In: Statistics and Probability
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
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.