91影视

In a sample of 5924 patients treated with Eliquis drug, 153 patients suffered from the adverse reaction of nausea.

(a)

The best point estimate of the proportion of patients who took Eliquis and developed nausea is computed below:

$$\begin{gathered} \hat{p}=\frac{153}{5924} \\ =0.026 \end{gathered}$$

Thus, the sample proportion equal to 0.026 is the best point estimate of the proportion of patients who took Eliquis and developed nausea.

(b)

The confidence level is equal to 99%. Thus, the corresponding level of significance is equal to 0.01.

From the standard normal distribution table, the right-tailed value of $z_{\frac{伪}{2}}$for $伪=0.01$is equal to 2.5758.

The margin of error is calculated below:

$$\begin{gathered} E=2.5758脳\sqrt{\frac{0.026脳0.9741}{5924}} \\ =0.0053 \end{gathered}$$

Thus, the margin of error is equal to 0.0053.

(c)

The formula for computing the confidence interval estimate of the population proportion is written below:

$CI=\left(\hat{p}-E,\hat{p}+E\right)$

The 95% confidence interval becomes equal to:

$$\begin{gathered} CI=\left(0.026-0.0053,0.026+0.0053\right) \\ =\left(0.021,0.031\right) \end{gathered}$$

Therefore, the 99% confidence interval estimate of the population proportion of patients who took Eliquis and developed nausea is equal to (0.021, 0.031).

(d)

There is 99% confidence that the true proportion ofpatients whotook Eliquis anddeveloped nauseawill lie between the values, 0.021 and 0.031.