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In a survey of 1010 adult Americans, the Gallup Organization asked, "Are you worried or not worried about having enough money for retirement?" Of the 1010 surveyed, 606 stated that they were worried about having enough money for retirement. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who are worried about having enough money for retirement.

Short Answer

Expert verified
The 90% confidence interval is (0.5747, 0.6253).

Step by step solution

01

State the sample proportion

Calculate the sample proportion of respondents who are worried about having enough money for retirement. The sample proportion is given by \( \ \hat{p} = \frac{606}{1010}. \)
02

Calculate the sample proportion

Divide 606 by 1010: \( \ \hat{p} = \frac{606}{1010} \approx 0.6. \)
03

Find the standard error

Compute the standard error for the sample proportion using the formula: \( \ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}, \) where \( \hat{p} \approx 0.6 \) and \( n = 1010. \)
04

Calculate the standard error

Substitute the values into the standard error formula: \( \ SE = \sqrt{0.6 \times 0.4 / 1010} \approx 0.0154. \)
05

Determine the critical value

For a 90% confidence interval, the critical value corresponding to \( z_{\alpha/2} \) (where \( \alpha = 1 - 0.90 = 0.10 \ \alpha/2 = 0.05 \)) can be found from the standard normal distribution (Z table). The critical value is approximately 1.645.
06

Calculate the margin of error

Multiply the critical value by the standard error: \( \ ME = 1.645 \times 0.0154 \approx 0.0253. \)
07

Construct the confidence interval

Add and subtract the margin of error from the sample proportion to obtain the confidence interval:\( \ CI: (\hat{p} - ME, \hat{p} + ME), \) which is \( \ (0.6 - 0.0253, 0.6 + 0.0253) = (0.5747, 0.6253). \)

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Key Concepts

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

sample proportion
When conducting surveys or sampling from a larger population, researchers often talk about the sample proportion. The sample proportion, denoted as \( \hat{p} \), is simply the ratio of the count of favorable outcomes to the total number of observations. In our exercise, 606 out of 1010 respondents worried about retirement, so the sample proportion is calculated as follows:
\(\hat{p} = \frac{606}{1010} \approx 0.6\)
This means that approximately 60% of the surveyed adults are worried about having enough money for retirement. This proportion serves as an estimate of the true proportion of all adult Americans who share this concern.
standard error
The standard error measures the variability of a sample statistic, like the sample proportion, from sample to sample. It gives us an idea of how much the sample proportion might fluctuate by chance alone.
The standard error (SE) for the sample proportion is calculated using the formula:
\( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \)
where \( \hat{p} \approx 0.6 \) is the sample proportion and \( n = 1010 \) is the sample size.
So for our exercise:
\( SE = \sqrt{0.6 \times 0.4 / 1010} \approx 0.0154 \)
The smaller the standard error, the closer the sample proportion should be to the true population proportion.
critical value
When calculating confidence intervals, we need to decide on a level of confidence. This level, such as 90%, 95%, or 99%, represents how sure we are that the true population parameter lies within the calculated interval. To find this interval, we use a critical value from the Z-distribution (standard normal distribution).
For a 90% confidence interval, we look for the critical value \( z_{\alpha/2} \), which corresponds to the odds of being in the tail ends of the normal curve. In our case, with \(\alpha = 0.10\) (since \( 1 - 0.90 = 0.10 \)), the critical value \( z_{\alpha/2} \) is approximately 1.645.
The critical value tells us how many standard errors to extend on either side of the sample proportion to capture the true population proportion within the desired confidence level.

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Most popular questions from this chapter

A Penny for Your Thoughts A researcher for the U.S. Department of the Treasury wishes to estimate the percentage of Americans who support abolishing the penny. What size sample should be obtained if he wishes the estimate to be within 2 percentage points with \(98 \%\) confidence if (a) he uses a June 2004 estimate of \(23 \%\) obtained from a Harris Poll? (b) he does not use any prior estimate?

Suppose the arrival of cars at Burger King's drive-through follows a Poisson process with \(\mu=4\) cars every 10 minutes. (a) Simulate obtaining 30 samples of size \(n=35\) from this population. (b) Obtain the sample mean and standard deviation for each of the 30 samples. (c) Construct \(90 \% ~ t\) -intervals for each of the 30 samples. (d) How many of the intervals do you expect to include the population mean? How many actually contain the population mean?

A Gallup poll conducted January \(17-\) February \(6,2005,\) asked 1028 teenagers aged 13 to \(17,\) "Typically, how many hours per week do you spend watching TV?" Survey results indicate that \(\bar{x}=13.0\) hours and \(s=2.3\) hours. Construct a \(95 \%\) confidence interval for the number of hours of TV teenagers watch each week. Interpret the interval.

A simple random sample of size \(n\) is drawn from a population whose population standard deviation, \(\sigma,\) is known to be \(3.8 .\) The sample mean, \(\bar{x}\), is determined to be \(59.2 .\) (a) Compute the \(90 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is 45 (b) Compute the \(90 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(55 .\) How does increasing the sample size affect the margin of error, \(E ?\) (c) Compute the \(98 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(45 .\) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the size of the margin of error, \(E ?\) (d) Can we compute a confidence interval about \(\mu\) based on the information given if the sample size is \(n=15 ?\) Why? If the sample size is \(n=15,\) what must be true regarding the population from which the sample was drawn?

Load the confidence interval for a mean (the impact of a confidence level) applet. (a) Set the shape to normal with mean \(=50\) and Std. Dev. \(=10 .\) Construct at least 1000 confidence intervals with \(n=10 .\) What proportion of the \(95 \%\) confidence intervals contain the population mean? What proportion did you expect to contain the population mean? (b) Repeat part (a). Did the same proportion of intervals contain the population mean? (c) Set the shape to normal with mean \(=50\) and Std. Dev. \(=10 .\) Construct at least 1000 confidence intervals with \(n=50 .\) What proportion of the \(95 \%\) confidence intervals contain the population mean? What proportion did you expect to contain the population mean? Does sample size have any impact on the proportion of intervals that capture the population mean? (d) Compare the width of the intervals for the samples of size \(n=50\) obtained in part (c) to the width of the intervals for the samples of size \(n=10\) obtained in part (a). Which are wider? Why? (b) Repeat part (a). Did the same proportion of intervals contain the population mean?

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