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A researcher wishes to estimate the proportion of households that have broadband 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 2009 estimate of 0.635 obtained from the National Telecommunications and Information Administration? (b) she does not use any prior estimates?

Short Answer

Expert verified
(a) 1004 (b) 1843

Step by step solution

01

Understand the Problem

The researcher wants to estimate the proportion of households with broadband Internet access, aiming for a margin of error of 0.03 with 99% confidence.
02

Identify the Formula for Sample Size

The formula to determine the sample size for estimating a proportion is: where n is the sample size, Z is the Z-score corresponding to the desired confidence level, p is the estimated proportion, and (1-p) is the complement of the proportion, and E is the margin of error.
03

Gather Z-score for 99% Confidence Level

For a 99% confidence interval, the Z-score is approximately 2.576.
04

Calculate Sample Size for Part (a)

Using the 2009 estimate of 0.635: \[ n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right) \] Substituting the values: \[ n = \left( \frac{(2.576)^2 \cdot 0.635 \cdot (1 - 0.635)}{(0.03)^2} \right) \] \[ n = \left( \frac{6.635 \cdot 0.635 \cdot 0.365}{0.0009} \right) \] Solving this, we get n
05

Calculate Sample Size for Part (b)

If no prior estimate is available, assume p = 0.5 for maximum variability: \[ n = \left( \frac{Z^2 \cdot 0.5 \cdot (1 - 0.5)}{E^2} \right) \] Substituting the values: \[ n = \left( \frac{(2.576)^2 \cdot 0.25}{(0.03)^2} \right) \] \[ n = \left( \frac{6.635 \cdot 0.25}{0.0009} \right) \] Solving this, we get n .

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Key Concepts

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

Proportion Estimates
Proportion estimates play a crucial role in statistics, especially when you want to make decisions or predictions about a large population based on a sample. In this context, the researcher aims to estimate the proportion of households with broadband Internet access. This is represented as a percentage or a proportion (p). If we can get a reliable estimate of this proportion, we can draw meaningful conclusions about the broader population.
In the exercise, the researcher uses a previous estimate from 2009, which is 0.635 (or 63.5%). This initial estimate helps in calculating the sample size required for accurate future studies. If there’s no prior estimate, a common practice is to use 0.5, representing the maximum variability, because this ensures that our sample size calculation is conservative and adequate.
Margin of Error
The margin of error (E) quantifies the range within which we expect the true population parameter to lie, given our sample estimate. In simpler terms, it's a measure of the uncertainty associated with the sample estimate. In the given exercise, the researcher wants a margin of error of 0.03. This means that the estimate should be accurate within 3 percentage points.
The formula for margin of error in proportion estimates is linked directly with the desired confidence level and the variability in the population. It ensures that our estimate isn’t too far off from the actual population proportion. A smaller margin of error requires a larger sample size, as seen in the calculation steps.
Confidence Intervals
Confidence intervals provide a range of values, derived from the sample data, within which we expect the true population parameter to fall. For instance, a 99% confidence interval means we are 99% sure that the true proportion lies within this range.
In our case, the researcher opts for a 99% confidence interval, which requires a specific Z-score in the calculations. A wider confidence interval requires a larger sample size but provides more certainty. It's a trade-off: higher confidence usually means the interval will be wider, or you will need a bigger sample to maintain the same margin of error.
Z-score
The Z-score in the context of confidence intervals represents how many standard deviations away from the mean a data point is. For our exercise, since the researcher wants a 99% confidence level, the corresponding Z-score is approximately 2.576.
This Z-score is crucial in determining the sample size because it affects how wide or narrow our confidence interval will be. Using the formula:ewline ewline ewline ewline ewlineewlineewline ewlineewlineewlineewlineewlineewline ewlineewline ewline ewlineewline ewlineewlineewlineewlineewlineewlineewline ewline ewline ewlineewlineewlineewline ewline ewline indicates that the Z-score was used alongside other variables like the proportion estimate (p) and the margin of error (E) to calculate the final required sample size. The higher the Z-score, the more confident you are about the interval, but it also means you'll require more data (a larger sample size).

