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Chapter 11: Problem 10

Construct a confidence interval for \(p_{1}-p_{2}\) at the given level of confidence. \(x_{1}=804, n_{1}=874, x_{2}=892, n_{2}=954,95 \%\) confidence

Short Answer

Expert verified
The 95% confidence interval for \( p_1 - p_2 \) is (-0.0389, 0.0069).

Step by step solution

01

Calculate Sample Proportions

Calculate the sample proportions for both groups. Use the formula \( \hat{p} = \frac{x}{n} \ \). For the first group, \( \hat{p}_{1} = \frac{804}{874} = 0.919 \ \). For the second group, \( \hat{p}_{2} = \frac{892}{954} = 0.935 \ \).
02

Determine the Difference in Sample Proportions

Subtract the sample proportions to find the difference: \( \hat{p}_{1} - \hat{p}_{2} = 0.919 - 0.935 = -0.016 \ \).
03

Calculate the Standard Error

Use the formula for the standard error of the difference between two proportions: \[ \text{SE} = \sqrt{ \frac{ \hat{p}_1 (1 - \hat{p}_1)}{n_1} + \frac{ \hat{p}_2 (1 - \hat{p}_2)}{n_2}} \ \]. Substitute the values: \[ \text{SE} = \sqrt{ \frac{0.919(1 - 0.919)}{874} + \frac{0.935(1 - 0.935)}{954}} \approx 0.0117 \ \].
04

Find the Z-Score for the Given Confidence Level

For a 95% confidence level, the Z-score is approximately 1.96.
05

Calculate the Margin of Error

Multiply the Z-score by the standard error to find the margin of error: \( \text{ME} = 1.96 \times 0.0117 = 0.0229 \ \).
06

Construct the Confidence Interval

Add and subtract the margin of error from the difference in sample proportions to find the confidence interval: \[ -0.016 \pm 0.0229 = (-0.0389, 0.0069) \ \].

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Key Concepts

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

sample proportions
Sample proportions are estimates that represent the proportion of a particular outcome in a sample of data. To calculate a sample proportion, use the formula: \( \hat{p} = \frac{x}{n} \ \). Here, \(x\ \) is the number of successes (or occurrences of the event being measured) and \(n\ \) is the total number of observations in the sample.
In the provided exercise, we calculated the sample proportions for two groups:
  • First group: \( \hat{p}_{1} = \frac{804}{874} = 0.919 \ \).
  • Second group: \( \hat{p}_{2} = \frac{892}{954} = 0.935 \ \).
These sample proportions are then used to determine the difference between the proportions of the two groups.
standard error
Standard Error (SE) quantifies the amount of variability in a sample statistic, such as a sample proportion, when estimating the population parameter. For the difference between two sample proportions, the standard error is calculated using:
\[ \text{SE} = \sqrt{ \frac{ \hat{p}_{1}(1 - \hat{p}_{1})}{n_{1}} + \frac{ \hat{p}_{2}(1 - \hat{p}_{2})}{n_{2}}} \ \].
In the exercise, substituting the values gave us:
  • \( \text{SE} = \sqrt{ \frac{0.919(1 - 0.919)}{874} + \frac{0.935(1 - 0.935)}{954}} \ \approx 0.0117 \ \ \).
This calculated standard error helps in determining the margin of error for constructing a confidence interval.
margin of error
The margin of error provides a range within which the true difference between population proportions is likely to fall. It is determined by multiplying the Z-score (associated with the desired confidence level) by the standard error (SE).
In the exercise, the Z-score for a 95% confidence level is approximately 1.96. To calculate the margin of error:
\( \text{ME} = 1.96 \times 0.0117 \ = 0.0229 \ \).
This value indicates the range around the sample proportion difference within which the true population proportion difference is likely to be found with 95% confidence.
Z-score
A Z-score is a measure of how many standard deviations an element is from the mean. In constructing confidence intervals, the Z-score corresponds to the desired confidence level.
For a 95% confidence interval, the Z-score is approximately 1.96. This value is determined from standard normal distribution tables (or normal Z-tables) and represents the number of standard deviations you must extend from the mean to encompass 95% of the data.
In this exercise, the Z-score of 1.96 is crucial for calculating the margin of error, which then helps establish the confidence interval around the difference in sample proportions.

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Most popular questions from this chapter

An educator wants to determine the difference between the proportion of males and females who have completed 4 or more years of college. What sample size should be obtained if she wishes the estimate to be within two percentage points with \(90 \%\) confidence, assuming that (a) she uses the 1999 estimates of \(27.5 \%\) male and \(23.1 \%\) female from the U.S. Census Bureau? (b) she does not use any prior estimates?

Construct a confidence interval for \(p_{1}-p_{2}\) at the given level of confidence. \(x_{1}=28, n_{1}=254, x_{2}=36, n_{2}=301,95 \%\) confidence

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Suppose that in clinical trials for treatment of a skin disorder 642 of 2105 patients receiving the current standard treatment were cured of the disorder and 697 of 2115 patients receiving a new proposed treatment were cured of the disorder. (a) Does the new procedure cure a higher percentage of patients at the \(\alpha=0.05\) level of significance? (b) Do you think that the difference in success rates is practically significant? What factors might influence your decision?
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