Questions: At a certain university, 16% of students fail general chemistry on their first attempt. Professor Brown teaches at this university and believes that the rate of first-time failure in his general chemistry classes is 33%. He samples 96 students from last semester who were first-time enrollees in general chemistry and finds that 15 of them failed his course. Using α=0.05, can you conclude that the percentage of failures differs from 33%?

At a certain university, 16% of students fail general chemistry on their first attempt. Professor Brown teaches at this university and believes that the rate of first-time failure in his general chemistry classes is 33%. He samples 96 students from last semester who were first-time enrollees in general chemistry and finds that 15 of them failed his course. Using α=0.05, can you conclude that the percentage of failures differs from 33%?

Transcript text: At a certain university, $16 \%$ of students fail general chemistry on their first attempt. Professor Brown teaches at this university and believes that the rate of first-time failure in his general chemistry classes is $33 \%$. He samples 96 students from last semester who were first-time enrollees in general chemistry and finds that 15 of them failed his course. Using $\alpha=0.05$, can you conclude that the percentage of failures differs from $33 \%$?

Solution

Solution Steps

Step 1: Hypothesis Formulation

We are testing the following hypotheses:

Null Hypothesis ($H_0$): $p = 0.33$ (The percentage of failures is 33%)
Alternative Hypothesis ($H_a$): $p \neq 0.33$ (The percentage of failures differs from 33%)

Step 2: Sample Proportion Calculation

The sample proportion ($\hat{p}$) is calculated as: \[ \hat{p} = \frac{\text{Number of failures}}{\text{Sample size}} = \frac{15}{96} \approx 0.15625 \]

Step 3: Test Statistic Calculation

The test statistic ($Z$) is calculated using the formula: \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] Substituting the values: \[ Z = \frac{0.15625 - 0.33}{\sqrt{\frac{0.33(1 - 0.33)}{96}}} \approx -3.6205 \]

Step 4: P-value Calculation

The P-value associated with the test statistic $Z = -3.6205$ is: \[ \text{P-value} \approx 0.0003 \]

Step 5: Critical Region Determination

For a significance level of $\alpha = 0.05$ in a two-tailed test, the critical region is defined as: \[ Z < -1.96 \quad \text{or} \quad Z > 1.96 \]

Step 6: Conclusion

Since the calculated test statistic $Z = -3.6205$ falls within the critical region and the P-value $0.0003$ is less than $\alpha = 0.05$, we reject the null hypothesis.