Question

Grades and AM/PM Section of Stats: There were two large sections of statistics this term at State College, an 8:00 (AM) section and a 1:30 (PM) section. The final grades for both sections are summarized in the contingency table below.

Observed Frequencies: O_i's

	A	B	C	D	F	Totals
AM	6	11	17	18	20	72
PM	19	19	17	12	9	76
Totals	25	30	34	30	29	148

The Test: Test for a significant dependent relationship between grades and the section of the course. Conduct this test at the 0.05 significance level.(a) What is the null hypothesis for this test?

H₀: The section (AM/PM) of a course and the grades are independent variables.

H₀: The section (AM/PM) of a course and the grades are dependent variables.     

(b) What is the value of the test statistic? Round to 3 decimal places unless your software automatically rounds to 2 decimal places.

χ² = ?

(c) Use software to get the P-value of the test statistic. Round to 4 decimal places unless your software automatically rounds to 3 decimal places.
P-value = ?

(d) What is the conclusion regarding the null hypothesis?

reject H₀

fail to reject H₀    

(e) Choose the appropriate concluding statement.

We have proven that grades and section of the course are independent.

The evidence suggests that there is a significant dependent relationship between grades and the section of the course.    

There is not enough evidence to conclude that there is a significant dependent relationship between grades and the section of the course

Answer 1

Given table data is as below

MATRIX	col1	col2	col3	col4	col5	TOTALS
row 1	6	11	17	18	20	72
row 2	19	19	17	12	9	76
TOTALS	25	30	34	30	29	N = 148

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calculation formula for E table matrix

E-TABLE	col1	col2	col3	col4	col5
row 1	row1*col1/N	row1*col2/N	row1*col3/N	row1*col4/N	row1*col5/N
row 2	row2*col1/N	row2*col2/N	row2*col3/N	row2*col4/N	row2*col5/N

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expected frequecies calculated by applying E - table matrix formulae

E-TABLE	col1	col2	col3	col4	col5
row 1	12.1622	14.5946	16.5405	14.5946	14.1081
row 2	12.8378	15.4054	17.4595	15.4054	14.8919

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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

Oi	Ei	Oi-Ei	(Oi-Ei)^2	(Oi-Ei)^2/Ei
6	12.1622	-6.1622	37.9727	3.1222
11	14.5946	-3.5946	12.9211	0.8853
17	16.5405	0.4595	0.2111	0.0128
18	14.5946	3.4054	11.5967	0.7946
20	14.1081	5.8919	34.7145	2.4606
19	12.8378	6.1622	37.9727	2.9579
19	15.4054	3.5946	12.9211	0.8387
17	17.4595	-0.4595	0.2111	0.0121
12	15.4054	-3.4054	11.5967	0.7528
9	14.8919	-5.8919	34.7145	2.3311
				ᴪ^2 o = 14.1681

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set up null vs alternative as
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
level of significance, alpha = 0.05
from standard normal table, chi square value at right tailed, χ^2 alpha/2 =9.4877
since our test is right tailed,reject Ho when ᴪ^2 o > 9.4877
we use test statistic χ^2 o = Σ(Oi-Ei)^2/Ei
from the table , χ^2 o = 14.1681
critical value
the value of | χ^2 alpha| at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 5 - 1 ) = 1 * 4 = 4 is 9.4877
we got | χ^2| =14.1681 & | χ^2 alpha | =9.4877
make decision
hence value of | χ^2 o | > | χ^2 alpha| and here we reject Ho
χ^2 p_value =0.0068
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null, Ho: the section (am/pm) of a course and the grades are independent variables.
alternative, H1: the section (AM/PM) of a course and the grades are dependent variables.
test statistic: 14.17
p-value:0.0068
decision: reject Ho          
the evidence suggests that there is a significant dependent relationship
between grades and the section of the course.