Question

a doctor is concerned that nausea may be a side - effect of tamiflu, but is not certain because nausea is common for people who have the flu. she researched some past studies, and found that about 33% of people who get the flu and do not use tamiflu experience nausea. she then collected data on 2256 patients who were taking tamiflu, and found that 703 experienced nausea. use a 0.05 significance level to test the claim that the percentage of people who take tamiflu for the relief of flu symptoms and experience nausea is 33%.
a. what type of test will be used in this problem?
select an answer
b. enter the null hypothesis for this test.
$h_0$: ? ?
c. enter the alternative hypothesis for this test.
$h_1$: ? ?
d. is the original claim located in the null or alternative hypothesis?
select an answer
e. what is the test statistic for the given statistics?

f. what is the p - value for this test?

Explanation:

Step1: Identify test type

We are testing a claim about a proportion, so a one - proportion z - test will be used.

Step2: Formulate null hypothesis

The null hypothesis is that the proportion $p$ of people who take Tamiflu and experience nausea is equal to the known proportion of non - Tamiflu flu patients with nausea. So $H_0:p = 0.33$.

Step3: Formulate alternative hypothesis

The alternative hypothesis is two - tailed since we are just testing if the proportion is different from 0.33. So $H_1:p
eq0.33$.

Step4: Determine claim location

The original claim is $p = 0.33$, which is in the null hypothesis.

Step5: Calculate sample proportion

The sample size $n = 2256$ and the number of successes $x = 703$. The sample proportion $\hat{p}=\frac{x}{n}=\frac{703}{2256}\approx0.3116$.

Step6: Calculate test statistic

The formula for the one - proportion z - test statistic is $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}$. Substituting $p = 0.33$, $\hat{p}=0.3116$ and $n = 2256$:
\[

$$\begin{align*} z&=\frac{0.3116 - 0.33}{\sqrt{\frac{0.33\times(1 - 0.33)}{2256}}}\\ &=\frac{- 0.0184}{\sqrt{\frac{0.33\times0.67}{2256}}}\\ &=\frac{-0.0184}{\sqrt{\frac{0.2211}{2256}}}\\ &=\frac{-0.0184}{\sqrt{0.000098}}\\ &=\frac{-0.0184}{0.0099}\\ &\approx - 1.86 \end{align*}$$

\]

Step7: Calculate p - value

Since it's a two - tailed test, the p - value is $2\times P(Z\lt|z|)$. For $z=-1.86$, $P(Z\lt - 1.86)=0.0314$, so the p - value is $2\times0.0314 = 0.0628$.

Answer:

a. One - proportion z - test
b. $H_0:p = 0.33$
c. $H_1:p
eq0.33$
d. Null hypothesis
e. $z\approx - 1.86$
f. $p - value\approx0.0628$