Use the sample data and confidence level given below to complete parts (a) through (d) A drug is used to help prevent blood clots in certain patients In clinical trials, among 4170 patients treated with the drug, 153 developed the adverse reaction of nausea. Constr proportion of adverse reactions. a) Find the best point estimate of the population proportion $$ p $$. $$ 0.037 $$ (Round to three decimal places as needed ) b) Identify the value of the margin of error E $$ E=0006 $$ (Round to three decimal places as needed ) c) Construct the confidence interval. $$ \square $$ $$

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**(a)**  
The best point estimate for the population proportion $$ p $$ is the sample proportion  
$$
\hat{p} = \frac{153}{4170} \approx 0.0367,
$$  
which rounds to  
$$
\hat{p} \approx 0.037.
$$

**(b)**  
Assuming a 95% confidence level, the margin of error $$ E $$ is calculated by  
$$
E = z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}},
$$  
where $$ z_{\alpha/2} \approx 1.96 $$, $$ \hat{p} \approx 0.0367 $$, and $$ n = 4170 $$. First compute the standard error:  
$$
\sqrt{\frac{0.0367(1-0.0367)}{4170}} = \sqrt{\frac{0.0367 \times 0.9633}{4170}} \approx \sqrt{\frac{0.03536}{4170}} \approx \sqrt{8.486 \times 10^{-6}} \approx 0.00291.
$$  
Then,  
$$
E \approx 1.96 \times 0.00291 \approx 0.00571,
$$  
which rounds to  
$$
E \approx 0.006.
$$

**(c)**  
The $$ 95\% $$ confidence interval for $$ p $$ is given by  
$$
\hat{p} \pm E.
$$  
Thus,  
$$
\text{Lower limit} = 0.037 - 0.006 = 0.031,
$$  
$$
\text{Upper limit} = 0.037 + 0.006 = 0.043.
$$  
So the confidence interval is  
$$
0.031 < p < 0.043.
$$

**(d)**  
Interpretation: We are $$ 95\% $$ confident that the true population proportion of patients who experience the adverse reaction of nausea when treated with the drug is between  
$$
0.031 \text{ and } 0.043.
$$
**(a)** The best estimate for the proportion of patients who experienced nausea is $$ 0.037. $$ **(b)** The margin of error is $$ 0.006. $$ **(c)** The confidence interval is $$ 0.031 < p < 0.043. $$ **(d)** We are $$ 95\% $$ confident that the true proportion of patients experiencing nausea is between $$ 0.031 \text{ and } 0.043. $$

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