In: Statistics and Probability
31. An instructor wants to test whether attending class
influences how students perform on an
exam. There are 54 students in the class. There are 25 students who
attended the class and passed
the exam, 6 students who attended class and failed the exam, 8
students who skipped the class and
passed the exam, 15 students who skipped the class and failed
class. Please perform a statistical
test and indicate whether attending class influence the exam
performance.
a) Parametric or nonparametric hypotheses?
b) Z distribution, t distribution, chi-square test or hypothesis
test of a proportion
c) Please indicate the null hypotheses.
d) Please indicate the alternative hypotheses
e) Please calculate the stand error? (If it is a chi-square test,
type NA for this question)
f) If z table will be used, type NA for this question. If t table
will be used, indicate the degree of
freedom.
g) Please calculate the statistical value.
h) Hypothesis is supported or no supported, and what is your
conclusion?
(a)
Correct option:
Nonparametric hypotheses
because the population distribution is unknown.
(b)
chi-square test
because both the variables are categorical variables.
(c)
H0: Null Hypothesis: Attending class does not influence the exam performance
(d)
HA: Alternative Hypothesis: Attending class influences the exam performance (Claim)
(e)
Correct option:
NA
since it is chi square test
(f)
Degrees of Freedom = (r - 1) X (c - 1)= (2 - 1) X (2 - 1) = 1
(g)
Observed Frequencies:
Attend class | Did not attend class | Total | |
Pass | 25 | 8 | 33 |
Fail | 6 | 15 | 21 |
Total | 31 | 23 | 54 |
Expected Frequencies:
Attend class | Did not attend class | Total | |
Pass | 31X33/54=18.944 | 23X21/54=14.056 | 33 |
Fail | 31X21/54=12.056 | 23X21/54=8,944 | 21 |
Total | 31 | 23 | 54 |
Test Statistic () is got as follows:
Observed (O) | Expected (E) | (O - E)2/E |
25 | 18.944 | 1.936 |
8 | 14.056 | 2.609 |
6 | 12.056 | 3.042 |
15 | 8.944 | 4.100 |
Total = = | 11.686 |
So,
the statistical value. = 11.686
(h)
By Technology, Critical Value of = 3.841
Since calculated value of = 11.686 is greater than critical value of = 3.841, the difference is significant. Reject null hypothesis.
Conclusion:
The data support the claim that attending class influences the exam
performance.