Question

Question 12 (3 points) 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 form the web, and 15% will read in 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. Use the framework below to guide your work. Hypotheses:: Ho: The actual distribution follow's the professor's predictions in that p(hard copy) =60%, p(print) = 25% and plonline) = 15% Ha: The actual distribution does not follow that of the professor's predictions Blank #1: With the assumption of Ho, the expected count of each cell is (enter your answer as hard copy E,print E,online E with no extra spaces and no rounding) Blank #2: The degrees of freedom for the GOF test are Blank #3: Test statistic = (round to one decimal place) Blank #4: p-value = (round to four decimal places) Blank #5: Test decision: We decide to Ho (reject or do not reject) Blank #6: Conclusion back into the words of problem: The evidence _(favors or does not favor) that the actual distribution does not follow that of the professor's predictions

Answer 1

Solution :

Null and alternative hypotheses :

The null and alternative hypotheses would be as follows:

H_{​​​​​​0} : The actual follows the professor's prediction in that P(hard copy) = 60%, P(Print) = 25% and P(Web) = 15%.

H_{​​​​​​a} : The actual prediction does not follow that of the professor's prediction.

Blank #1 :

With the assumption of H_{​​​​​​0}, the expected count of each cell are as follows :

Hard copy E = (126 × 0.60) = 75.6

Print E = (126 × 0.25) = 31.5

Web E = (126 × 0.15) = 18.9

Blank #2 :

Degrees of freedom = (c - 1)

Where, c is number of categories in which prediction has been made.

We have, c = 3

The degrees of freedom for the GOE test are (3 - 1) = 2.

Blank #3 :

To test the hypothesis we shall use chi-square test of goodness of fit. The test statistic is given as follows :

2 (Ο; – e;)2 e; i=1

Where, O_{​​​​​​i}​​​​​'s are observed counts and e_{​​​​​​i}​​​​​'s are expected counts.

Oi ei (Oi - ei)2 (Oi - ei)?/ei 21.16 0.2799 71 75.6 30 31.5 2.25 0.0714 25 18.9 37.21 1.9688 2.3201

From the above table we get,

[(Ο; – e;)? Σ = 2.3201 ei

$\large \therefore \chi^{2} = 2.3201$

The value of the test statistic is 2.3201.

Blank #4 :

The p-value is given as follows :

P-value = P(χ^{​​​​​​2} > value of the test statistic)

P-value = P(χ² > 2.3201)

P-value = 0.3135

The p-value is 0.3135.

Blank #5 :

In our question the significance level is not given. Generally significance level 0.05 or 0.01 is used. We shall use significance level 0.05

Significance level = 0.05

P-value = 0.3135

(0.3135 > 0.05)

Since, p-value is greater than the significance level of 0.05, therefore we shall be fail to reject the null hypothesis (H_{​​​​​​0}) at 0.05 significance level.

We decide to do not reject H_{​​​​​​0}.

Blank #6 :

The evidence does not favor that the actual distribution does not follow that of the professor's prediction.

Please rate the answer. Thank you.