Question

Suppose the local state university wants to determine whether there is a relationship between a student’s gender and a student’s major in college. The registrar was asked to randomly select 55 students and record their gender and major. The majors were grouped into categories of Natural Science (NS), Social Sciences (SS), and Humanities (H). Answer the following questions based on the results in the table below.

   ns    ss    H total
men    11    9 3    23   

women 9    13    10    32

total 20    22 13 55

part I

1. Determine the expected frequency for each of the cells within the table.

2. Compute the sample chi-square statistic from the contingency table.

3. Conduct a chi-square test of independence to determine whether there is a relationship between gender and college majors. Show all of your work to support your chi-square test.

4. What conclusion can be determined from the results of the chi-square test?

Part II:

Suppose we are only interested in the college majors of the women in our study. We would like to compare our sample to the national percentage of women majoring in each of the categories (NS, SS, and H) and determine whether the sample distribution fits the national distribution. Suppose the national percentage of women majoring in Natural Sciences is 22%, majoring in Social Sciences is 28%, and majoring in the Humanities is 30%.

NS    SS H Total
9    13    10 32
1. Conduct a chi-square goodness-of-fit test to determine whether our sample data fits the national distribution. Show all of your work to support your chi-square test.

2. What conclusion can be determined from the results of the chi-square test?

Answer 1

	ns	ss	h	total
men	11	9	3	23
women	9	13	10	32
total	20	22	13	55

1)Expected frequencies of each table is given by, Expected frequencies are computed by, (row total*column total)/n

expected frequencies	ns	ss	h	total
men	8.368	9.29	5.465	22
women	11.612	12.813	7.568	33
total	20	22	13	55

2) Pearson's chi-square test statistic is given by,

  

where, O_i is the observed frequency and E_i is the expected frequency.

(11-8)²/8 + (9-12)²/12 + .....+ (10-8)²/8 = 3.175

3) Chi-square test of independence,

test statistic is given by,

  

degrees of freedom = (3-1)*(2-1) = 2*1=2

where n= 55,

f_ij  is the frequency of (i,j)th cell and f_i0 and f_0j are the marginal total of ith rows and jth column.

= 11²/(23*20)+9²/(23*22)+.....+10²/(32*13)

= 1.0602268

= 55*(1.0602268-1) = 3.3125

at = 0.05 , =5.991

As we accept the null hypothesis i.e. Gender and college majors are independent of each other.

4) We conclude from both chi-square test that there is no relationship between gender and college majors.

Part II

ns	ss	h	total
9	13	10	32

1) Given that, The percentage of women majoring in natural science is 22% and in social science is 28% and in humanities 30%.

i.e. p₁= 0.22, p₂= 0.28, p₃= 0.30

here n= 32

Chi-square goodness-of-fit test :

H₀ : p₁= 0.22, p₂= 0.28, p₃= 0.30

the test statistic is given by,

with degrees of freedom, 3-1=2

where f_i are cell frequency and p_i are corresponding probabilities.

= 9²/(32*0.22) + 13²/(32*0.28) + 10²/(32*0.30) = 40.78396

= 40.78396 - 32 = 8.78

, as at = 0.05 , =5.991.

Hence we reject the null hypothesis.

again, at , as  at = 0.01 , =9.210.

Here we failed to reject the null hypothesis.

2) Our sample data doesn't fits the national distribution at 5% significance level but fitsat 1% significance level.

As a result , we consider 5% significance level. so, our sample data doesn't fit the national distribution.

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