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Chapter 9: Problem 7

Construct a confidence interval of the population proportion at the given level of confidence. $$x=120, n=500,99 \% \text { confidence }$$

Short Answer

Expert verified
The 99% confidence interval for the population proportion is \(0.189 \le p \le 0.291\).

Step by step solution

01

Calculate the sample proportion

The sample proportion \(\hat{p}\) is calculated using the formula: \[ \hat{p} = \frac{x}{n} \] Substituting the given values: \[ \hat{p} = \frac{120}{500} = 0.24 \]
02

Find the margin of error

The margin of error (E) for a population proportion is given by: \[ E = Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \] For a 99% confidence interval, the critical value \(Z_{\alpha/2}\) is approximately 2.576. Now substitute the values: \[ E = 2.576 \cdot \sqrt{\frac{0.24 \cdot (1 - 0.24)}{500}} = 0.051 \]
03

Construct the confidence interval

The confidence interval for the population proportion \(p\) is given by: \[ \hat{p} - E \le p \le \hat{p} + E \] Substituting the values: \[ 0.24 - 0.051 \le p \le 0.24 + 0.051 \] Therefore, \[ 0.189 \le p \le 0.291 \]

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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, denoted as \( \hat{p} \), is an important concept in statistics. It represents the proportion of individuals in a sample that possess a certain characteristic.
To find the sample proportion, use the formula:
\[ \hat{p} = \frac{x}{n} \]
In this exercise, \( x \) is the number of successes (120) and \( n \) is the total sample size (500).
By plugging in these values, we get:
\[ \hat{p} = \frac{120}{500} = 0.24 \]
This means that 24% of the sample possesses the characteristic we are studying. Understanding how to calculate \( \hat{p} \) is fundamental since it is a building block for more advanced calculations.
Margin of Error
The margin of error (E) gives us an idea of the uncertainty around our sample proportion. It tells us how much we can expect the population proportion (\( p \)) to differ from our sample proportion (\( \hat{p} \)).
The formula to find E is:
\[ E = Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]
Here, \( Z_{\alpha/2} \) is the critical value that comes from the standard normal distribution. For a 99% confidence interval, \( Z_{\alpha/2} \) is approximately 2.576.
Substituting the values, we get:
\[ E = 2.576 \cdot \sqrt{\frac{0.24 \cdot (1 - 0.24)}{500}} = 0.051 \]
So, the margin of error here is 0.051.
99% Confidence Interval
A confidence interval provides a range of values which is likely to contain the population proportion \( p \) with a certain level of confidence. For a 99% confidence interval, we are saying that we are 99% sure the true population proportion lies within this range.
To construct the confidence interval, use the formula:
\[ \hat{p} - E \le p \le \hat{p} + E \]
With our sample proportion \( \hat{p} \) as 0.24 and margin of error E as 0.051, we get:
\[ 0.24 - 0.051 \le p \le 0.24 + 0.051 \]
Which simplifies to:
\[ 0.189 \le p \le 0.291 \]
Therefore, we can say with 99% confidence that the true population proportion is between 0.189 and 0.291.

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Most popular questions from this chapter

(a) Find the \(t\) -value such that the area in the right tail is 0.10 with 25 degrees of freedom. (b) Find the \(t\) -value such that the area in the right tail is 0.05 with 30 degrees of freedom. (c) Find the \(t\) -value such that the area left of the \(t\) -value is 0.01 with 18 degrees of freedom. [Hint: Use symmetry. (d) Find the critical \(t\) -value that corresponds to \(90 \%\) confidence. Assume 20 degrees of freedom.

A researcher wishes to estimate the mean number of miles on 3-year-old Chevy Cavaliers. (a) How many cars should be in a sample to estimate the mean number of miles within 2000 miles with 98\% confidence, assuming that \(\sigma=16,100 ?\) (b) How many cars should be in a sample to estimate the mean number of miles within 1000 miles with 98\% confidence, assuming that \(\sigma=16,100 ?\) (c) What effect does doubling the required accuracy have on the sample size? Why is this the expected result?

Suppose the arrival of cars at Burger King's drive-through follows a Poisson process with \(\mu=4\) cars every 10 minutes. (a) Simulate obtaining 30 samples of size \(n=40\) from this population. (b) Construct \(90 \%\) confidence intervals for each of the 30 samples. [Note: \(\sigma=\sqrt{\mu}\) in a Poisson process.] (c) How many of the intervals do you expect to include the population mean? How many actually contain the population mean?

Own a Gun? In a Harris Poll conducted in May 2000 , \(39 \%\) of the people polled answered yes to the following question: "Do you happen to have in your home or garage any guns or revolvers?" The margin of error in the poll was \(\pm 3 \%=\pm 0.03,\) and the estimate was made with \(95 \%\) confidence. How many people were surveyed?

Pepcid A study of 74 patients with ulcers was conducted in which they were prescribed \(40 \mathrm{mg}\) of Pepcid \(^{\mathrm{TM}}\). After 8 weeks, 58 reported ulcer healing. (a) Obtain a point estimate for the proportion of patients with ulcers receiving Pepcid who will report ulcer healing. (b) Verify that the requirements for constructing a confidence interval about \(\hat{p}\) are satisfied. (c) Construct a \(99 \%\) confidence interval for the proportion of patients with ulcers receiving Pepcid who will report ulcer healing. (d) Interpret the confidence interval.
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