/*! 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

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?

A study suggests that \(60 \%\) of college student spend 10 or more hours per week communicating with others online. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. You randomly sample 160 students from your dorm and find that \(70 \%\) spent 10 or more hours a week communicating with others online. A friend of yours, who offers to help you with the hypothesis test, comes up with the following set of hypotheses. Indicate any errors you see. $$ \begin{array}{l} H_{0}: \hat{p}<0.6 \\ H_{A}: \hat{p}>0.7 \end{array} $$

Determine whether the following statement is true or false, and explain your reasoning: "With large sample sizes, even small differences between the null value and the observed point estimate can be statistically significant."

Of all freshman at a large college, \(16 \%\) made the dean's list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. (a) What is this distribution called? (b) Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning. (c) Calculate the variability of this distribution. (d) What is the formal name of the value you computed in (c)? (e) Suppose the students decide to sample again, this time collecting 90 students per sample, and they again collect 1,000 samples. They build a new distribution of sample proportions. How will the variability of this new distribution compare to the variability of the distribution when each sample contained 40 observations?

In each part below, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios. (a) The standard error of \(\hat{p}\) when (I) \(n=125\) or (II) \(n=500\). (b) The margin of error of a confidence interval when the confidence level is (I) \(90 \%\) or (II) \(80 \%\). (c) The p-value for a Z-statistic of 2.5 calculated based on a (I) sample with \(n=500\) or based on a (II) sample with \(n=1000\). (d) The probability of making a Type 2 Error when the alternative hypothesis is true and the significance level is (I) 0.05 or (II) 0.10 .

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