Questions: An associate at a law firm conducts a hypothesis test at the 0.05 level of significance to determine if the proportion, p, of all people who experience nausea after taking a certain antibiotic is greater than 12%. Here are the null hypothesis H0 and the alternative hypothesis H1 for the test. H0: p=0.12 H1: p>0.12 (a) The result of the associate's hypothesis test will be to either reject or not reject the null hypothesis, H0. Based on whether H0 is in fact true or is in fact false, this result can be correct, a Type I error, or a Type II error. Complete the table below to show when the result will be correct, a Type I error, or a Type II error. (Choose one) (Choose one) V (Choose one) (Choose one) V Type I error Correct (Choose one) V Correct (b) The associate takes a random sample and determines it is appropriate to use a Z test. The value of the test statistic is z=1.991 (rounded to three decimal places). Draw the appropriate figure for the test. Standard Normal Distribution Step 1: Select one-tailed or two-tailed. One-tailed Two-tailed

An associate at a law firm conducts a hypothesis test at the 0.05 level of significance to determine if the proportion, p, of all people who experience nausea after taking a certain antibiotic is greater than 12%. Here are the null hypothesis H0 and the alternative hypothesis H1 for the test. H0: p=0.12 H1: p>0.12 (a) The result of the associate's hypothesis test will be to either reject or not reject the null hypothesis, H0. Based on whether H0 is in fact true or is in fact false, this result can be correct, a Type I error, or a Type II error. Complete the table below to show when the result will be correct, a Type I error, or a Type II error. (Choose one) (Choose one) V (Choose one) (Choose one) V Type I error Correct (Choose one) V Correct (b) The associate takes a random sample and determines it is appropriate to use a Z test. The value of the test statistic is z=1.991 (rounded to three decimal places). Draw the appropriate figure for the test. Standard Normal Distribution Step 1: Select one-tailed or two-tailed. One-tailed Two-tailed
Transcript text: An associate at a law firm conducts a hypothesis test at the 0.05 level of significance to determine if the proportion, $p$, of all people who experience nausea after taking a certain antibiotic is greater than $12 \%$. Here are the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ for the test. \[ \begin{array}{l} H_{0}: p=0.12 \\ H_{1}: p>0.12 \end{array} \] (a) The result of the associate's hypothesis test will be to either reject or not reject the null hypothesis, $H_{0}$. Based on whether $H_{0}$ is in fact true or is in fact false, this result can be correct, a Type I error, or a Type II error. Complete the table below to show when the result will be correct, a Type I error, or a Type II error. \begin{tabular}{|l|l|l|l|} \hline & (Choose one) & $\mathbf{V}$ & (Choose one) \\ \hline (Choose one) & $\mathbf{V}$ & Type I error & \\ \hline & & Correct \\ \hline (Choose one) & $\mathbf{V}$ & Correct & \\ \hline \end{tabular} (b) The associate takes a random sample and determines it is appropriate to use a $Z$ test. The value of the test statistic is $z=1.991$ (rounded to three decimal places). Draw the appropriate figure for the test. \begin{tabular}{|l|l|} \hline Standard Normal Distribution & \\ Step 1: Select one-tailed or two-tailed. & \\ One-tailed \\ Two-tailed & \end{tabular}
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Solution

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Solution Steps

Step 1: Hypothesis Setup

We are conducting a hypothesis test to determine if the proportion p p of people who experience nausea after taking a certain antibiotic is greater than 12% 12\% . The hypotheses are defined as follows:

\begin{align_} H_{0}: & \quad p = 0.12 \\ H_{1}: & \quad p > 0.12 \end{align_}

Step 2: Critical Value Calculation

For a one-tailed test at a significance level of α=0.05 \alpha = 0.05 , we calculate the critical Z value. The critical Z value is determined using the formula:

Z=Φ−1(1−α)=Φ−1(0.95) Z = \Phi^{-1}(1 - \alpha) = \Phi^{-1}(0.95)

The calculated critical Z value is:

Zcritical=1.96 Z_{critical} = 1.96

Step 3: Test Statistic and Decision

The calculated test statistic from the sample is:

Z=1.991 Z = 1.991

Since Z=1.991>Zcritical=1.96 Z = 1.991 > Z_{critical} = 1.96 , we reject the null hypothesis H0 H_{0} .

Step 4: P-value Calculation

The p-value associated with the test statistic Z=1.991 Z = 1.991 is calculated as follows:

p-value=1−Φ(Z)≈0.0232 p\text{-value} = 1 - \Phi(Z) \approx 0.0232

Step 5: Conclusion

Given that the p-value 0.0232 0.0232 is less than the significance level α=0.05 \alpha = 0.05 , we reject the null hypothesis H0 H_{0} .

Final Answer

The decision is to reject the null hypothesis H0 H_{0} . Thus, we conclude that there is sufficient evidence to suggest that the proportion of people who experience nausea after taking the antibiotic is greater than 12% 12\% .

Reject H0 \boxed{\text{Reject } H_{0}}

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