A professor using an open source introductory statistics book predicts that 60%
ID: 3230558 • Letter: A
Question
A professor using an open source introductory statistics book predicts that 60% of the students will purchase a hard copy of the book, 25% will print it out from the web, and 15% will read it online. At the end of the semester he asks his students to complete a survey where they indicate what format of the book they used. Of the 126 students, 71 said they bought a hard copy of the book, 30 said they printed it out from the web, and 25 said they read it online.
What is the 2 statistic for the hypothesis test of {H0: the observed proportions/counts = professor’s predicted proportions/counts; HA: the observed proportions/counts professor’s predicted proportions/counts}?
Answer:___________________________
Following up with the question above, what is the p-value for this hypothesis test?
Select one:
a. Between 0.02 and 0.05
b. Between 0.05 and 0.1
c. Between 0.1 and 0.2
d. Between 0.2 and 0.3
e. Greater than 0.3
Explanation / Answer
H0: the observed proportions/counts = professor’s predicted proportions/counts;
HA: the observed proportions/counts professor’s predicted proportions/counts
2 = (Oi – Ei )2 / Ei
Observed (Oi)
Expected (Ei)
chi-squ
purchase a hard copy
71
126*0.60=75.6
0.279894
printout
30
126*0.25=31.5
0.071429
read online
25
126*0.15=18.9
1.968783
Total
126
126
2.320106
2 = 2.320106
d.f = 3-1 = 2
= 0.05
P-value is obtained using software.
p-value = 0.3135
p values is e) greater than 0.3
Since p value is greater than 0.05, there is no evidence to reject the null hypothesis. Hence we conclude that the observed proportions/counts = professor’s predicted proportions/counts.
Observed (Oi)
Expected (Ei)
chi-squ
purchase a hard copy
71
126*0.60=75.6
0.279894
printout
30
126*0.25=31.5
0.071429
read online
25
126*0.15=18.9
1.968783
Total
126
126
2.320106
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