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.
Adentify the null and alternative hypotheses for this test Choose the correct answer below.
A.

C.

D.

Identify the test statistic for this hypothesis test.
The test statistic for this hypothesis test is
(Round to two 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. Adentify the null and alternative hypotheses for this test Choose the correct answer below. A. $$ H_{0}: p=0.20 $$ $$ H_{1}: p 
eq 0.20 $$ $$ H_{0}: p=0.20 $$ $$ H_{1}: p>0.20 $$ C. $$ H_{0}: p>0.20 $$ $$ H_{1}: p=0.20 $$ D. $$ H_{0}: p=0.20 $$ $$ H_{1}: p<0.20 $$ Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is (Round to two decimal places as needed.)

UpStudy AI · Accepted Answer

**Step 1. State the Hypotheses**

The claim is that more than $$20\%$$ of users develop nausea. This implies that the alternative hypothesis should be $$ p > 0.20 $$. The null hypothesis is conventionally the equality hypothesis. Thus, we have:

$$
H_0: p = 0.20 \quad \text{and} \quad H_1: p > 0.20
$$

So, the correct answer is:

$$
H_0: p = 0.20 \quad \text{and} \quad H_1: p > 0.20
$$

**Step 2. Identify the Test Statistic**

The test statistic for a single proportion is given by:

$$
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}
$$

where  
$$\hat{p}$$ is the sample proportion,  
$$p_0$$ is the hypothesized proportion, and  
$$n$$ is the sample size.

Given:  
$$ n = 227 $$  
Number of subjects with nausea $$ = 50 $$  
Therefore, the sample proportion is

$$
\hat{p} = \frac{50}{227} \approx 0.2203
$$

The standard error (SE) is:

$$
SE = \sqrt{\frac{0.20 \times 0.80}{227}} = \sqrt{\frac{0.16}{227}} \approx \sqrt{0.000704} \approx 0.02653
$$

Now, compute the test statistic:

$$
z = \frac{0.2203 - 0.20}{0.02653} \approx \frac{0.0203}{0.02653} \approx 0.76
$$

Thus, the test statistic is approximately:

$$
z \approx 0.76
$$
**Hypotheses:** - $$ H_0: p = 0.20 $$ - $$ H_1: p > 0.20 $$ **Test Statistic:** $$ z \approx 0.76 $$

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