Question

Q: Assistance to understand clearly and solve this example from Probability and Statistical Inference with the steps of the solution to better understand, thanks.

**Please give the step by step with details to completely see how the solution came about, plenty of thanks.

1) To test two methods of instruction, 50 students are selected at random from each of two groups. At the end of the instruction period, each student is assigned a grade (A, B, C, D, or F) by an evaluating team. The data are recorded as following:

Grades

	A	B	C	D	F	Totals
Group I	8	13	16	10	3	50
Group II	6	10	14	14	6	50

Test the null hypothesis H₀ that the two methods of instruction made no difference at the 5% significance level.

Answer 1

Null hypothesis:Ho: two methods of instruction made no difference

Alternate hypothesis:Ha: two methods of instruction made significant difference

degree of freedom(df) =(rows-1)*(columns-1)=

4

for 4 df and 0.05 level of signifcance critical region       χ2=

9.488

Applying chi square test :

Ei=row total*column total/grand total	A	B	C	D	E	Total
Group1	7.000	11.500	15.000	12.000	4.500	50
Group2	7.000	11.500	15.000	12.000	4.500	50
total	14	23	30	24	9	100
=(Oi-Ei)2/Ei	A	B	C	D	E	Total
Group1	0.1429	0.1957	0.0667	0.3333	0.5000	1.239
Group2	0.1429	0.1957	0.0667	0.3333	0.5000	1.239
total	0.286	0.391	0.133	0.667	1.000	2.4770

Asa test statistics X² =2.477 does not fall in rejection region; we can not reject null hypothesis

we do not have evidence at 0.05 level to conclude that two methods of instruction made significant difference