Question

In: Statistics and Probability

Sixty-four students in an introductory college economics class were asked how many credits they had earned...

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 = .01, is the degree of certainty independent of credits earned?

Credits Earned Very
Uncertain
Somewhat
Certain
Very
Certain
Row Total
0 – 9 11 6 5 22
10 – 59 10 5 6 21
60 or more 1 8 12 21
Col Total 22 19 23 64

(a) At α = .01, 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 = .01. (Round your answer to 2 decimal places.)

Critical value               

(d) We can reject the null hypotheses and find independence.

  • No

  • Yes

Solutions

Expert Solution

Observed Frequencies
Very Uncertain Somewhat Certain Very Certain Total
0 - 9 11 6 5 22
10 - 59 10 5 6 21
60 or more 1 8 12 21
Total 22 19 23 64
Expected Frequencies
Very Uncertain Somewhat Certain Very Certain Total
0 - 9 22 * 22 / 64 = 7.5625 19 * 22 / 64 = 6.5313 23 * 22 / 64 = 7.9063 22
10 - 59 22 * 21 / 64 = 7.2188 19 * 21 / 64 = 6.2344 23 * 21 / 64 = 7.5469 21
60 or more 22 * 21 / 64 = 7.2188 19 * 21 / 64 = 6.2344 23 * 21 / 64 = 7.5469 21
Total 22 19 23 64
(fo-fe)²/fe
0 - 9 (11 - 7.5625)²/7.5625 = 1.5625 (6 - 6.5313)²/6.5313 = 0.0432 (5 - 7.9063)²/7.9063 = 1.0683
10 - 59 (10 - 7.2188)²/7.2188 = 1.0716 (5 - 6.2344)²/6.2344 = 0.2444 (6 - 7.5469)²/7.5469 = 0.3171
60 or more (1 - 7.2188)²/7.2188 = 5.3573 (8 - 6.2344)²/6.2344 = 0.5 (12 - 7.5469)²/7.5469 = 2.6276

Null and Alternative hypothesis:

Ho: Factors are independent.

H1: Factor are dependent.

a) Yes

b) Test statistic:

χ² = ∑ ((fo-fe)²/fe) = 12.79

df = (r-1)(c-1) = 4

p-value = CHISQ.DIST.RT(12.79, 4) = 0.0123

c) Critical value:

χ²α = CHISQ.INV.RT(0.01, 4) = 13.28

d) Decision:

p-value > α, Do not reject the null hypothesis.

No.


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