Question

In: Statistics and Probability

31. An instructor wants to test whether attending class influences how students perform on an exam....

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?

Solutions

Expert Solution

(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.


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