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Chapter 9: Problem 17

High-Speed Internet Access A researcher wishes to estimate the proportion of adults who have high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.03 with 99\% confidence if (a) she uses a 2004 estimate of 0.44 obtained from a Harris poll? (b) she does not use any prior estimates?

Short Answer

Expert verified
a) 1825, b) 1844

Step by step solution

01

Determine the formula for sample size estimation

The formula to estimate the sample size required for a proportion is given by \[ n = \frac{{Z^2 \times p \times (1-p)}}{E^2} \] where \( Z \) is the z-value corresponding to the confidence level, \( p \) is the estimated proportion, and \( E \) is the margin of error.
02

Identify the z-value for 99% confidence level

For a 99% confidence level, the z-value (\( Z \)) is approximately 2.576.
03

Calculate the sample size using the prior estimate (Part a)

Given \( p = 0.44 \) and \( E = 0.03 \), substitute the values into the formula: \[ n = \frac{{2.576^2 \times 0.44 \times (1-0.44)}}{0.03^2} \] Calculating this, \[ n \approx \frac{{6.635 \times 0.44 \times 0.56}}{0.0009} \approx \frac{{1.642}}{0.0009} \approx 1824.44 \] The required sample size is 1825 (rounding up to the next whole number).
04

Calculate the sample size without any prior estimates (Part b)

When no prior estimates are available, use \( p = 0.5 \) for maximum variability. Substitute the values into the formula: \[ n = \frac{{2.576^2 \times 0.5 \times (1-0.5)}}{0.03^2} \] Calculating this, \[ n \approx \frac{{6.635 \times 0.25}}{0.0009} \approx \frac{{1.65875}}{0.0009} \approx 1843.05 \] The required sample size is 1844 (rounding up to the next whole number).

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Key Concepts

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

Confidence Level
The confidence level helps us understand how sure we can be about our estimate. Imagine it like a safety net for your guess. In this exercise, the researcher wants to be 99% confident. This means there鈥檚 only a 1% chance the true proportion does not fall within our calculated interval. Common confidence levels are 90%, 95%, and 99%. The higher the confidence level, the more certain we want to be but this often requires a larger sample size. For a 99% confidence level, we use a z-value (a special number from a statistical table) of 2.576.
Margin of Error
The margin of error tells us how accurate our estimate is, or the range within which we believe the true proportion lies. If our margin of error is small, say 3% (which it is in this exercise), our estimate is quite precise. If the margin of error is larger, our estimate is less precise. The formula we use in sample size calculations helps ensure our estimate is within this margin. This means our final answer has to fall within 3 percentage points of the true population proportion.
Proportion Estimation
Proportion estimation is figuring out what part of a whole supports a particular characteristic. In this case, it鈥檚 the proportion of adults with high-speed Internet access. In our calculations, if we have a prior estimate (from earlier polls or studies) we include it. For instance, the researcher had a 2004 estimate of 44%, which helps make the calculation more accurate. If no estimates are available, we use 50%, assuming maximum uncertainty to cover all possibilities.
Z-Value
The z-value is a statistic that tells us how many standard deviations an element is from the mean in a standard normal distribution. It鈥檚 like finding out how far a student鈥檚 score is from the average in a class. For our 99% confidence level, the z-value is 2.576. This means we鈥檙e nearly 3 standard deviations away from the mean, which supports the high confidence level. The z-value changes with different confidence levels (e.g., for 95%, it鈥檚 about 1.96). Every time we set a confidence level, we look up the corresponding z-value.
Prior Estimates in Sampling
When conducting research, prior estimates can greatly influence our calculations. These are previous data points or survey results. They help refine our sample size determination. If we know, for example, that 44% of adults had high-speed Internet in 2004, we use this information (p = 0.44) to make a more educated estimate in our sample size formula. When no prior estimates are available, we use 0.5 for p. This maximizes our sample size because it assumes maximum variability, ensuring we have enough data to cover all variations. This safe approach ensures our estimate is valid even without prior data.

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Most popular questions from this chapter

A Gallup poll conducted January \(23,2003-\) February \(10,2003,\) asked American teens (aged 13 to 17 ) how much time they spent each week using the Internet. How many subjects are needed to estimate the time American teens spend on the Internet each week within 0.5 hour with \(95 \%\) confidence? Initial survey results indicate that \(\sigma=6.6\) hours.

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?

IQ scores as measured by the Stanford-Binet IQ test are normally distributed with \(\mu=100\) and \(\boldsymbol{\sigma}=16\) (a) Simulate obtaining 20 samples of size \(n=15\) from this population. (b) Construct \(95 \%\) confidence intervals for each of the 20 samples. (c) How many of the intervals do you expect to include the population mean? How many actually contain the population mean?

Home Ownership An urban economist wishes to estimate the proportion of Americans who own their house. What size sample should be obtained if he wishes the estimate to be within 0.02 with 90\% confidence if (a) he uses an estimate of 0.675 from the fourth quarter of 2000 obtained from the U.S. Census Bureau? (b) he does not use any prior estimates?

A researcher wishes to estimate the mean number of miles on 4 -year-old Saturn SCIs. (a) How many cars should be in a sample to estimate the mean number of miles within 1000 miles with \(90 \%\) confidence, assuming that \(\sigma=19,700 ?\) (b) How many cars should be in a sample to estimate the mean number of miles within 500 miles with \(90 \%\) confidence, assuming that \(\sigma=19,700 ?\) (c) What effect does doubling the required accuracy have on the sample size? Why is this the expected result?
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