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Most popular questions from this chapter

Blood Alcohol Concentration A random sample of 51 fatal crashes in 2013 in which the driver had a positive blood alcohol concentration (BAC) from the National Highway Traffic Safety Administration results in a mean BAC of 0.167 gram per deciliter \((\mathrm{g} / \mathrm{dL})\) with a standard deviation of \(0.010 \mathrm{~g} / \mathrm{dL}\) (a) A histogram of blood alcohol concentrations in fatal accidents shows that BACs are highly skewed right. Explain why a large sample size is needed to construct a confidence interval for the mean BAC of fatal crashes with a positive \(\mathrm{BAC}\) (b) In \(2013,\) there were approximately 25,000 fatal crashes in which the driver had a positive BAC. Explain why this, along with the fact that the data were obtained using a simple random sample, satisfies the requirements for constructing a confidence interval. (c) Determine and interpret a \(90 \%\) confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC. (d) All 50 states and the District of Columbia use a BAC of \(0.08 \mathrm{~g} / \mathrm{dL}\) as the legal intoxication level. Is it possible that the mean BAC of all drivers involved in fatal accidents who are found to have positive BAC values is less than the legal intoxication level? Explain.

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. $$ n=9 ; \text { Correlation }=0.997 $$

A simple random sample of size \(n\) is drawn from a population that is known to be normally distributed. The sample variance, \(s^{2},\) is determined to be 19.8 (a) Construct a \(95 \%\) confidence interval for \(\sigma^{2}\) if the sample size, \(n,\) is 10 (b) Construct a \(95 \%\) confidence interval for \(\sigma^{2}\) if the sample size, \(n\), is \(25 .\) How does increasing the sample size affect the width of the interval? (c) Construct a \(99 \%\) confidence interval for \(\sigma^{2}\) if the sample size, \(n\), is \(10 .\) Compare the results with those obtained in part (a). How does increasing the level of confidence affect the width of the confidence interval?

The following small data set represents a simple random sample from a population whose mean is \(50 .\) $$ \begin{array}{llllll} \hline 43 & 63 & 53 & 50 & 58 & 44 \\ \hline 53 & 53 & 52 & 41 & 50 & 43 \\ \hline \end{array} $$ (a) A normal probability plot indicates that the data could come from a population that is normally distributed with no outliers. Compute a \(95 \%\) confidence interval for this data set. (b) Suppose that the observation, 41 , is inadvertently entered into the computer as 14 . Verify that this observation is an outlier (c) Construct a \(95 \%\) confidence interval on the data set with the outlier. What effect does the outlier have on the confidence interval? (d) Consider the following data set, which represents a simple random sample of size 36 from a population whose mean is 50\. Verify that the sample mean for the large data set is the same as the sample mean for the small data set from part (a). $$ \begin{array}{llllll} \hline 43 & 63 & 53 & 50 & 58 & 44 \\ \hline 53 & 53 & 52 & 41 & 50 & 43 \\ \hline 47 & 65 & 56 & 58 & 41 & 52 \\ \hline 49 & 56 & 57 & 50 & 38 & 42 \\ \hline 59 & 54 & 57 & 41 & 63 & 37 \\ \hline 46 & 54 & 42 & 48 & 53 & 41 \\ \hline \end{array} $$ (e) Compute a \(95 \%\) confidence interval for the large data set. Compare the results to part (a). What effect does increasing the sample size have on the confidence interval? (f) Suppose that the last observation, 41 , is inadvertently entered as 14 . Verify that this observation is an outlier. (g) Compute a \(95 \%\) confidence interval for the large data set with the outlier. Compare the results to part (e). What effect does an outlier have on a confidence interval when the data set is large?

Construct a confidence interval of the population proportion at the given level of confidence. \(x=120, n=500,99 \%\) confidence

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