Question

4. I have reviewed the midterm grades, and I am trying to predict how well the class will do on the final exam. The frequency values are actual grades for the Summer II midterm:
Grade	Expected	Frequency (Observed)
Greater than A	5	7
A	6	9
B	8	4
C	3	4
D	1	0
F	2	1

a. What will be my degrees of freedom?

b. What is O-E (difference) for grades of B?

c. What is O-E^2 for grades of A?

d. What is (O-E)^2/E for grades of F?

e. What is O-E (difference) for grades of B?

f. What is the chi square (χ2) test statistic?

g. What is the p-value?

h. At the 5% significance level, what is your decision?

i. What is the reason for the decision?

j. What is the conclusion in full sentences?

Answer 1

Solution:

We have to test if distribution of midterm grades is fitted with expected grades of final examination.

Thus hypothesis of the study are:

H0:  The distribution of midterm grades is fitted with expected grades of final examination.

Vs

Ha:  The distribution of midterm grades does not fitted with expected grades of final examination.

We are given following table of expected and observed grades.

Grade	E: Expected	O: Frequency (Observed)
Greater than A	5	7
A	6	9
B	8	4
C	3	4
D	1	0
F	2	1

Part a. What will be my degrees of freedom?

Degrees of freedom = k - 1

Where k = Number of categories of grades = 6

Degrees of freedom = 6 - 1

Degrees of freedom = 5

Part b. What is O-E (difference) for grades of B?

O-E for B grade = 4 - 8

O-E for B grade = - 4

Part c. What is O-E^2 for grades of A?

(O-E)² for A grade = ( 9 - 6 )² = (3)² = 9

Part d. What is (O-E)^2/E for grades of F?

(O-E)² / E = (1-2)² / 2 = (-1)² / 2 = 1 / 2 = 0.5

Part e. What is O-E (difference) for grades of B?

O-E for B grade = 4 - 8

O-E for B grade = - 4

Part f. What is the chi square (χ2) test statistic?

Thus we need to make following table:

Grade	E: Expected	O: Frequency (Observed)	(O - E)	(O - E)2	(O - E)2 / E
Greater than A	5	7	2	4	0.8
A	6	9	3	9	1.5
B	8	4	-4	16	2
C	3	4	1	1	0.333
D	1	0	-1	1	1
F	2	1	-1	1	0.5

Part g. What is the p-value?

Look in Chi-square table for df = 5 row and find the interval in which fall, then find corresponding right tail area.

fall between 1.610 and 9.236, corresponding right tail area is between 0.100 and 0.900

thus p-value is between 0.100 and 0.900

That is: 0.100 < p-value < 0.900

To get exact p-value we use excel command:

=CHISQ.DIST.RT( x² , df )

=CHISQ.DIST.RT( 6.133 , 5 )

=0.2935

Thus p-value = 0.2935

Part h. At the 5% significance level, what is your decision?

At the 5% significance level, we fail to reject null hypothesis H0.

Part i. What is the reason for the decision?

We fail to reject null hypothesis H0. since p-value > 0.05 significance level.

Part j. What is the conclusion in full sentences?

Since we failed to reject null hypothesis H0, The distribution of midterm grades is fitted with expected grades of final examination.