Suppose 227 subjects are treated with a drug that is used to treat pain and 50 of them developed nausea. Use a 0.01 significance level to test the claim that more than of users develop nausea.
A.

B.

C.

D.

Identify the null and altemative hypotheses for this test. Choose the correct answer below.
The test statistic for this hypothesis ter this hypothesis test is 0.76
Round to two decimal places as needed.)
Identify the P -value for this hypothesis test
The P-value for this hypothesis test is
(Round to three decimal places as needed )

Question

Suppose 227 subjects are treated with a drug that is used to treat pain and 50 of them developed nausea. Use a 0.01 significance level to test the claim that more than $$ 20 \% $$ of users develop nausea. A. $$ \mathrm{H}_{0}: \mathrm{p}=0.20 $$ $$ \mathrm{H}_{1}: \mathrm{p} 
eq 0.20 $$ B. $$ \mathrm{H}_{0}: \mathrm{p}=0.20 $$ $$ \mathrm{H}_{1}: \mathrm{p}>0.20 $$ C. $$ \mathrm{H}_{0}: \mathrm{p}>0.20 $$ $$ \mathrm{H}_{1}: \mathrm{p}=0.20 $$ D. $$ \mathrm{H}_{0}: \mathrm{p}=0.20 $$ $$ \mathrm{H}_{1}: \mathrm{p}<0.20 $$ Identify the null and altemative hypotheses for this test. Choose the correct answer below. The test statistic for this hypothesis ter this hypothesis test is 0.76 Round to two decimal places as needed.) Identify the P -value for this hypothesis test The P-value for this hypothesis test is (Round to three decimal places as needed )

UpStudy AI · Accepted Answer

**Step 1. Identify the Claim and Write the Hypotheses**

The claim is that more than $$20\%$$ of users develop nausea. This implies:
$$
\mathrm{H}_0: p = 0.20 \quad\text{and}\quad \mathrm{H}_1: p > 0.20.
$$
Thus, the correct choice is **B**.

**Step 2. Compute the Sample Proportion**

The number of subjects who experienced nausea is $$x = 50$$ out of $$n = 227$$. The sample proportion is:
$$
\hat{p} = \frac{x}{n} = \frac{50}{227} \approx 0.2203.
$$

**Step 3. Calculate the Test Statistic**

Under the null hypothesis ($$p_0 = 0.20$$), the test statistic is calculated using:
$$
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}.
$$
Substitute the values:
$$
z = \frac{0.2203 - 0.20}{\sqrt{\frac{0.20 \times 0.80}{227}}}.
$$
First, compute the standard error:
$$
\text{Standard Error} = \sqrt{\frac{0.16}{227}} \approx \sqrt{0.000704} \approx 0.02655.
$$
Now, compute the $$z$$-value:
$$
z = \frac{0.0203}{0.02655} \approx 0.76.
$$

**Step 4. Determine the P-value**

Since the alternative hypothesis is one-sided ($$p > 0.20$$), the P-value is:
$$
\text{P-value} = 1 - \Phi(0.76),
$$
where $$\Phi(z)$$ is the cumulative distribution function for the standard normal distribution.

Using standard normal tables or a calculator:
$$
\Phi(0.76) \approx 0.776.
$$
Thus, the P-value is:
$$
\text{P-value} \approx 1 - 0.776 = 0.224.
$$

**Summary of Answers**

- **Null and Alternative Hypotheses:**  
  $$\mathrm{H}_0: p = 0.20$$ and $$\mathrm{H}_1: p > 0.20$$ (Choice B).

- **Test Statistic:**  
  $$z \approx 0.76$$.

- **P-value:**  
  Approximately $$0.224$$.
**Null and Alternative Hypotheses:** - **Null Hypothesis ($$\mathrm{H}_0$$):** The proportion of users developing nausea is $$20\%$$. - **Alternative Hypothesis ($$\mathrm{H}_1$$):** The proportion of users developing nausea is greater than $$20\%$$. **Test Statistic:** - $$z \approx 0.76$$ **P-value:** - Approximately $$0.224$$

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