Question

low levels of testosterone in adult males may be treated using medication in clinical studies of 225 adult males who were being treated with a particular medication, it was found that 28 saw their prostate - specific antigen (psa) elevated the psa is a protein produced by cells of the prostate gland complete parts (a) through (c)
(a) determine a 95% confidence interval for the proportion of adult males treated with this medication who will experience elevated levels of psa
lower bound (square). upper bound (square)
(round to three decimal places as needed )
(b) determine a 99% confidence interval for the proportion of adult males treated with this medication who will experience elevated levels of psa
lower bound (square). upper bound. (square)
(round to three decimal places as needed )
(c) what is the impact of increasing the level of confidence on the margin of error? choose the correct answer
○ a. as the level of confidence increases, the margin of error decreases
○ b. as the level of confidence increases, the margin of error increases
○ c. the margin of error is constant for all levels of confidence
○ d. the margin of error changes independently of the level of confidence

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 28$ and $n=225$. So, $\hat{p}=\frac{28}{225}\approx0.124$.

Step2: For 95% confidence interval

The critical value $z_{\alpha/2}$ for 95% confidence level is $z_{0.025}=1.96$.
The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=1.96\sqrt{\frac{0.124\times(1 - 0.124)}{225}}\approx1.96\times0.022\approx0.043$.
The lower bound is $\hat{p}-E=0.124 - 0.043=0.081$.
The upper bound is $\hat{p}+E=0.124+0.043 = 0.167$.

Step3: For 99% confidence interval

The critical value $z_{\alpha/2}$ for 99% confidence level is $z_{0.005}=2.576$.
The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=2.576\sqrt{\frac{0.124\times(1 - 0.124)}{225}}\approx2.576\times0.022\approx0.057$.
The lower bound is $\hat{p}-E=0.124-0.057 = 0.067$.
The upper bound is $\hat{p}+E=0.124 + 0.057=0.181$.

Step4: Impact of confidence level on margin of error

The formula for margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. As the confidence level increases, $z_{\alpha/2}$ increases (e.g., from 1.96 for 95% to 2.576 for 99%), so the margin of error increases.

Answer:

(a) Lower bound $0.081$, Upper bound $0.167$
(b) Lower bound $0.067$, Upper bound $0.181$
(c) B. As the level of confidence increases, the margin of error increases.