91Ó°ÊÓ

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. (b) How many students did the professor expect to buy the book, print the book, and read the book exclusively online? (c) This is an appropriate setting for a chi-square test. List the conditions required for a test and verify they are satisfied. (d) Calculate the chi-squared statistic, the degrees of freedom associated with it, and the p-value. (e) Based on the p-value calculated in part (d), what is the conclusion of the hypothesis test? Interpret your conclusion in this context.

Step by step solution

01

State the Hypotheses

We want to test whether the professor's predictions about the formats students use to access the book match the observations. Therefore, we set up the null and alternative hypotheses as follows: - Null Hypothesis (\(H_0\)): The distribution of formats is as the professor predicted: 60\% hard copy, 25\% printed, and 15\% online.- Alternative Hypothesis (\(H_a\)): The distribution of formats is not as the professor predicted.

02

Calculate Expected Counts

To find the expected number of students for each format based on the professor's predictions:- Expected number buying hard copy \(= 126 \times 0.60 = 75.6\)- Expected number printing out \(= 126 \times 0.25 = 31.5\)- Expected number reading online \(= 126 \times 0.15 = 18.9\)

03

Verify Chi-Square Test Conditions

The conditions that must be met to perform a chi-square test are: 1. The samples should be randomly selected. 2. The data should be in counts and categorized. 3. Expected counts should be greater than or equal to 5 for each category. From Step 2, all expected counts (75.6, 31.5, 18.9) exceed 5, so the chi-square test conditions are met.

04

Calculate Chi-Square Statistic

The chi-square statistic is calculated using:\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i},\]where \(O_i\) is the observed count and \(E_i\) is the expected count. Calculating for each category:- Hard copy: \(\frac{(71 - 75.6)^2}{75.6} = 0.279\)- Printed: \(\frac{(30 - 31.5)^2}{31.5} = 0.071\)- Online: \(\frac{(25 - 18.9)^2}{18.9} = 1.985\)The total chi-square statistic is: \(0.279 + 0.071 + 1.985 = 2.335\).

05

Determine Degrees of Freedom and P-value

Degrees of freedom for a chi-square test are calculated as the number of categories minus 1. In this case, \(3 - 1 = 2\).Using a chi-square distribution table or calculator, with 2 degrees of freedom and a chi-square statistic of 2.335, we look up the p-value.

06

Conclusion of Hypothesis Test

Given the p-value from Step 5 (which we assume is greater than 0.05), we fail to reject the null hypothesis. Interpretation: There is not enough evidence to say that the professor's predictions about book format usage were inaccurate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis

In the context of a chi-square test, the null hypothesis is a statement suggesting no effect or no difference between expected and observed results. It assumes that the observed frequencies in the data follow a specific distribution given by prior predictions or observations.
For example, in our exercise, the null hypothesis (\( H_0 \)) is that the students' use of book formats matches exactly what the professor predicted: 60% buying hard copies, 25% printing, and 15% reading online.

• The null hypothesis acts as a baseline for the study.
• If the null hypothesis is true, any difference between observed and expected counts is due to random chance.
• A chi-square test ultimately assesses whether these small differences are just random or if they indicate a significant departure from what was expected.

Rejecting the null hypothesis suggests that the professor's predictions are inaccurate, while failing to reject implies they are consistent with the data observed from the survey.

Expected Counts

Expected counts are calculated based on predictions or previous data. These counts represent the number of observations we would expect in each category if the null hypothesis were true. It's like setting a benchmark before we assess how reality compares to it.
To find these counts:

• Multiply the total number of observations by the predicted probability for each category.
• For example, the expected students buying a book are computed as 126 students times 60%, resulting in 75.6.
• For printing, it is 126 times 25%, equalling 31.5.
• And for online reading, it is 126 times 15%, yielding 18.9.

If all expected counts are greater than 5, the chi-square test is valid because this condition ensures that the approximation to the chi-square distribution is appropriate. Accurate computation of expected counts is crucial as it directly influences the chi-square statistic calculation.

Degrees of Freedom

Degrees of freedom is a concept that tells us the number of values that have the freedom to vary when calculating a statistic. For a chi-square test, the degrees of freedom is computed as the number of categories minus one.
In our example:

• There are three categories (hard copy, printed, online reading).
• Thus, degrees of freedom = 3 - 1 = 2.

Degrees of freedom impact the critical value from the chi-square distribution, which in turn influences the p-value.
Having more categories increases the complexity and, thereby, usually increases the degrees of freedom. Calculating degrees of freedom correctly is essential because it determines which chi-square distribution to use when looking up critical values or p-values.

p-value

The p-value is a key result of a hypothesis test that helps us make a decision regarding the null hypothesis. It quantitatively measures how likely it is to observe the data, or something more extreme, if the null hypothesis is true.

• A low p-value (typically less than 0.05) indicates that such observed results are unlikely under the null hypothesis, leading to its rejection.
• A high p-value suggests that the observed data is consistent with the null hypothesis.

For the current chi-square test: The p-value is derived from the chi-square distribution under the calculated degrees of freedom.
In our scenario, the calculated p-value is assumed to be greater than 0.05, meaning the null hypothesis isn't rejected.
This indicates that there is not enough statistical evidence to conclude the professor’s predictions were inaccurate. It underlines an important point: failure to reject a null hypothesis doesn’t prove it true; it merely suggests there's insufficient evidence against it.