Question

Sixty-four students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. At α = .02, is the degree of certainty independent of credits earned? Credits Earned Very Uncertain Somewhat Certain Very Certain Row Total 0 – 9 15 7 2 24 10 – 59 10 5 6 21 60 or more 4 10 14 28 Col Total 29 22 22 73 Click here for the Excel Data File

(a) At α = .02, the hypothesis for the given issue is H0: Credits Earned and Certainty of Major are independent. Yes No

(b) Calculate the chi-square test statistic, degrees of freedom, and the p-value. (Round your test statistic value to 2 decimal places and the p-value to 4 decimal places.) Test statistic d.f. p-value

(c) Find the critical value of the chi-square for α = .02. (Round your answer to 2 decimal places.) Critical value

(d) We can reject the null hypotheses and find independence. No Yes

Answer 1

Answer:

Sixty-four students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. At α = .02, is the degree of certainty independent of credits earned?

Solution:

a) Null and Alternative hypothesis:

H0: Credits Earned and Certainty of Major are independent.

Ha: Credits Earned and Certainty of Major are not independent.

b) Calculate the chi-square test statistic, degrees of freedom, and the p-value.

Credit Earns	Very Uncertain	Somewhat Uncertain	Very Certain	Row Total
0 - 9	15	7	2	24
10 - 59	10	5	6	21
60 or more	4	10	14	28
Total	29	22	22	73

Expected frequency, E

E = (row total)(column total)/Grand total

E = (24*29)/73 = 9.53

E = (21*29)/73 = 8.34

E = (24*22)/73 = 7.23

Same calculation for all rows and columns,and we get

Formula for chi-square test statistic, χ​​​​​​2

χ​​​​​​2 = (obser. freq - expec. freq)2 / expec. freq.

Observed Frequency (O)	Expected Frequency (E)	(O - E)2/E
15	9.53	3.14
7	7.23	0.007
2	7.23	3.78
10	8.34	0.33
5	6.33	0.28
6	6.33	0.008
4	11.12	4.56
10	8.44	0.29
14	8.44	3.66
		16.055

Therefore, chi-square test statistic, χ​​​​​​2 =16.055

Degree of freedom, df

df = (row -1)(column -1)

df = (3 - 1)(3 - 1)

df = 4

p value from test statistic χ​​​​​​2 = 16.055 and DF = 4

p value =0.0029

c) Find the critical value of the chi-square for

α = .02:

At = 0.02 significance level and degree of freedom = 4

Critical value = 11.689

(from critical value of χ​​​​​​2 table)

d) We can reject the null hypotheses and find independence. No Yes:

As test statistic is greater than critical value (16.055 > 11.689), we reject the null hypothesis at 0.02 significance level. Therefore, there is sufficient evidence to conclude that Credits Earned and Certainty of Major are not independent.