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Chapter 7: Problem 20

The drug OxyContin (oxycodone) is used to treat pain, but it is dangerous because it is addictive and can be lethal. In clinical trials, 227 subjects were treated with OxyContin and 52 of them developed nausea (based on data from Purdue Pharma L.P.). a. Construct a \(95 \%\) confidence interval estimate of the percentage of OxyContin users who develop nausea.b. Compare the result from part (a) to this \(95 \%\) confidence interval for 5 subjects who developed nausea among the 45 subjects given a placebo instead of OxyContin: \(1.93 \%

Short Answer

Expert verified
The 95% confidence interval for OxyContin users is 17.4% to 28.4%. This indicates a higher likelihood of developing nausea compared to the placebo group.

Step by step solution

01

Understand the Given Data

There are 227 subjects treated with OxyContin, and 52 developed nausea. We need to construct a 95% confidence interval for the percentage of OxyContin users who develop nausea.
02

Calculate the Sample Proportion

The sample proportion (\( \hat{p} \)) is the number of subjects who developed nausea divided by the total number of subjects. \[ \hat{p} = \frac{52}{227} = 0.229 \]
03

Determine the Standard Error

The standard error (SE) of the sample proportion is calculated using the formula \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \] where \( n \) is the sample size. Here, \[ SE = \sqrt{\frac{0.229 (1 - 0.229)}{227}} = 0.028 \]
04

Find the Z-value for 95% Confidence

For a 95% confidence interval, the Z-value is 1.96. This value comes from standard normal distribution tables.
05

Calculate the Margin of Error

The margin of error (MOE) is the product of the Z-value and the standard error. \[ MOE = 1.96 \times 0.028 = 0.055 \]
06

Construct the Confidence Interval

The 95% confidence interval is given by \[ \hat{p} \pm MOE \] Thus, the confidence interval is \[ 0.229 - 0.055 < p < 0.229 + 0.055 \] Simplifying, \[ 0.174 < p < 0.284 \]
07

Interpret the Confidence Interval

The 95% confidence interval estimate of the percentage of OxyContin users who develop nausea is between 17.4% and 28.4%.
08

Compare with Given Placebo Interval

The given 95% confidence interval for subjects using a placebo is 1.93% to 20.3%. While the placebo interval is significantly lower, compare the overlap. The OxyContin interval (17.4% to 28.4%) overlaps the upper bound of the placebo interval (20.3%).
09

Conclusion

The interval for OxyContin users is significantly higher, indicating a greater likelihood of developing nausea compared to the placebo group. The overlapping area suggests that the difference might not be due to random chance alone.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values used to estimate a population parameter. In this case, we want to estimate the percentage of OxyContin users who develop nausea. A 95% confidence interval implies that if the same sampling process were repeated many times, 95% of the resulting intervals would be expected to contain the true population percentage.
To construct this interval, we start with the sample proportion, then calculate the margin of error and use these to form the interval around the sample proportion.
Sample Proportion
The sample proportion (\(\backslash\thet {\backslash} chat{p}\)\backslash) represents the fraction of the sample that exhibits the characteristic we are studying. Here, we are interested in nausea among OxyContin users. The proportion is calculated as the number of subjects who developed nausea divided by the total number of subjects.
The formula is \(\backslash\thet{\backslash}hat {p} = \backslash frac{ \backslash number subjects with characteristic }{\backslash total number of subjects}\)\backslash.
For example, with 52 out of 227 subjects experiencing nausea, the sample proportion is \(\backslash\thet {0.229}\)\backslash .
Standard Error
The standard error (SE) measures the variability of a sample proportion. It helps us understand how much the sample proportion might differ from the true population proportion, due to random sampling error. The formula for SE is:
\(\backslash \thet{ SE = \backslash sqrt{ \backslash\thet(\backslash hat { p} (1 - \backslash \thet(1 - \backslash hat{ p})) / n}}\)\backslash,where \(\backslash\thet{n}\)\backslash is the sample size.
In our case, it results in \(\backslash\thet {SE=0.028}\)\backslash.
Margin of Error
The margin of error (MOE) quantifies the amount of random sampling error in the survey's results. It combines the standard error with the Z-value associated with the desired level of confidence. For a 95% confidence interval, the Z-value is 1.96. The formula is:
\(\backslash \theta{ MOE= Z \backslash\thet (\times SE)}\)\backslash. In our example,\(\backslash\theta { MOE = 1.96 \times 0.028 = 0.055}\)\backslash. This indicates how far we expect the sample proportion to be from the true population proportion.
Statistical Comparison
Statistical comparison involves examining confidence intervals to see if there's an overlap, which can indicate a lack of significant difference between groups. In this exercise, we compare the nausea confidence interval for OxyContin users to those given a placebo. The OxyContin interval (\(\backslash\theta {17.4\backslash 鈥�28.4\backslash\theta %}\) overlaps slightly with the placebo interval (\(\backslash\theta {1.93鈥揬backslash 20.3\backslash\theta %}\)). While both intervals include the value around 20.3%, the OxyContin confidence interval lies mostly above the placebo interval, suggesting a higher likelihood of nausea among OxyContin users compared to those on a placebo.

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Most popular questions from this chapter

Example 2 showed how the statistics of \(n=22\) and \(s=14.3\) result in this \(95 \%\) confidence interval estimate of \(\sigma: 11.0<\sigma<20.4\). That confidence interval can also be expressed as \((11.0,20.4)\), but it cannot be expressed as \(15.7 \pm 4.7\). Given that \(15.7 \pm 4.7\) results in values of \(11.0\) and \(20.4\), why is it wrong to express the confidence interval as \(15.7 \pm 4.7 ?\)

Assume that all grade-point averages are to be standardized on a scale between 0 and 4 . How many grade-point averages must be obtained so that the sample mean is within \(0.01\) of the population mean? Assume that a \(95 \%\) confidence level is desired. If we use the range rule of thumb, we can estimate \(\sigma\) to be range \(/ 4=(4-0) / 4=1\). Does the sample size seem practical?

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You are the operations manager for American Airlines and you are considering a higher fare level for passengers in aisle seats. You want to estimate the percentage of passengers who now prefer aisle seats. How many randomly selected air passengers must you survey? Assume that you want to be \(95 \%\) confident that the sample percentage is within \(2.5\) percentage points of the true population percentage. a. Assume that nothing is known about the percentage of passengers who prefer aisle seats. b. Assume that a prior survey suggests that about \(38 \%\) of air passengers prefer an aisle seat (based on a \(3 \mathrm{M}\) Privacy Filters survey).

Here is a sample of measured radiation emissions \((\mathrm{cW} / \mathrm{kg})\) for cell phones (based on data from the Environmental Working Group): \(38,55,86,145\). Here are ten bootstrapsamples: \(\\{38,145,55,86\\},\\{86,38,145,145\\},\\{145,86,55,55\\},\\{55,55,55,145\\}\), \(\\{86,86,55,55\\},\\{38,38,86,86\\},\\{145,38,86,55\\},\\{55,86,86,86\\},\\{145,86,55,86\\}\), \(\\{38,145,86,55\\}\) a. Using only the ten given bootstrap samples, construct an \(80 \%\) confidence interval estimate of the population mean. b. Using only the ten given bootstrap samples, construct an \(80 \%\) confidence interval estimate of the population standard deviation.
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