Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 5754 patients treated with this drug,
157 developed the adverse reaction of nausea. Use a 0.10 significance level to test the claim that of users develop nausea. Does
nausea appear to be a problematic adverse reaction?
B.

C.

D.

Identify the test statistic for this hypothesis test.
The test statistic for this hypothesis test is -1.21.
(Round to two decimal places as needed.)
Identify the -value for this hypothesis test.
The -value for this hypothesis test is .
(Round to three decimal places as needed.)

Ask by Logan Parry. in the United States

Mar 13,2025

Question

Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 5754 patients treated with this drug, 157 developed the adverse reaction of nausea. Use a 0.10 significance level to test the claim that $$ 3 \% $$ of users develop nausea. Does nausea appear to be a problematic adverse reaction? B. $$ H_{0}: p=0.03 $$ $$ H_{1}: p<0.03 $$ C. $$ H_{0}: p 
eq 0.03 $$ $$ H_{1}: p=0.03 $$ D. $$ H_{0}: p=0.03 $$ $$ H_{1}: p>0.03 $$ Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is -1.21. (Round to two decimal places as needed.) Identify the $$ P $$-value for this hypothesis test. The $$ P $$-value for this hypothesis test is $$ \square $$. (Round to three decimal places as needed.)

UpStudy AI · Accepted Answer

We begin by computing the sample proportion. With  
$$
\hat{p}=\frac{157}{5754}\approx0.0273,
$$  
we compare this with the claimed proportion $$ p_0=0.03 $$.

The test being considered is one‐tailed with  
$$
H_{0}:p=0.03\quad \text{and} \quad H_{1}:p>0.03.
$$

The test statistic for a proportion is given by  
$$
z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}.
$$

1. Compute the standard error:
   $$
   SE=\sqrt{\frac{0.03\times0.97}{5754}}.
   $$
   First, calculate $$0.03\times0.97=0.0291$$. Then,
   $$
   SE=\sqrt{\frac{0.0291}{5754}}\approx\sqrt{0.00000506}\approx0.00225.
   $$

2. Compute the test statistic:
   $$
   z=\frac{0.0273-0.03}{0.00225}\approx\frac{-0.0027}{0.00225}\approx -1.21.
   $$
   (Rounded to two decimal places, the test statistic is $$-1.21$$.)

Since our alternative hypothesis is that $$ p>0.03 $$, the rejection region is in the right tail of the standard normal distribution. However, our test statistic is negative, indicating that the sample proportion is lower than $$0.03$$.

The $$ P $$-value for the test is computed as the probability of obtaining a $$ z $$-value at least as extreme as $$-1.21$$ in the direction of the alternative hypothesis. Formally, since the alternative is $$ p>0.03 $$, we compute:
$$
P\text{-value}=P(Z\geq -1.21).
$$
Using the symmetry of the normal distribution:
$$
P(Z\geq -1.21)=1-P(Z<-1.21)=1-0.113\approx0.887.
$$
(Rounded to three decimal places, the $$ P $$-value is $$ 0.887 $$.)

Because the $$ P $$-value is much larger than the significance level $$ \alpha = 0.10 $$, we fail to reject the null hypothesis. Therefore, we do not have evidence to conclude that more than $$ 3\% $$ of users develop nausea. In other words, nausea does not appear to be a problematic adverse reaction for this drug.
The test statistic is -1.21, and the $$ P $$-value is 0.887.

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