Suppose a teacher gives her statistics class a four-question, multiple-choice quiz at the beginning of the semester to measure now well prepared they were questions correct. Using a = 0.05, perform a chi-square test to determine if the number of correct answers per student follows the binomial probability distribution. Click the icon to view the data table. class. The accompanying table shows the number of students wno had 0, 1, 2, What is the null hypothesis, Ho? O A. Hn: The mean number of correct answers per student is equal to 0. O B. Hn: The mean number of correct answers per student is not equal to 0. OC. Ho: The number of correct answers per student follows the binomial probability distribution. O D. Hn: The number of correct answers per student does not follow the binomial probability distribution. What is the alternative hypothesis, H, ? O A. H,: The number of correct answers per student does not follow the binomial probability distribution. O B. H,: The mean number of correct answers per student is equal to 0. OC. H,: The mean number of correct answers per student is not equal to 0. O D. H;: The number of correct answers per student follows the binomial probability distribution. What is the probability of success, p, for this binomial distribution? p= (Type an integer or a decimal.) Calculate the test statistic. (Round to two decimal places as needed.) Determine the p-value for the test statistic:

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**Hypothesis Testing at the 5% Significance Level**

At the 5% significance level, there is a decision to be made regarding whether there is enough evidence to conclude that the distribution of the number of correct answers per student does not follow the claimed, or expected, distribution.

**Table: Number of Correct Answers per Student**

| Number of Correct Answers per Student | Frequency |
|---------------------------------------|-----------|
| 0                                     | 3         |
| 1                                     | 6         |
| 2                                     | 12        |
| 3                                     | 8         |
| 4                                     | 11        |
| **Total**                             | **40**    |

**Decision Options:**

1. Reject \( H_0 \)
2. Do not reject \( H_0 \)

**Conclusion Options:**

- (1) ______ At the 5% significance level, there (2) ______ enough evidence to conclude that the distribution does not follow the claimed distribution.

Choose from:
- (1) Reject
- Do not reject

Choose from:
- (2) is
- is not

This setup is commonly used in hypothesis testing to determine whether observed data fits an expected distribution, often analyzed using chi-square tests or other statistical methods.
Transcribed Image Text:**Hypothesis Testing at the 5% Significance Level** At the 5% significance level, there is a decision to be made regarding whether there is enough evidence to conclude that the distribution of the number of correct answers per student does not follow the claimed, or expected, distribution. **Table: Number of Correct Answers per Student** | Number of Correct Answers per Student | Frequency | |---------------------------------------|-----------| | 0 | 3 | | 1 | 6 | | 2 | 12 | | 3 | 8 | | 4 | 11 | | **Total** | **40** | **Decision Options:** 1. Reject \( H_0 \) 2. Do not reject \( H_0 \) **Conclusion Options:** - (1) ______ At the 5% significance level, there (2) ______ enough evidence to conclude that the distribution does not follow the claimed distribution. Choose from: - (1) Reject - Do not reject Choose from: - (2) is - is not This setup is commonly used in hypothesis testing to determine whether observed data fits an expected distribution, often analyzed using chi-square tests or other statistical methods.
**Statistical Hypothesis Testing Using Chi-Square Test**

A teacher administered a four-question, multiple-choice quiz to her statistics class at the start of the semester. The aim was to assess the students' preparedness for the class. The data table provided (not shown here) lists the number of students who scored 0, 1, 2, 3, and 4 correct answers.

The task involves using a significance level (\(\alpha\)) of 0.05 to perform a chi-square test. The objective is to determine if the distribution of correct answers per student aligns with the binomial probability distribution.

**Hypotheses:**

1. **Null Hypothesis (\(H_0\)):**
   - \(H_0: \) The number of correct answers per student follows the binomial probability distribution.

2. **Alternative Hypothesis (\(H_1\)):**
   - \(H_1: \) The number of correct answers per student does not follow the binomial probability distribution.

**Steps to Solve the Problem:**

1. **Probability of Success:**
   - Determine \(p\), the probability of success, for this binomial distribution. Enter this as a decimal or integer.

     \(p = \underline{\hspace{3cm}}\)

2. **Calculating the Test Statistic:**
   - Compute the chi-square test statistic (\(\chi^2\)). This involves comparing observed and expected frequencies from your data.

     \(\chi^2 = \underline{\hspace{3cm}}\) (Round to two decimal places as needed.)

3. **Determining the p-value:**
   - Calculate the p-value corresponding to the chi-square test statistic obtained in the previous step.

     p-value = \underline{\hspace{3cm}}\) (Round to three decimal places as needed.)

4. **Conclusion:**
   - Based on the p-value, determine whether to reject the null hypothesis. This conclusion should be stated clearly.

Reflect on the results to draw meaningful interpretations about the conformity of student performance to the expected binomial distribution model.
Transcribed Image Text:**Statistical Hypothesis Testing Using Chi-Square Test** A teacher administered a four-question, multiple-choice quiz to her statistics class at the start of the semester. The aim was to assess the students' preparedness for the class. The data table provided (not shown here) lists the number of students who scored 0, 1, 2, 3, and 4 correct answers. The task involves using a significance level (\(\alpha\)) of 0.05 to perform a chi-square test. The objective is to determine if the distribution of correct answers per student aligns with the binomial probability distribution. **Hypotheses:** 1. **Null Hypothesis (\(H_0\)):** - \(H_0: \) The number of correct answers per student follows the binomial probability distribution. 2. **Alternative Hypothesis (\(H_1\)):** - \(H_1: \) The number of correct answers per student does not follow the binomial probability distribution. **Steps to Solve the Problem:** 1. **Probability of Success:** - Determine \(p\), the probability of success, for this binomial distribution. Enter this as a decimal or integer. \(p = \underline{\hspace{3cm}}\) 2. **Calculating the Test Statistic:** - Compute the chi-square test statistic (\(\chi^2\)). This involves comparing observed and expected frequencies from your data. \(\chi^2 = \underline{\hspace{3cm}}\) (Round to two decimal places as needed.) 3. **Determining the p-value:** - Calculate the p-value corresponding to the chi-square test statistic obtained in the previous step. p-value = \underline{\hspace{3cm}}\) (Round to three decimal places as needed.) 4. **Conclusion:** - Based on the p-value, determine whether to reject the null hypothesis. This conclusion should be stated clearly. Reflect on the results to draw meaningful interpretations about the conformity of student performance to the expected binomial distribution model.
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