A teacher figures that final grades in the statistics department are distributed as:

A, 25%; B, 25%; C, 40%; D, 5%; E, 5%. At the end of a randomly selected semester, the following number of grades were recorded.

Grade | A | B | C | D | E |
---|---|---|---|---|---|

Number | 36 | 42 | 60 | 14 | 8 |

Determine if the grade distribution for the department is different than expected. Use $\alpha = 0.01$.

#### Solution

The observed data is

Grade | Obs. Freq.$(O)$ | Prop. |
---|---|---|

A | 36 | 0.25 |

B | 42 | 0.25 |

C | 60 | 0.4 |

D | 14 | 0.05 |

F | 8 | 0.05 |

##### Step 1 Setup the hypothesis

The null and alternative hypothesis are as follows:

`$H_0:p_{A}=0.25, p_{B} =0.25, p_{C} =0.40, p_{D}= 0.05, p_{F} = 0.05$`

(i.e., The distribution of grade for the department is same as expected.)

$H_1:$ The distribution of grade for the department is different than expected.

##### Step 2 Test statistic

The test statistic for testing above hypothesis is

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

##### 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 $df=k-1=5-1 =4$.

The critical value of $\chi^2$ for $df=4$ and $\alpha=0.01$ level of significance is $\chi^2 =13.2767$.

##### Step 5 Test Statistic

The expected frequencies can be calculated as

` $$ \begin{equation*} E_{i} =N*p_i \end{equation*} $$ `

For example, $E_{1}$ is given by

` $$ \begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 160*0.25\\ &=&40. \end{eqnarray*} $$ `

Grade | Obs. Freq.$(O)$ | Prop. $p_i$ | Expe.Freq.$(E)$ | $(O-E)^2/E$ |
---|---|---|---|---|

A | 36 | 0.25 | 40 | 0.4 |

B | 42 | 0.25 | 40 | 0.1 |

C | 60 | 0.4 | 64 | 0.25 |

D | 14 | 0.05 | 8 | 4.5 |

F | 8 | 0.05 | 8 | 0 |

The test statistic is

` $$ \begin{eqnarray*} \chi^2&=& \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ &=&\frac{(36-40)^2}{40}+\cdots + \frac{(8-8)^2}{8}\\ &=& 0.4 +\cdots + 0\\ &=& 5.25. \end{eqnarray*} $$ `

##### Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =5.25$ which falls $outside$ the critical region bounded by the critical value $13.2767$, we $\textit{fail to reject}$ the null hypothesis.

**OR**

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

The p-value is $P(\chi^2_{4}>5.25) =0.26259$.

As the p-value $0.2626$ is $\textit{greater than}$ the significance level of $\alpha = 0.01$, we $\textit{fail to reject}$ the null hypothesis.

There is not sufficient evidence to support the claim that the grades are different than expected.

#### Further Reading

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