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Low levels of testosterone in adult males may be treated using AndroGel \(1.62 \% .\) In clinical studies of 234 adult males who were being treated with AndroGel \(1.62 \%,\) it was found that 26 saw their prostate-specific antigen (PSA) elevated. The PSA is a protein produced by cells of the prostate gland. (a) Determine a \(95 \%\) confidence interval for the proportion of adult males treated with AndroGel \(1.62 \%\) who will experience elevated levels of PSA. (b) Determine a \(99 \%\) confidence interval for the proportion of adult males treated with AndroGel \(1.62 \%\) who will experience elevated levels of PSA. (c) What is the impact of increasing the level of confidence on the margin of error?

Short Answer

Expert verified
(a) 95% CI: (0.0691, 0.1529). (b) 99% CI: (0.0559, 0.1661). (c) Increasing confidence level increases margin of error.

Step by step solution

01

- Determine Sample Proportion

Calculate the sample proportion \(\hat{p}\) using the formula: \(\hat{p} = \frac{26}{234} \). The value is approximately 0.111.
02

- Calculate Standard Error

Calculate the standard error (SE) using the formula: \[SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]. Plugging in the values: \[SE = \sqrt{\frac{0.111(1 - 0.111)}{234}} \approx 0.0214. \]
03

- Find z-Score for 95% Confidence Interval

For a 95% confidence interval, the z-score is approximately 1.96.
04

- Calculate Margin of Error for 95% Confidence Interval

Calculate the margin of error (ME) using the formula: \[ME = z \times SE \]. Substituting the values: \[ME = 1.96 \times 0.0214 \approx 0.0419. \]
05

- Determine 95% Confidence Interval

Compute the confidence interval using \[\hat{p} \pm ME \]. The interval is approximately \[0.111 \pm 0.0419 \] or (0.0691, 0.1529).
06

- Find z-Score for 99% Confidence Interval

For a 99% confidence interval, the z-score is approximately 2.576.
07

- Calculate Margin of Error for 99% Confidence Interval

Calculate the margin of error (ME) using the formula: \[ME = z \times SE \]. Substituting the values: \[ME = 2.576 \times 0.0214 \approx 0.0551. \]
08

- Determine 99% Confidence Interval

Compute the confidence interval using \[\hat{p} \pm ME \]. The interval is approximately \[0.111 \pm 0.0551 \] or (0.0559, 0.1661).
09

- Impact of Increasing Confidence Level

Increasing the level of confidence from 95% to 99% increases the margin of error. A higher confidence level means a broader range to ensure greater certainty.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

sample proportion
The sample proportion is a way to express the fraction of the sample that has a certain characteristic. For example, in the exercise above, we are interested in the proportion of adult males who experienced elevated PSA levels after being treated with AndroGel 1.62%. We calculate the sample proportion (denoted as \(\begin{split} \hat{p} \end{split}\)) with the formula: \(\begin{split} \hat{p} = \frac{26}{234} \end{split}\). This gives us a sample proportion of approximately 0.111. This value means that roughly 11.1% of the sample saw elevated PSA levels. Understanding this fraction is crucial for further calculations in developing our confidence intervals.
standard error
Standard error measures the variability of the sample proportion. It helps to understand how much the sample proportion you have found might vary from the actual population proportion. For our exercise, the formula for standard error is: \(\begin{split} SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \end{split}\), where \(\begin{split} n \end{split}\) is the sample size. Plugging in our sample proportion (0.111) and sample size (234), we get: \(\begin{split} SE = \sqrt{\frac{0.111(1 - 0.111)}{234}} \approx 0.0214 \end{split}\). This value tells us how much the proportion calculated from our sample might differ from the true population proportion.
z-score
The z-score is a critical value used to determine the confidence intervals. It reflects how many standard deviations an element is from the mean. For different confidence levels, different z-scores are used. For example, a 95% confidence level corresponds to a z-score of approximately 1.96. This means that we are 95% confident that our interval captures the true population parameter. For a 99% confidence interval, the z-score is approximately 2.576, meaning we are almost 99% certain. These z-scores are essential for calculating the margin of error and the width of the confidence interval.
margin of error
The margin of error gives us a range within which we expect the true population proportion to lie. It is calculated by multiplying the z-score by the standard error. For a 95% confidence interval in our exercise: \(\begin{split} ME = 1.96 \times 0.0214 \approx 0.0419 \end{split}\), and for a 99% confidence interval: \(\begin{split} ME = 2.576 \times 0.0214 \approx 0.0551 \end{split}\). A larger margin of error indicates a wider confidence interval. As a result, higher confidence levels mean broader intervals, ensuring more certainty that the true proportion lies within this range but also reflecting increased uncertainty about its exact value.

