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Chapter 22: Problem 20

An opinion poll asks an SRS of 100 college seniors how they view their job prospects. In all, 53 say Good. The large-sample \(95 \%\) confidence interval for estimating the proportion of all college seniors who think their job prospects are good is (a) \(0.530 \pm 0.082 .\) (b) \(0.530 \pm 0.098 .\) (c) \(0.530 \pm 0.049\).

Short Answer

Expert verified
The correct interval is (b) \(0.530 \pm 0.098 \).

Step by step solution

01

Identify the Sample Proportion

First, determine the sample proportion \( \hat{p} \) of college seniors who say their job prospects are good. Since 53 out of 100 seniors say 'Good', \( \hat{p} = \frac{53}{100} = 0.530 \).
02

Calculate the Standard Error

The standard error of the sample proportion is given by \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( n \) is the sample size. So, \( SE = \sqrt{\frac{0.530 \times 0.470}{100}} \approx 0.0497 \).
03

Find the Z-score for 95% Confidence

For a 95% confidence interval, the Z-score is approximately 1.96 (from standard normal distribution tables).
04

Calculate the Margin of Error

The margin of error \( ME \) is calculated as \( ME = Z \times SE = 1.96 \times 0.0497 \approx 0.0975 \).
05

Determine the Confidence Interval

The 95% confidence interval is \( \hat{p} \pm ME = 0.530 \pm 0.0975 \). Approximating, the confidence interval is \( 0.530 \pm 0.098 \), which matches option (b).

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Key Concepts

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

Sample Proportion
When we talk about the sample proportion, we're referring to the fraction of individuals in a sample who hold a specific attribute we're interested in. For example, if we surveyed a group of students and found that some of them rated their job prospects as "Good," the sample proportion would reflect this finding numerically.
The sample proportion is often represented as \( \hat{p} \). It's calculated by dividing the number of successes (in our case, students who say their job prospects are "Good") by the total number of respondents in the sample.
  • In this example, out of 100 college seniors, 53 think their job prospects are good. So, the sample proportion \( \hat{p} \) is \( \frac{53}{100} = 0.530 \).
This simple calculation helps us estimate how a larger population might feel, based on our smaller, representative sample.
Standard Error
The standard error (SE) is crucial in statistics as it measures how much the sample proportion is expected to vary from the actual population proportion. This accounts for possible sampling variability.
The formula for calculating standard error of the sample proportion is:
  • \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \)
where \( \hat{p} \) is the sample proportion, and \( n \) is the size of the sample.
In our example:
  • \( SE = \sqrt{\frac{0.530 \times 0.470}{100}} \approx 0.0497 \)
This result tells us the typical distance between the sample proportion and the true population proportion.
Z-score
The Z-score is a statistical measurement that tells us how many standard deviations an element is from the mean. In the context of confidence intervals, it's used to find the probability that a sample statistic falls within a certain range.
For creating a 95% confidence interval, the Z-score is determined using a standard normal distribution table.
  • The common Z-score for a 95% confidence level is 1.96.
This score represents a confidence level where 95% of sample proportions would fall within \( 1.96 \cdot SE \) of the actual population proportion. This Z-score is multiplied by the standard error to calculate the margin of error, expanding or contracting the range of our confidence interval.
Margin of Error
The margin of error quantifies the uncertainty involved in sampling and helps us express the confidence interval.
  • It's calculated by multiplying the Z-score with the standard error.
  • In our example: \( ME = 1.96 \times 0.0497 \approx 0.0975 \).
A higher margin of error means more uncertainty, whereas a lower margin represents more precision in the estimate.
The margin of error defines an interval around the sample proportion, giving us a range where the true population proportion is likely to be.
  • Thus, the 95% confidence interval becomes \( 0.530 \pm 0.0975 \), or simply \( 0.530 \pm 0.098 \).
This means we're 95% confident that the true proportion of all college seniors who think their job prospects are "Good" falls in this interval.

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Most popular questions from this chapter

A random digit dialing telephone survey of 880 drivers asked, "Recalling the last 10 traffic lights you drove through, how many of them were red when you entered the intersections?" Of the 880 respondents, 171 admitted that at least one light had been red. \({ }^{24}\) (a) Give a \(95 \%\) confidence interval for the proportion of all drivers who ran one or more of the last 10 red lights they met. (b) Nonresponse is a practical problem for this survey-only \(21.6 \%\) of calls that reached a live person were completed. Another practical problem is that people may not give truthful answers. What is the likely direction of the bias: do you think more or fewer than 171 of the 880 respondents really ran a red light? Why?

Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. Only half of all mice complete one particular maze in less than 18 seconds. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures the proportion of 40 mice that completed the maze in less than 18 seconds with noise as a stimulus. The proportion of mice that completed the maze in less than 18 seconds is \(\mathrm{p}^{\wedge \hat{p}}=0.7\). The hypotheses for a test to answer the researcher's question are (a) \(H_{0}: p=0.5, H_{a}: p>0.5\). (b) \(H_{0}: p=0.5, H_{a}: p<0.5\). (c) \(H_{0}: p=0.5, H_{a}: p \neq 0.5\).

Explain whether we can use the \(z\) test for a proportion in these situations. (a) You toss a coin 10 times in order to test the hypothesis \(H_{0}: p=0.5\) that the coin is balanced. (b) A local candidate contacts an SRS of 900 of the registered voters in his district to see if there is evidence that more than half support the bill he is sponsoring. (c) A college president says, " \(99 \%\) of the alumni support my firing of Coach Boggs." You contact an SRS of 200 of the college's 15,000 living alumni to test the hypothesis \(H_{0}: p=0.99\).

In the National AIDS Behavioral Surveys sample of 2673 adult heterosexuals, \(0.2 \%\) (that's \(0.002\) as a decimal fraction) had both received a blood transfusion and had a sexual partner from a group at high risk of AIDS. Explain why we can't use the large-sample confidence interval to estimate the proportion \(p\) in the population who share these two risk factors.

Some shrubs have the useful ability to resprout from their roots after their tops are destroyed. Fire is a particular threat to shrubs in dry climates because it can injure the roots as well as destroy the aboveground material. One study of resprouting took place in a dry area of Mexico. \({ }^{32}\) The investigators clipped the tops of samples of several species of shrubs. In some cases, they also applied a propane torch to the stumps to simulate a fire. Of 12 specimens of the shrub Krameria cytisoides, five resprouted after fire. Estimate with \(90 \%\) confidence the proportion of all shrubs of this species that will resprout after fire.
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