/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 In Exercise \(5.19,\) we learned... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercise \(5.19,\) we learned that a Rasmussen Reports survey of \(1,000 \mathrm{US}\) adults found that \(42 \%\) believe raising the minimum wage will help the economy. Construct a \(99 \%\) confidence interval for the true proportion of US adults who believe this.

Short Answer

Expert verified
The 99% confidence interval is [0.3801, 0.4599].

Step by step solution

01

Identify the Sample Proportion

The sample proportion \( \hat{p} \) is the percentage given in the survey. Since \(42\%\) of the surveyed 1,000 US adults believe raising the minimum wage will help the economy, we have \( \hat{p} = 0.42 \).
02

Determine the Confidence Level

The exercise asks for a \(99\%\) confidence interval. This means that we are using a confidence level of \(0.99\).
03

Find the Critical Value

For a \(99\%\) confidence level, we need the critical value \(z\) for a standard normal distribution. This value can be found using a Z-table or calculator and is approximately \(z \approx 2.576\).
04

Calculate Standard Error

The standard error (SE) is calculated using the formula \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \(n\) is the sample size. Therefore, \(SE = \sqrt{\frac{0.42 \times 0.58}{1000}} \approx 0.0155\).
05

Compute the Margin of Error

The margin of error (ME) is calculated as \( ME = z \times SE \). Substituting the known values, \( ME = 2.576 \times 0.0155 \approx 0.0399 \).
06

Construct the Confidence Interval

The confidence interval is given by \( \hat{p} \pm ME \). Therefore, the \(99\%\) confidence interval is \(0.42 - 0.0399\) to \(0.42 + 0.0399\), which simplifies to \( [0.3801, 0.4599] \).

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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, often denoted as \( \hat{p} \), is a pivotal concept in statistics that reflects how a sample behaves with respect to a particular attribute or characteristic. In the context of this exercise, \( \hat{p} \) is the fraction of survey participants that believe raising the minimum wage will help the economy. The survey states that \(42\%\) of the 1,000 adults hold this belief, so \( \hat{p} = 0.42 \).

Understanding the sample proportion helps us estimate the true proportion in the entire population. It is essential to keep in mind that the sample proportion is subject to sampling variability, meaning it might differ if a different sample is taken. This is why further statistical tools, like confidence intervals, are needed to make inferences about the full population.
Standard Error
The standard error (SE) is a measure that captures the variability of the sample proportion estimated from different possible samples of the same size. Essentially, it tells us how much the sample proportion \( \hat{p} \) would vary if we were to repeatedly sample from the population under the same conditions.

The formula for standard error is given by \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]where \( \hat{p} \) is the sample proportion and \( n \) is the sample size. In the exercise, substituting \( \hat{p} = 0.42 \) and \( n = 1000 \), we find \( SE \approx 0.0155 \).

The standard error plays a crucial role in the calculation of the confidence interval, as it helps to determine the margin of error.
Critical Value
The critical value is a factor used in calculating the margin of error and constructing the confidence interval. It depends on the chosen confidence level and the sampling distribution. For a normal distribution and a high confidence level of \(99\%\), the critical value corresponds to a Z-score that leaves \(0.5\%\) in each tail of the standard normal distribution.

In our case, this critical value is approximately \( z = 2.576 \). This means that a \(99\%\) confidence interval corresponds to the range within which \(99\%\) of the sample proportions would fall if we repeatedly took samples of the same size.

Accurately determining the critical value is vital because it ensures the robustness of the confidence interval, keeping the interval width appropriate for the desired level of confidence.
Margin of Error
The margin of error (ME) quantifies the maximum expected difference between the actual population proportion and the observed sample proportion. It's a crucial component of the confidence interval, indicating the extent of possible sampling error.

The formula for margin of error is:\[ ME = z \times SE \]where \( z \) is the critical value and \( SE \) is the standard error. For this exercise, inserting the known figures \( z = 2.576 \) and \( SE \approx 0.0155 \), we get \( ME \approx 0.0399 \).

Thus, the margin of error helps create the bounds of the confidence interval, allowing us to say with confidence (\(99\%\) in this case) that the true population proportion lies within this range from the observation \( \hat{p} \).

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Most popular questions from this chapter

For each of the following situations, state whether the parameter of interest is a mean or a proportion. It may be helpful to examine whether individual responses are numerical or categorical. (a) In a survey, one hundred college students are asked how many hours per week they spend on the Internet. (b) In a survey, one hundred college students are asked: "What percentage of the time you spend on the Internet is part of your course work?" (c) In a survey, one hundred college students are asked whether or not they cited information from Wikipedia in their papers. (d) In a survey, one hundred college students are asked what percentage of their total weekly spending is on alcoholic beverages. (e) In a sample of one hundred recent college graduates, it is found that 85 percent expect to get a job within one year of their graduation date.

A USA Today / Gallup poll asked a group of unemployed and underemployed Americans if they have had major problems in their relationships with their spouse or another close family member as a result of not having a job (if unemployed) or not having a full-time job (if underemployed). \(27 \%\) of the 1,145 unemployed respondents and \(25 \%\) of the 675 underemployed respondents said they had major problems in relationships as a result of their employment status. (a) What are the hypotheses for evaluating if the proportions of unemployed and underemployed people who had relationship problems were different? (b) The p-value for this hypothesis test is approximately \(0.35 .\) Explain what this means in context of the hypothesis test and the data.

As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. (a) What population is under consideration in the data set? (b) What parameter is being estimated? (c) What is the point estimate for the parameter? (d) What is the name of the statistic can we use to measure the uncertainty of the point estimate? (e) Compute the value from part (d) for this context. (f) The historical rate of defects is \(10 \%\). Should the engineer be surprised by the observed rate of defects during the current week? (g) Suppose the true population value was found to be \(10 \%\). If we use this proportion to recompute the value in part (e) using \(p=0.1\) instead of \(\hat{p},\) does the resulting value change much?

Write the null and alternative hypotheses in words and then symbols for each of the following situations. (a) A tutoring company would like to understand if most students tend to improve their grades (or not) after they use their services. They sample 200 of the students who used their service in the past year and ask them if their grades have improved or declined from the previous year. (b) Employers at a firm are worried about the effect of March Madness, a basketball championship held each spring in the US, on employee productivity. They estimate that on a regular business day employees spend on average 15 minutes of company time checking personal email, making personal phone calls, etc. They also collect data on how much company time employees spend on such non-business activities during March Madness. They want to determine if these data provide convincing evidence that employee productivity changed during March Madness.

It is believed that nearsightedness affects about \(8 \%\) of all children. In a random sample of 194 children, 21 are nearsighted. Conduct a hypothesis test for the following question: do these data provide evidence that the \(8 \%\) value is inaccurate?

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