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Most popular questions from this chapter

A simple random sample of size \(n<30\) for \(a\) quantitative variable has been obtained. Using the normal probability plot, the correlation between the variable and expected z-score, and the boxplot, judge whether a t-interval should be constructed. $$ n=9 ; \text { Correlation }=0.997 $$

Sleep apnea is a disorder in which you have one or more pauses in breathing or shallow breaths while you sleep. In a cross-sectional study of 320 individuals who suffer from sleep apnea, it was found that 192 had gum disease. Note: In the general population, about \(17.5 \%\) of individuals have gum disease. (a) What does it mean for this study to be cross-sectional? (b) What is the variable of interest in this study? Is it qualitative or quantitative? Explain. (c) Estimate the proportion of individuals who suffer from sleep apnea who have gum disease with \(95 \%\) confidence. Interpret your result.

The following small data set represents a simple random sample from a population whose mean is \(50 .\) $$ \begin{array}{llllll} \hline 43 & 63 & 53 & 50 & 58 & 44 \\ \hline 53 & 53 & 52 & 41 & 50 & 43 \\ \hline \end{array} $$ (a) A normal probability plot indicates that the data could come from a population that is normally distributed with no outliers. Compute a \(95 \%\) confidence interval for this data set. (b) Suppose that the observation, 41 , is inadvertently entered into the computer as 14 . Verify that this observation is an outlier (c) Construct a \(95 \%\) confidence interval on the data set with the outlier. What effect does the outlier have on the confidence interval? (d) Consider the following data set, which represents a simple random sample of size 36 from a population whose mean is 50\. Verify that the sample mean for the large data set is the same as the sample mean for the small data set from part (a). $$ \begin{array}{llllll} \hline 43 & 63 & 53 & 50 & 58 & 44 \\ \hline 53 & 53 & 52 & 41 & 50 & 43 \\ \hline 47 & 65 & 56 & 58 & 41 & 52 \\ \hline 49 & 56 & 57 & 50 & 38 & 42 \\ \hline 59 & 54 & 57 & 41 & 63 & 37 \\ \hline 46 & 54 & 42 & 48 & 53 & 41 \\ \hline \end{array} $$ (e) Compute a \(95 \%\) confidence interval for the large data set. Compare the results to part (a). What effect does increasing the sample size have on the confidence interval? (f) Suppose that the last observation, 41 , is inadvertently entered as 14 . Verify that this observation is an outlier. (g) Compute a \(95 \%\) confidence interval for the large data set with the outlier. Compare the results to part (e). What effect does an outlier have on a confidence interval when the data set is large?

A simple random sample of size \(n\) is drawn from a population that is normally distributed. The sample mean, \(\bar{x},\) is found to be \(50,\) and the sample standard deviation, \(s,\) is found to be \(8 .\) (a) Construct a \(98 \%\) confidence interval for \(\mu\) if the sample size, \(n,\) is 20 (b) Construct a \(98 \%\) confidence interval for \(\mu\) if the sample size, \(n\), is \(15 .\) How does decreasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(95 \%\) confidence interval for \(\mu\) if the sample size, \(n\), is 20. Compare the results to those obtained in part (a). How does decreasing the level of confidence affect the margin of error, \(E\) ? (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Why?

Explain why the \(t\) -distribution has less spread as the number of degrees of freedom increases.

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