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It is believed that \(25 \%\) of U.S. homes have a direct satellite television receiver. How large a sample is necessary to estimate the true population of homes that do with \(95 \%\) confidence and within 3 percentage points? How large a sample is necessary if nothing is known about the proportion?

Short Answer

Expert verified
With a known proportion, the sample size is 801. When the proportion is unknown, the sample size is 1068.

Step by step solution

01

Understanding the problem

We need to determine the sample size required to estimate a proportion with a specified level of confidence and margin of error. We are given a known estimated proportion and are also asked to consider a situation where the proportion is unknown.
02

Define the known values

From the problem, the estimated proportion \( p = 0.25 \), and we need a confidence level of \(95\%\) with a margin of error of \(3\%\) or \(0.03\). The Z-score for a \(95\%\) confidence level is approximately \(1.96\).
03

Use the proportion formula to find the sample size

The formula for sample size with a known proportion is: \[ n = \left(\frac{Z^2 \cdot p \cdot (1 - p)}{E^2}\right) \]Here, \( Z = 1.96 \), \( p = 0.25 \), and \( E = 0.03 \).Substitute the values to get:\[ n = \frac{(1.96)^2 \cdot 0.25 \cdot 0.75}{(0.03)^2} \]
04

Calculate the sample size

Calculate \[ n = \frac{3.8416 \cdot 0.25 \cdot 0.75}{0.0009} \approx \frac{0.7206}{0.0009} \approx 800.67 \] Round up to the next whole number, so \( n = 801 \).
05

When the proportion is unknown

If the proportion is unknown, we use \( p = 0.5 \) for maximizing sample size:\[ n = \frac{(1.96)^2 \cdot 0.5 \cdot 0.5}{(0.03)^2} \]\[ n = \frac{3.8416 \cdot 0.25}{0.0009} = \frac{0.9604}{0.0009} \approx 1067.11 \]Again round up, so \( n = 1068 \).

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Key Concepts

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

Understanding Confidence Levels
A confidence level, often expressed as a percentage, tells us how confident we can be about our statistical estimations. In simple terms, if you have a 95% confidence level, it indicates that if you repeated a study multiple times (say 100), 95 of those times, the result would include the true population parameter.

Confidence levels typically use a Z-score, which is a statistical measurement denoting how far away a particular point is from the mean (average) in standard deviation units. For common confidence levels, the Z-scores are usually:
  • 90% confidence level: Z-score approximately 1.645
  • 95% confidence level: Z-score approximately 1.96
  • 99% confidence level: Z-score approximately 2.576
For our exercise, a confidence level of 95% was used, meaning we expect our estimation to reflect the true proportion 95 times out of 100 trials.
Importance of Margin of Error
Margin of error signifies the amount of random sampling error in a survey's results and is a critical component of statistical analysis when estimating proportions. It's usually expressed as a percentage and gives a buffer range where the true population value may lie.

For instance, in our exercise, the margin of error was set at 3%, which means that the proportion could vary by this amount either way. Smaller margins of error require larger sample sizes, as a higher precision (meaning smaller error) needs more data to accurately predict the population parameter. By understanding the margin of error, you grasp how much 'wiggle room' there is around the estimate, helping in making educated decisions based on data findings.
Proportion Estimation Techniques
Proportion estimation involves determining the percentage of a population that shares a particular characteristic. Often, the formula for estimating population proportions when given a margin of error and confidence level is used actively in research work.

The challenge is greater when nothing is known about the likely proportion. Here, the most conservative approach is to use an estimated proportion of 50% (p = 0.5). This choice maximizes sample size, ensuring that regardless of the true population proportion, the sample size will be adequate to provide a meaningful estimate within the margin of error.

Therefore, the sample size for unknown proportions is generally larger as seen in our exercise, where using 0.5 as the proportion increased the needed sample size to 1068 compared to 801 when the proportion was known.
Decoding Z-Scores
Z-score is an important statistical tool that helps in the standardization of data, enabling predictions and comparisons across different datasets. A Z-score tells us how many standard deviations an element is from the mean.

In the context of our problem, it helps ascertain how confident our estimate is. Specifically, the Z-score of 1.96 for a 95% confidence level shows that our range is adequate to capture the true proportion about 95 times out of 100. This means that the sample mean (or the estimate) lies within 1.96 standard deviations from the true mean. Calculating sample size involves squaring this Z-score to emphasize how the spread of data influences confidence in estimates. Z-scores streamline this process, providing a straightforward way to link probability distributions with sample sizes and errors.

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Most popular questions from this chapter

A random sample of 205 college students was asked if they believed that places could be haunted, and 65 responded yes. Estimate the true proportion of college students who believe in the possibility of haunted places with \(99 \%\) confidence. According to Time magazine, \(37 \%\) of all Americans believe that places can be haunted.

In 2014 six percent of the cars sold had a manual transmission. A random sample of college students who owned cars revealed the following: out of 122 cars, 26 had manual transmissions. Estimate the proportion of college students who drive cars with manual transmissions with \(90 \%\) confidence.

What assumption must be made when computing confidence intervals for variances and standard deviations?

The number of carbohydrates (in grams) per 8 -ounce serving of yogurt for each of a random selection of brands is listed below. Estimate the true population variance and standard deviation for the number of carbohydrates per 8 -ounce serving of yogurt with \(95 \%\) confidence. Assume the variable is normally distributed. \(\begin{array}{lllllllll}17 & 42 & 41 & 20 & 39 & 41 & 35 & 15 & 43 \\ 25 & 38 & 33 & 42 & 23 & 17 & 25 & 34 & \end{array}\)

Assume that all variables are approximately normally distributed. A meteorologist who sampled 13 randomly selected thunderstorms found that the average speed at which they traveled across a certain state was 15.0 miles per hour. The standard deviation of the sample was 1.7 miles per hour. Find the \(99 \%\) confidence interval of the mean. If a meteorologist wanted to use the highest speed to predict the times it would take storms to travel across the state in order to issue warnings, what figure would she likely use?

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