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.

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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.