sixty four students... Sixty-four students in an introductory college economics
ID: 3176651 • Letter: S
Question
sixty four students...
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 alpha = .01, is the degree of certainty independent of credits earned? (a) At alpha = .01, the hypothesis for the given issue is H_0: Credits Earned and Certainty of Major are independent. Yes No (b) Calculate the chi-square test statistic, degrees of freedom, and the p-value. (c) Find the critical value of the chi-square for alpha = .01. (d) We can reject the null hypotheses and find independence. YesExplanation / Answer
here we use chi-square test and chi-square=sum(O-E)2/E with (r-1)(c-1) df
chi-square=12.66 with (2-1)(2-1)=1 df
critical chi-square chi(0.01,1)=6.63
(a) Yes,
the null hypothesis will be there is no association between credit urns and certaintiy
(b) chi square statistic=12.66
df=1
p-value=0.0004 ( using ms-excel command =chidist(12.66,1))
(c) critical chi-square(0.01,1)=6.63
(d) we reject H0 and conclude that there is no indpendence between credit unrs and certaintity of major
since p-value is less than the alpha=0.1 ( or test stastistc 12.66 is more than critical chi-square 6.63)
very uncertain some what certain very certail row total 0 to 9 11 10 3 24 10 to 59 10 5 9 24 60 or more 2 9 11 22 col total 23 24 23 70 Expected(E) Observed (O-E) (O-E)2/E E(11)= 24*23/70 7.885714286 11 3.1142857 1.229917184 E(10)= 24*24/70 8.228571429 10 1.7714286 0.381349206 E(3)= 24*23/70 7.885714286 3 -4.8857143 3.027018634 E(10)= 24*23/70 7.885714286 10 2.1142857 0.566873706 E(5)= 24*24/70 8.228571429 5 -3.2285714 1.266765873 E(9)= 24*23/70 7.885714286 9 1.1142857 0.157453416 E(2)= 22*23/70 7.228571429 2 -5.2285714 3.781931112 E(9)= 22*24/70 7.542857143 9 1.4571429 0.281493506 E(11)= 22*23/70 7.228571429 11 3.7714286 1.967701863 total 70 70 0 12.6605045Related Questions
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