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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 largesample \(95 \%\) 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 confidence interval is option b: \(0.530 \pm 0.098\).

Step by step solution

01

Define the Sample Proportion

Calculate the sample proportion \( \hat{p} \) using the formula \( \hat{p} = \frac{x}{n} \), where \( x = 53 \) is the number of successes (students who said "Good") and \( n = 100 \) is the total sample size. Thus, \( \hat{p} = \frac{53}{100} = 0.530 \).
02

Determine the Confidence Level and Corresponding Z-Score

For a \(95\%\) confidence level, we use the standard normal distribution table to find the Z-score, often looked up to be approximately \(Z = 1.96\).
03

Calculate the Standard Error

The standard error (SE) of the sample proportion is given by \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \). Substituting known values gives \( SE = \sqrt{\frac{0.530 \times 0.470}{100}} = 0.0497 \).
04

Compute the Margin of Error

The margin of error (ME) is calculated using \( ME = Z \times SE \). Substituting the Z-score and SE gives \( ME = 1.96 \times 0.0497 \approx 0.0975 \), which rounds to \(0.098\).
05

Write the Confidence Interval

The confidence interval is given by \( \hat{p} \pm \text{ME} \). For this exercise, it is \( 0.530 \pm 0.098 \).

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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 crucial starting point in statistics, especially when dealing with survey data or polls. It helps us estimate the proportion of the population with a certain characteristic based on a sample.
To find the sample proportion, use the formula:
  • \( \hat{p} = \frac{x}{n} \)
where \( x \) is the number of individuals in the sample with the trait of interest (in this case, those who said their job prospects are "Good"), and \( n \) is the total number of individuals in the sample. By applying this formula, we found that in the college seniors' survey, \( \hat{p} = \frac{53}{100} = 0.530 \).
This means 53% of the surveyed students viewed their job prospects favorably. By observing this sample proportion, we can make inferences about the entire population of college seniors with some level of confidence.
Standard Error
The standard error is a measure of the variability or dispersion of the sample proportion from the actual population proportion. It gives us an idea of how much the sample results might differ from one sample to another. Standard error is crucial because it forms the basis for all further calculations in estimating the true population proportion.
The formula for the standard error of the sample proportion is given by:
  • \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \)
In the practice exercise, using \( \hat{p} = 0.530 \) and \( n = 100 \), the standard error is calculated as:
\( SE = \sqrt{\frac{0.530 \times 0.470}{100}} \approx 0.0497 \).
The standard error helps quantify the margin by which our sample proportion might vary due to random sampling variability.
Margin of Error
The margin of error represents the range within which we expect the true population proportion to lie, at a certain confidence level. It's a key component of constructing a confidence interval and is derived from the standard error. The formula for the margin of error involves the Z-score (which relates to the desired confidence level) multiplied by the standard error.
The formula is:
  • \( ME = Z \times SE \)
For a 95% confidence level, our Z-score is approximately 1.96. Therefore, in the example given:
\( ME = 1.96 \times 0.0497 \approx 0.0975 \) rounded to 0.098.
This margin of error tells us that the true proportion of all college seniors who view their job prospects positively is likely within 9.8% points of the sample proportion.
Z-Score
The Z-score is a statistical measure that helps standardize data points, indicating how many standard deviations a value is from the mean. In the context of confidence intervals, the Z-score corresponds to the level of confidence we wish to have about our estimate falling within a certain range.
For example:
  • A 95% confidence level, commonly used in surveys and polling, correlates to a Z-score of approximately 1.96.
This Z-score tells us how wide the confidence interval should be, based on the standard error. By multiplying this Z-score by the standard error, we calculate the margin of error. It's important because it adjusts the width of the confidence interval, balancing the precision of our estimate with the level of confidence we require.

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Most popular questions from this chapter

The IRS Plans an SRS. The Internal Revenue Service plans to examine an SRS of individual federal income tax returns from each state. One variable of interest is the proportion of returns claiming itemized deductions. The total number of tax returns in a state varies from almost 30 million in California to approximately 500,000 in Wyoming. a. Will the margin of error for estimating the population proportion change from state to state if an SRS of 2000 tax returns is selected in each state? Explain your answer. b. Will the margin of error change from state to state if an SRS of \(1 \%\) of all tax returns is selected in each state? Explain your answer.

Stopping Traffic with a Smile! Throughout Europe, more than 8000 pedestrians are killed each year in road accidents, with approximately \(25 \%\) of these dying when using a pedestrian crossing. Although failure to stop for pedestrians at a pedestrian crossing is a serious traffic offense in France, more than half of drivers do not stop when a pedestrian is waiting at a crosswalk. In this experiment, a female research assistant was instructed to stand at a pedestrian crosswalk and stare at the driver's face as a car approached the crosswalk. In 400 trials, the research assist ant maintained a neutral expression, and in a second set of 400 trials, the research assistant was instructed to smile. The order of smiling or not smiling was randomized, and several pedestrian crossings were used in a town on the coast in the west of France. The research assistant was dressed in normal attire for her age (jeans, t-shirt, and sneakers). \({ }^{24}\) a. In the 400 trials in which the assistant maintained a neutral expression, the driver stopped in 229 out of the 400 trials. Find a 95\% confidence interval for the proportion of drivers who would stop when a neutral expression is maintained. b. In the 400 trials in which the assistant smiled at the driver, the driver stopped in 277 out of the 400 trials. Find a \(95 \%\) confidence interval for the proportion of drivers who would stop when the assistant is smiling. c. What do your results in parts (a) and (b) suggest about the effect of a smile on a driver stopping at a pedestrian crosswalk? Explain briefly. (In Chapter 23 , we will consider formal methods for comparing two proportions.)

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 \(\widehat{p}=0.7\). 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. \(H_{0}: p=0.5, H_{a}: p \neq 0.5\).

A Gallup Poll in November 2019 found that \(55 \%\) of the people in the sample said they wanted to lose weight. The poll's margin of error for \(95 \%\) confidence was \(4 \%\). This means that a. the poll used a method that gets an answer within \(4 \%\) of the truth about the population \(95 \%\) of the time. b. we can be sure that the percentage of all adults who want to lose weight is between \(50 \%\) and \(58 \%\). c. if Gallup takes another poll using the same method, the results of the second poll will lie between \(51 \%\) and \(59 \%\).

Do Smokers Know That Smoking is Bad for Them? The Harris Poll asked a sample of smokers, "Do you believe that smoking will probably shorten your life, or not?" Of the 1010 people in the sample, 848 said "yes." a. Harris called residential telephone numbers at random in an attempt to contact an SRS of smokers. Based on what you know about national sample surveys, what is likely to be the biggest weakness in the survey? b. We will nonetheless act as if the people interviewed are an SRS of smokers. Give a \(95 \%\) confidence interval for the percent of smokers who agree that smoking will probably shorten their lives.
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