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Chapter 9: Problem 17

In a survey of 1008 adult Americans, the Gallup organization asked, "When you retire, do you think you will have enough money to live comfortably or not?" Of the 1008 surveyed, 526 stated that they were worried about having enough money to live comfortably in retirement. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who are worried about having enough money to live comfortably in retirement.

Short Answer

Expert verified
The 90% confidence interval is approximately 0.4960 to 0.5476.

Step by step solution

01

Identify the Sample Proportion

The sample proportion (p-hat) is the number of successes divided by the total number of respondents. Here, successes are the 526 adults who stated they were worried about having enough money. Compute it as follows: \(\hat{p} = \frac{526}{1008}\)
02

Calculate the Sample Proportion and Complement

The calculated value of \(\hat{p}\) is: \(\hat{p} = 0.5218\). The complement of \(\hat{p}\) is found by \(q = 1 - \hat{p}\): \(q = 1 - 0.5218 = 0.4782\)
03

Find the Z-Score for the Confidence Level

For a 90% confidence interval, the Z-score (critical value) is 1.645. This value is found using Z-tables or standard normal distribution tables.
04

Calculate the Standard Error

The standard error (SE) is calculated using the formula: \(SE = \sqrt{\frac{\hat{p}q}{n}}\). Here: \(SE = \sqrt{\frac{(0.5218)(0.4782)}{1008}} \approx 0.0157\)
05

Determine the Margin of Error

The margin of error (ME) is obtained by multiplying the Z-score by the standard error: \( ME = Z \times SE = 1.645 \times 0.0157 \approx 0.0258\)
06

Construct the Confidence Interval

The confidence interval is found by adding and subtracting the margin of error from the sample proportion: \(CI = \hat{p} \pm ME\). Therefore: \(0.5218 - 0.0258 = 0.4960\) and \(0.5218 + 0.0258 = 0.5476\). Thus, the confidence interval is \(0.4960 < p < 0.5476\)

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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 key concept in statistics. It represents the proportion of individuals in a sample with a particular characteristic.
For example, consider a survey of 1008 adult Americans where 526 expressed concern about having enough money for retirement. The sample proportion is calculated using: \(\hat{p} = \frac{526}{1008}\rightarrow 0.5218\).
This means approximately 52.18% of the surveyed adults are worried.
The sample proportion is essential because it helps estimate the population proportion and informs other calculations.
Margin of Error
The margin of error indicates the range of values above and below a sample statistic and provides an estimate of the potential error in the inference.
It is determined by multiplying the Z-score associated with the desired confidence level by the standard error (SE). For instance, the margin of error (ME) in our case is calculated as: \(\text{ME} = Z \times \text{SE} \rightarrow 1.645 \times 0.0157 \approx 0.0258\).
Thus, our sample proportion estimate could vary by about 卤0.0258.
Standard Error
The standard error (SE) measures the statistical accuracy of the sample proportion and helps in constructing the confidence interval.
It quantifies the variability of the sample proportion and is calculated using the formula: \(\text{SE} = \sqrt{\frac{\hat{p}q}{n}}\鈫� \sqrt{\frac{(0.5218)(0.4782)}{1008}} \approx 0.0157\).
This value tells us how much the sample proportion might deviate from the true population proportion.
Z-score
The Z-score (or Z-value) for a confidence level quantifies the number of standard errors you have to go away from the mean to capture the desired confidence level.
For a 90% confidence interval, the Z-score is 1.645.
This value can be obtained from Z-tables or standard normal distribution tables. The Z-score helps generate the margin of error and ultimately, the confidence interval for our sample proportion.

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Most popular questions from this chapter

Suppose the following data represent the amount of time (in hours) a random sample of students enrolled in College Algebra spent working on a homework assignment: \(3.2,4.1,1.2,0.6,\) and 2.5. Below are three bootstraps samples. For each bootstrap sample, determine the bootstrap sample mean. Bootstrap Sample 1: 1.2,0.6,3.2,3.2,1.2 Bootstrap Sample 2: 0.6,4.1,4.1,0.6,4.1 Bootstrap Sample 3: 4.1,3.2,3.2,0.6,1.2

A simple random sample of size \(n<30\) for \(a\) quantitative variable has been obtained. Using the normal probability plot, the correlation between the variable and expected z-score, and the boxplot, judge whether a t-interval should be constructed. $$ n=9 ; \text { Correlation }=0.997 $$

The exponential probability distribution can be used to model waiting time in line or the lifetime of electronic components. Its density function is skewed right. Suppose the wait-time in a line can be modeled by the exponential distribution with \(\mu=\sigma=5\) minutes. (a) Use StatCrunch, Minitab, or some other statistical software to generate 100 random samples of size \(n=4\) from this population. (b) Construct \(95 \% t\) -intervals for each of the 100 samples found in part (a). (c) How many of the intervals do you expect to include the population mean? How many of the intervals actually contain the population mean? Explain what your results mean. (d) Repeat parts (a)-(c) for samples of size \(n=15\) and \(n=25\). Explain what your results mean.

As the number of degrees of freedom in the \(t\) -distribution increases, the spread of the distribution ________ (increases/decreases).

A recent Gallup poll asked Americans to disclose the number of books they read during the previous year. Initial survey results indicate that \(s=16.6\) books. (a) How many subjects are needed to estimate the number of books Americans read the previous year within four books with \(95 \%\) confidence? (b) How many subjects are needed to estimate the number of books Americans read the previous year within two books with \(95 \%\) confidence? (c) What effect does doubling the required accuracy have on the sample size? (d) How many subjects are needed to estimate the number of books Americans read the previous year within four books with \(99 \%\) confidence? Compare this result to part (a). How does increasing the level of confidence in the estimate affect sample size? Why is this reasonable?
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