Questions: A researcher claims that at most 76% of college students enrolled in a course that uses a free textbook pass the class. One professor thinks the proportion is actually higher. She surveys 120 students in classes that use a free textbook and finds the 103 of those students ultimately pass the class. Perform a hypothesis test using a 1% level of significance. Step 1: State the null and alternative hypotheses. Ha : p > 0.76 (So we will be performing a right-tailed test.) Step 2: Assuming the null hypothesis is true, identify the sampling distribution. The sampling distribution is a normal distribution with mean 0.76 and standard error 0.03899 Step 3: Find the p-value.

A researcher claims that at most 76% of college students enrolled in a course that uses a free textbook pass the class. One professor thinks the proportion is actually higher. She surveys 120 students in classes that use a free textbook and finds the 103 of those students ultimately pass the class. Perform a hypothesis test using a 1% level of significance.

Step 1: State the null and alternative hypotheses.

Ha : p > 0.76

(So we will be performing a right-tailed test.)

Step 2: Assuming the null hypothesis is true, identify the sampling distribution.

The sampling distribution is a normal distribution with mean 0.76 and standard error

0.03899

Step 3: Find the p-value.

Transcript text: A researcher claims that at most $76 \%$ of college students enrolled in a course that uses a free textbook pass the class. One professor thinks the proportion is actually higher. She surveys 120 students in classes that use a free textbook and finds the 103 of those students ultimately pass the class. Perform a hypothesis test using a $1 \%$ level of significance. Step 1: State the null and alternative hypotheses. \[ \begin{array}{l} H_{a} \text { : pV } \cdot \text { >レ } 0.76 \end{array} \] (So we will be performing a $\square$ right-tailed test.) Step 2: Assuming the null hypothesis is true, identify the sampling distribution. The sampling distribution is a normal distribution $\vee \checkmark$ with mean $0.76 \quad \checkmark$ and standard error \[ 0.03899 \] Step 3: Find the $p$-value. \[ \begin{array}{l} P(\hat{p \vee v ?} \square) \\ \quad=P(? \vee ? \vee \square) \end{array} \]

Solution

Solution Steps

Step 1: State the Hypotheses

We are conducting a hypothesis test to evaluate the researcher's claim regarding the proportion of college students who pass a class using a free textbook. The hypotheses are stated as follows:

Null Hypothesis ($H_0$): $p \leq 0.76$
Alternative Hypothesis ($H_a$): $p > 0.76$

This indicates that we will perform a right-tailed test.

Step 2: Identify the Sampling Distribution

Assuming the null hypothesis is true, the sampling distribution of the sample proportion is approximately normal with:

Mean ($\mu$) = $0.76$
Standard Error ($SE$) = $0.03899$

Step 3: Calculate the Test Statistic and P-value

The test statistic ($Z$) is calculated using the formula:

\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]

Where:

$\hat{p} = \frac{103}{120} = 0.8583$
$p_0 = 0.76$
$n = 120$

Substituting the values, we find:

\[ Z = 2.5222 \]

The corresponding P-value for this test statistic is $0.0058$.

Step 4: Determine the Critical Region

For a significance level of $\alpha = 0.01$ in a right-tailed test, the critical value of $Z$ is approximately $2.3263$. Thus, the critical region is defined as:

\[ Z > 2.3263 \]

Step 5: Conclusion

Since the calculated test statistic $Z = 2.5222$ falls within the critical region ($Z > 2.3263$) and the P-value $0.0058$ is less than the significance level $\alpha = 0.01$, we reject the null hypothesis.