The high school teacher was also interested in whether there is a gender differe
ID: 2922572 • Letter: T
Question
The high school teacher was also interested in whether there is a gender difference in terms of student’s choice of one of the four type of majors. So she broke the data down by gender in the following table and conducted a Chi-square test for independence with a = .05.
Type of Major
Female
Male
Total
social sciences
12
8
20
natural sciences
7
8
15
liberal arts
8
2
10
engineering
2
3
5
What are the variables in this analysis? What type of variable is each (nominal, ordinal, or continuous)? (2 points total: 1 for each variable- .5 for variable name, .5 for variable type)
State the null and alternative hypotheses in words (2 points total: 1 for each hypothesis)
Calculate X2 statistic (3 points total: 1 for final answer, 1 for setting up correct table, 1 for calculation- deduct .5 for each calculation error)
Calculate the degree of freedom and then identify the critical value (2 points total, 1 for df, 1 for critical value)
Compare the X2 statistic with the critical value, then report the hypothesis test result, using “reject” or “fail to reject” the null hypothesis in the answer (1 point total, .5 for each answer)
Explain the conclusion in a sentence or two (1 point)
Type of Major
Female
Male
Total
social sciences
12
8
20
natural sciences
7
8
15
liberal arts
8
2
10
engineering
2
3
5
Explanation / Answer
Given table data is as below
------------------------------------------------------------------
calculation formula for E table matrix
------------------------------------------------------------------
expected frequecies calculated by applying E - table matrix formulae
------------------------------------------------------------------
calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above
------------------------------------------------------------------
set up null vs alternative as
null, Ho: no relation b/w X and Y OR X and Y are independent or there is a gender difference in terms of student’s choice
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent or no difference in terms of student’s choice
level of significance, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =7.815
since our test is right tailed,reject Ho when ^2 o > 7.815
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o =3.475
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 4 - 1 ) = 1 * 3 = 3 is 7.815
we got | ^2| =3.475 & | ^2 | =7.815
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0
ANSWERS
---------------
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
test statistic:3.475
critical value: 7.815
p-value:0
decision: do not reject Ho
we have evidence that there is a gender difference in terms of student’s choice
A 12 8 20 B 7 8 15 C 8 2 10 D 2 3 5 Totals 29 21 50Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.