# Solved (Free): A teacher figures that final grades in the statistics department are distributed as

#### ByDr. Raju Chaudhari

Apr 3, 2021

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.