Chapter 12.1, Problem 22E

Grade distribution: A statistics teacher claims that, on the average, 20% of her students get a grade of A, 35% get a B, 25% get a C, 10% get a D, and 10% get an F. The grades of a random sample of 100 students were recorded. The following table presents the results.

1. How many of the students in the sample got an A? How many got an F?
2. Compute the expected frequencies.
3. Which grades were given more often than expected? Which grades were given less often than expected?
4. What is the value of ${\chi }^2?$
5. How many degrees of freedom are there?
6. Test the hypothesis that the grades follow the distribution claimed by the teacher. Use the 0.05 level of significance.

To determine

To calculate: The number of students that got an A and an F.

Explanation of Solution

Given:

The data set is:

Category	Observed
A	29
B	42
C	20
D	5
F	4

From the provided table, it is clear that there are 29 students that got an A and 4 students got an F.

To determine

To calculate: The expected frequencies.

Explanation of Solution

The formula to calculate the expected frequency is:

  ${\left(\text{E}\right)}_i=n{p}_i$

Here, n can be calculated as:

  $\begin{array}{c}n=29+42+20+5+4\\ =100\end{array}$

Thus, the expected frequencies can be calculated as:

  $\begin{array}{c}{E}_1=n{p}_1\\ =100\left(0.2\right)\\ =20\\ E_2=n{p}_2\\ =100\left(0.35\right)\\ =35\\ E_3=n{p}_3\\ =100\left(0.25\right)\\ =25\\ E_4=n{p}_4\\ =100\left(0.10\right)\\ =10\\ E_5=n{p}_5\\ =100\left(0.10\right)\\ =10\end{array}$

To determine

To find: The grades that were given more often and less often than expected.

Explanation of Solution

By comparing the expected frequencies with observed frequencies it is clear that A and B grades were given more often than expected. While, C and D grades were given less often.

To determine

To find: The value of ${\chi }^2$ .

Answer to Problem 22E

The required value is 12.55.

Explanation of Solution

The provided data can be calculated as:

Category	$0$	$E$	$\left(0-E\right)$	${\left(0-E\right)}^2$	$\frac{{\left(0-E\right)}^2}{E}$
A	29	20	9	81	4.05
B	42	35	7	49	1.4
C	20	25	-5	25	1
D	5	10	-5	25	2.5
F	4	10	-6	36	3.6

Thus, the value of chi-square $\left({\chi }^2\right)$ can be calculated as:

  $\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(0-E\right)}^2}{E}}\\ =4.05+1.4+\mathrm{....}+3.6\\ =12.55\end{array}$

To determine

To find: The degree of freedom.

Answer to Problem 22E

The degree of freedom is 4.

Explanation of Solution

Number of categories are 5.

The degree of freedom can be calculated as:

  $\begin{array}{c}\text{Degree of freedom}=\text{Number of categories-1}\\ =5-1\\ =4\end{array}$

To determine

To check: Whether the grades are same as suggested by teacher by conducting the hypothesis at 5% level of significance.

Explanation of Solution

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0=p_1=0.20,p_2=0.35,p_3=0.25,p_4=0.10,p_5=0.10\\ H_a:\text{Some of the }{p}_i\text{ are different}\end{array}$

The critical value of chi-square at 5% level of significance and 4 degree of freedom is 9.488. And, the calculated value of chi-square is 1.2.55.

Since, the calculated value is greater than the critical chi-square value, the null hypothesis is rejected. Thus, there is insufficient evidence to conclude that grades do not follow the distribution as suggested by the teacher.