Question

6.41 Open source textbook: 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.
(a) State the hypotheses for testing if the professor's predictions were inaccurate.

• Ho: p_Buy = .6, p_Print=.25, p_Online=.15
  Ha: all of the claimed probabilities are different
• Ho: p_Buy = .6, p_Print=.25, p_Online=.15
  Ha: at least one of the claimed probabilities is zero
• Ho: p_Buy = .6, p_Print=.25, p_Online=.15
  Ha: at least one of the claimed probabilities is different

(b) How many students did the professor expect to buy the book, print the book, and read the book exclusively online? (please do not round)

	Observed	Expected
Buy Hard Copy	71
Print Out	30
Read Online	25

(c) Calculate the chi-squared statistic, the degrees of freedom associated with it, and the p-value.
The value of the test-statistic is:  (please round to two decimal places) The degrees of freedom associated with this test are:  The p-value associated with this test is:

• less than .01
• between .05 and .1
• greater than .1
• between .01 and .05

(e) Based on the p-value calculated in part (d), what is the conclusion of the hypothesis test?

• Since p ≥ α we do not have enough evidence to reject the null hypothesis
• Since p ≥ α we accept the null hypothesis
• Since p<α we fail to reject the null hypothesis
• Since p ≥ α we reject the null hypothesis and accept the alternative
• Since p<α we reject the null hypothesis and accept the alternative

Interpret your conclusion in this context.

• The data provide sufficient evidence to claim that the actual distribution differs from what the professor expected
• The data do not provide sufficient evidence to claim that the actual distribution differs from what the professor expected

Answer 1

a) Ho: p_Buy = .6, p_Print=.25, p_Online=.15
Ha: at least one of the claimed probabilities is zero

b)

		observed	Expected
category		Oi	Ei=total*p
1		71.000	75.600
2		30.000	31.500
3		25.000	18.900

c)

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	0.600	71.000	75.600	-0.53	0.280
2	0.250	30.000	31.500	-0.27	0.071
3	0.150	25.000	18.900	1.40	1.969
total	1.000	126	126		2.320
test statistic X2 =		2.32

degree of freedom =categories-1=

2

p value =

0.3135

p value greater than .1

Since p ≥ α we do not have enough evidence to reject the null hypothesis

he data do not provide sufficient evidence to claim that the actual distribution differs from what the professor expected