

The level of significance is $\alpha =0.01$.

**Step 4** Critical value of $\chi^2$

The level of significance is $\alpha =0.01$. Degrees of freedom $df=(r-1)(c-1)=(4-1)(4-1) =9$.

The critical value of $\chi^2$ for $df=9$ and $\alpha=0.01$ level of significance is $\chi^2_{0.01,9} =21.666$.

**Step 5** Computation of test Statistic

The expected frequency for $(i,j)^{th}$ cell is given by

` $$ \begin{equation*} E_{ij} =\frac{i^{th}\text{ row total }\times j^{th}\text{ column total}}{N} \end{equation*} $$ `

For example, $E_{11}$ is given by

` $$ \begin{eqnarray*} E_{11} & = &\frac{1^{st}\text{ row total }\times 1^{st}\text{ column total}}{N}\\ &=& \frac{61*70}{214}\\ &=&19.95. \end{eqnarray*} $$ `

Table of expected frequencies

OR A | OR B | OR C | OR Other | Sum | |
---|---|---|---|---|---|

ST A | 19.95 | 9.69 | 17.39 | 13.97 | 61.00 |

ST B | 17.66 | 8.58 | 15.39 | 12.36 | 53.99 |

ST C | 16.36 | 7.94 | 14.25 | 11.45 | 50.00 |

ST Other | 16.03 | 7.79 | 13.97 | 11.22 | 49.01 |

Sum | 70.00 | 34.00 | 61.00 | 49.00 | 214.00 |

The test statistic is

` $$ \begin{eqnarray*} \chi^2&=& \sum \sum \frac{(O_{ij} -E_{ij})^2}{E_{ij}} \sim \chi^2_{(r-1)(c-1)}\\ &=&\frac{(25-19.95)^2}{19.95}+\cdots + \frac{(20-11.22)^2}{11.22}\\ &=& 25.5545. \end{eqnarray*} $$ `

**Step 6** Decision (Traditional approach)

The test statistic is $\chi^2 =25.5545$ which falls $inside$ the critical region bonded by the critical value $\chi^2_{0.01,9}=21.666$, we $\textit{reject}$ the null hypothesis.

**OR**

**Step 6** Decision ($p$-value approach)

The p-value is $P(\chi^2_{9}>25.5545) =0.00242$.

As the p-value $0.0024$ is $\textit{less than}$ the significance level of $\alpha = 0.01$, we $\textit{reject}$ the null hypothesis.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators