91Ó°ÊÓ

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. (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 need to determine whether the observed distribution of book formats differs significantly from the professor's predictions. - **Null Hypothesis ( H_0 ):** The observed distribution matches the professor's predicted distribution. - **Alternative Hypothesis ( H_a ):** The observed distribution does not match the professor's predicted distribution.

02

Calculate Expected Counts

According to the professor's prediction: - **Hard copy:** 60% - **Printed from web:** 25% - **Read online:** 15% There are 126 students surveyed. - Expected hard copy: \(0.60 \times 126 = 75.6\) - Expected printed from web: \(0.25 \times 126 = 31.5\) - Expected read online: \(0.15 \times 126 = 18.9\)

03

Verify Chi-Square Test Conditions

There are two main conditions for the chi-square test: 1. The sample size must be large enough (usually at least 5 expected in each category). 2. The observations should be independent. - **Sample Size Check:** The expected counts are 75.6, 31.5, and 18.9, all of which are greater than 5. - **Independence Check:** Each student's choice is independent of another's, assuming no influence between them during the survey.

04

Calculate Chi-Square Statistic

The chi-square statistic is calculated using the formula: \[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\] Where \(O_i\) are observed values and \(E_i\) are expected values.- For hard copy: \(\frac{(71 - 75.6)^2}{75.6} = \frac{(-4.6)^2}{75.6} = 0.279\) - For printed: \(\frac{(30 - 31.5)^2}{31.5} = \frac{(-1.5)^2}{31.5} = 0.071\)- For online: \(\frac{(25 - 18.9)^2}{18.9} = \frac{(6.1)^2}{18.9} = 1.97\)Add these: \(\chi^2 = 0.279 + 0.071 + 1.97 = 2.32\)

05

Determine Degrees of Freedom and p-value

Degrees of freedom (df) for a chi-square test is \(\text{df} = k - 1\), where \(k\) is the number of categories. Here, \(k = 3\). So, \(\text{df} = 2\).Using a chi-square distribution table or calculator, find the p-value for \(\chi^2 = 2.32\) with \(\text{df} = 2\). The p-value is approximately 0.313.

06

Conclusion of Hypothesis Test

Since the p-value (0.313) is greater than the typical significance level of 0.05, we fail to reject the null hypothesis. **Conclusion:** There is not enough evidence to say that the observed distribution of book format usage differs from the professor's predictions. The observed format choices are consistent with the predictions.

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Key Concepts

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

Hypothesis Testing

Hypothesis testing is a fundamental concept in statistics used to make inferences about a population based on sample data. Here, it helps us determine whether the professor's predictions about book format usage among students hold true.

When conducting a hypothesis test, we start with two opposing statements:

• Null Hypothesis ( $H_0$): This assumes that there is no effect or difference. In this case, it suggests that the distribution of book formats used by students matches the professor's prediction.
• Alternative Hypothesis ( $H_a$): This claims that there is an effect or difference. Here, it implies that the students' book format choices deviate from what the professor anticipated.

We test these hypotheses to see which one is more supported by the data. A key goal is to determine whether any observed differences are due to random chance or whether they reflect actual disparities.

Expected Counts

Expected counts are the theoretical number of observations we predict for each category based on initial assumptions or predictions. They play a crucial role in conducting a chi-square test. Here, the expected counts are derived from the professor's predictions.

To calculate these:

• Hard copy: We expect 60% of the students to choose this format. The expected count: 60% of 126 = 75.6.
• Printed from the web: Predicted at 25%. The expected count: 25% of 126 = 31.5.
• Read online: Expected at 15%. The expected count: 15% of 126 = 18.9.

These anticipated numbers form a basis for comparison with the actual observations, highlighting any significant deviations.

Degrees of Freedom

Degrees of freedom is a concept that helps us understand the variability we expect in our statistical estimates, especially when testing hypotheses. In the context of a chi-square test, it influences the shape of the chi-square distribution used to determine the p-value.

For a chi-square test, degrees of freedom (\(df\)) are calculated based on the number of categories being analyzed. Specifically, \(\text{df} = k - 1\), where \(k\) is the number of categories. In our scenario, \(k = 3\) (hard copy, print, and read online). Thus, \(\text{df} = 2\).
This value is essential for looking up or calculating the p-value associated with our chi-square statistic, which in turn helps us interpret our hypothesis test results.

P-Value

The p-value is a probability measure that helps determine the significance of the test results in hypothesis testing. It indicates the likelihood of observing the data, or something more extreme, assuming the null hypothesis is true.

For our exercise, after calculating the chi-square statistic, we need the p-value to make conclusions. A low p-value (typically less than 0.05) suggests that the observed data is unlikely under the null hypothesis, prompting us to reject it. Conversely, a high p-value indicates that the data aligns well with the null hypothesis.
In this case, the chi-square statistic was 2.32, with $df = 2$. The corresponding p-value is approximately 0.313. Since this p-value exceeds the common threshold of 0.05, we do not have sufficient evidence to reject the null hypothesis, suggesting that the observed distribution is consistent with the predictions.