An undergraduate math course gives all the usual letter grades of A, B, C, D, or F. The department has determined that there should be 10% A’s given, 30% B’s given, 40% C’s, 10% D's, and 10% F's given as final grades for the course. For the spring 2020 semester, 500 students took the course and 100 students received A’s, 200 students received B’s, 100 students received C's, 50 students received D's, and 50 students received F’s, so those are my observed frequencies. We want to do a Chi-square Multinomial Goodness of Fit test to see if our grading guidelines match up with what is actually happening. What is the best description for the null hypothesis from the choices below?

Answer :

cchilabert

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the final grades for the current semester follow the  grading guidelines for the final grades given for the course.

The theoretical frequencies are:

PA= 0.10; PB= 0.30; PC= 0.40; PD= 0.10; PF= 0.10

The observed frequencies are:

A: 100 students

B: 200 students

C: 100 students

D: 50 students

F: 50 students

n= 500 students

To test this you have to apply a Chi Square goodness to fit test. If the variable of interest is X: Final grade that a student received for the course, categorized: A, B, C, D, F.

The hypotheses are:

H₀: PA= 0.10; PB= 0.30; PC= 0.40; PD= 0.10; PF= 0.10

H₁: The observed frequencies do not follow the theoretical distribution

α: 0.05

This test statistic has k-1 degrees of freedom (k= number of categories of the variable) and the rejection region is one-tailed to the right:

[tex]X^2_{k-1; 1-\alpha /2}= X^2_{4; 0.975}= 11.143[/tex]

The rejection region is then X₄² ≥ 11.143

The expected frequencies for each category is calculated using the formula:

Ei= n*Pi

[tex]E_A= n*PA= 500*0.10= 50[/tex]

[tex]E_B= n*PB= 500*0.3= 150\\[/tex]

[tex]E_C= n*PC= 500*0.40= 200[/tex]

[tex]E_D= n*PD= 500*0.10= 50[/tex]

[tex]E_F= n*PF= 500*0.10= 50[/tex]

[tex]X^2= sum (\frac{(O_i-E_i)^2}{E_i})~~X^2_{k-1}[/tex]

[tex]X^2_{H_0}= \frac{(O_A-E_A)^2}{E_A} +\frac{(O_B-E_B)^2}{E_B} +\frac{(O_C-E_C)^2}{E_C} +\frac{(O_D-E_D)^2}{E_D} +\frac{(O_F-E_F)^2}{E_F}[/tex]

[tex]X^2_{H_0}= \frac{(100-50)^2}{50} +\frac{(200-150)^2}{150} +\frac{(100-200)^2}{200} +\frac{(50-50)^2}{50} +\frac{(50-50)^2}{50}= \frac{350}{3}= 116.67[/tex]

The value of the statistic is greater than the critical value, the decision is to reject the null hypothesis.

At a 5% significance level, there is significant evidence to reject the null hypothesis, so you can conclude that the distribution of the final grades of this semesters course doesn't follow the grading guidelines.

I hope this helps!

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