In: Statistics and Probability
1)
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 she asks her 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 would be the appropriate set of hypotheses for testing if the professor's predictions were accurate?
HA: The actual proportion of students who purchase the hard copy of the book is higher than 60%.
HA: The actual proportion of students who purchase the hard copy of the book is lower than 60%.
HA: The distribution of the format of the book used by the students does not follow the professor's predictions.
2)
You are conducting a multinomial hypothesis test (αα = 0.05) for the claim that all 5 categories are equally likely to be selected. Complete the table.
Category | Observed Frequency |
Expected Frequency |
|
---|---|---|---|
A | 12 | ||
B | 20 | ||
C | 8 | ||
D | 25 | ||
E | 6 |
Report all answers accurate to three decimal places. But
retain unrounded numbers for future calculations.
What is the chi-square test-statistic for this data? (Report answer
accurate to three decimal places, and remember to use the unrounded
Pearson residuals in your calculations.)
χ2=χ2=
What are the degrees of freedom for this test?
d.f.=
What is the p-value for this sample? (Report answer accurate to
four decimal places.)
p-value =
The p-value is...
This test statistic leads to a decision to...
As such, the final conclusion is that...
1)
H0: The distribution of the format of the book used by the students follows the professor's predictions.
HA: The distribution of the format of the book used by the students does not follow the professor's predictions.
2)
applying chi square goodness of fit test: |
relative | observed | Expected | residual | Chi square | |
category | frequency(p) | Oi | Ei=total*p | R2i=(Oi-Ei)/√Ei | R2i=(Oi-Ei)2/Ei |
1 | 0.20 | 12.00 | 14.20 | -0.58 | 0.341 |
2 | 0.20 | 20.00 | 14.20 | 1.54 | 2.369 |
3 | 0.20 | 8.00 | 14.20 | -1.65 | 2.707 |
4 | 0.20 | 25.00 | 14.20 | 2.87 | 8.214 |
5 | 0.20 | 6.00 | 14.20 | -2.18 | 4.735 |
total | 1.000 | 71 | 71 | 18.3662 | |
test statistic X2 = | 18.366 |
degree of freedom =categories-1= | 4 |
p value = | 0.0010 |
The p-value is less than alpha
reject the null
There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.