A teacher expects that 50% of students will pass and 50% will fail a test. In a class of 40 students, 22 passed and 18 failed.

Perform a Chi-square goodness-of-fit test to see if the observed distribution of passes and fails differs from the expected distribution.

To determine whether the observed distribution of students who passed and failed differs significantly from the expected 50-50 distribution, we can perform a **Chi-square goodness-of-fit test**.

### Step 1: Define hypotheses
- **Null hypothesis \(H_0\)**: The observed distribution of pass/fail matches the expected 50-50 distribution.
- **Alternative hypothesis \(H_1\)**: The observed distribution of pass/fail differs from the expected 50-50 distribution.

### Step 2: Set up the expected frequencies
Since the teacher expects 50% of the students to pass and 50% to fail, and there are 40 students:
- Expected Passes = \( 40 \times 0.5 = 20 \)  
- Expected Fails = \( 40 \times 0.5 = 20 \)

### Step 3: Compute the Chi-square test statistic  
The formula for the Chi-square test statistic is:

\[
\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
\]

Where:
- \(O_i\) = Observed frequency  
- \(E_i\) = Expected frequency  

For this problem:
- Observed Passes (\(O_1\)) = 22  
- Observed Fails (\(O_2\)) = 18  
- Expected Passes (\(E_1\)) = 20  
- Expected Fails (\(E_2\)) = 20  

Now, calculate:

\[
\chi^2 = \frac{(22 - 20)^2}{20} + \frac{(18 - 20)^2}{20}
\]

\[
\chi^2 = \frac{4}{20} + \frac{4}{20} = 0.2 + 0.2 = 0.4
\]

### Step 4: Determine the degrees of freedom  
Degrees of freedom \(df\) for a Chi-square goodness-of-fit test is:

\[
df = k - 1
\]

where \(k\) is the number of categories. Here, \(k = 2\) (pass and fail), so:

\[
df = 2 - 1 = 1
\]

### Step 5: Find the critical value and p-value  
Using a Chi-square distribution table, the critical value for \(\alpha = 0.05\) and \(df = 1\) is:

\[
\chi^2_{critical} = 3.841
\]

We can also calculate the p-value for the observed \(\chi^2 = 0.4\).

Using a Chi-square distribution calculator or table:

\[
p \approx 0.527
\]

### Step 6: Make a decision  
- **If** \(\chi^2_{observed} > \chi^2_{critical}\), reject the null hypothesis.  
- **If** \(p \leq \alpha\), reject the null hypothesis.

Here:
- \(\chi^2_{observed} = 0.4 < 3.841\)  
- \(p = 0.527 > 0.05\)  

Since the observed test statistic is less than the critical value and the p-value is greater than 0.05, **we fail to reject the null hypothesis**.

### Conclusion
There is **no significant difference** between the observed distribution of pass/fail results and the expected 50-50 distribution.

---

Would you like more details or have any follow-up questions?  

Here are 5 related questions to explore:  
1. How is the p-value interpreted in hypothesis testing?  
2. What happens if the observed values are further from the expected ones?  
3. Can Chi-square tests be used for other types of categorical data?  
4. What is the effect of changing the significance level (\(\alpha\)) on the conclusion?  
5. How would the test change if there were more categories (e.g., A, B, C grades)?  

**Tip:** Always ensure the expected frequencies are at least 5 in each category for the Chi-square test to be valid.

A teacher expects that 50% of students will pass and 50% will fail a test. In a class of 40 students, 22 passed and 18 failed. Perform a Chi-square goodness-of-fit test to see if the observed distribution of passes and fails differs from the expected distribution.

Learn how to apply the Chi-square goodness-of-fit test to assess if the observed pass/fail distribution in a class test significantly differs from the expected 50-50 distribution. This problem includes statistical concepts and hypothesis testing for students in Grades 10-12.

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