Question

A teacher figures that final grades in the chemistry department are distributed as: A, 25%; B, 25%;C, 40%;D, 5%; F, 5%. At the end of a randomly selected semester, the following number of grades were recorded. Calculate the chi-square test statistic x^2 to determine if the grade distribution for the department is different than expected. Use α = 0.01.
Grade A B C D F
Number 36 42 60 14 8

a. 6.87
b. 0.6375
c. 5.25
d. 4.82

Answer 1

Answer:

$E_{A} =0.25*160=40$  

$E_{B} =0.25*160=40$

$E_{C} =0.4*160=64$

$E_{D} =0.05*160=8$  

$E_{F} =0.05*160=8$  

And now we can calculate the statistic:  

$\chi^2 = \frac{(36-40)^2}{40}+\frac{(42-40)^2}{40}+\frac{(60-64)^2}{64}+\frac{(14-8)^2}{8}+\frac{(8-8)^2}{8} =5.25$  

The answer would be:

c. 5.25

Step-by-step explanation:

The observed values are given by:

A: 36

B: 42

C: 60

D: 14

E: 8

Total =160

We need to conduct a chi square test in order to check the following hypothesis:  

H0: There is no difference in the proportions for the final grades

H1: There is a difference in the proportions for the final grades

The level of significance assumed for this case is $\alpha=0.01$  

The statistic to check the hypothesis is given by:  

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$  

Now we just need to calculate the expected values with the following formula $E_i = \% * total$  

And the calculations are given by:  

$E_{A} =0.25*160=40$  

$E_{B} =0.25*160=40$

$E_{C} =0.4*160=64$

$E_{D} =0.05*160=8$  

$E_{F} =0.05*160=8$  

And now we can calculate the statistic:  

$\chi^2 = \frac{(36-40)^2}{40}+\frac{(42-40)^2}{40}+\frac{(60-64)^2}{64}+\frac{(14-8)^2}{8}+\frac{(8-8)^2}{8} =5.25$  

The answer would be:

c. 5.25

Now we can calculate the degrees of freedom for the statistic given by:  

$df=(categories-1)=(5-1)=4$

And we can calculate the p value given by:  

$p_v = P(\chi^2_{4} >5.25)=0.263$  

The p value is higher than the significance so we have enough evidence to FAIL to reject the null hypothesis