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Suppose you wish to estimate a population mean based on a random sample of \(n\) observations, and prior experience suggests that \(\sigma=12.7\). If you wish to estimate \(\mu\) correct to within 1.6 , with probability equal to .95, how many observations should be included in your sample?

Short Answer

Expert verified
Answer: You should include 245 observations in your sample.

Step by step solution

01

Identify the Confidence Interval formula for the sample mean

For a population with known standard deviation (\(\sigma\)), the formula to calculate the Confidence Interval for the sample mean (\(\bar{x}\)) is: $$\bar{x} \, \pm \, Z_{\frac{\alpha}{2}} \cdot \frac{\sigma}{\sqrt{n}}$$ Where: - \(\bar{x}\): Sample mean - \(Z_{\frac{\alpha}{2}}\): Z-score corresponding to a given confidence level - \(n\): Sample size - \(\sigma\): Population standard deviation In this case, we don't need the entire formula; only the margin of error part: $$ E = Z_{\frac{\alpha}{2}} \cdot \frac{\sigma}{\sqrt{n}}$$ We are given the margin of error \(E = 1.6\), and we need to find the sample size (\(n\)) for a 95% confidence level.
02

Find the Z-score for the given Confidence Level

For a given confidence level, we can find the corresponding Z-score from the standard normal distribution table or using a calculator/online tool. For a 95% confidence level, the Z-score (\(Z_{\frac{\alpha}{2}}\)) is approximately 1.96. (We can verify this using any Z-table, calculator, or online tool.)
03

Solve for the sample size \(n\)

Now, we have: $$ E = Z_{\frac{\alpha}{2}} \cdot \frac{\sigma}{\sqrt{n}}$$ $$ 1.6 = 1.96 \cdot \frac{12.7}{\sqrt{n}}$$ Our objective is to find \(n\). First, let's isolate the \(\sqrt{n}\) term: $$ \sqrt{n} = \frac{1.96 \cdot 12.7}{1.6}$$ $$ \sqrt{n} \approx 15.6277 $$ Now, square both sides of the equation to find \(n\): $$ n \approx 15.6277^2 $$ $$ n \approx 244.06 $$
04

Round up to the nearest whole number

Since we can't have a fraction of an observation, we round up the calculated sample size to the nearest whole number. Thus, the sample size needed to estimate the population mean with a margin of error of 1.6 and 95% confidence level is: $$ n \approx 245$$ Answer: You should include 245 observations in your sample.

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Key Concepts

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

Confidence Interval
The confidence interval is a range of values that is used to estimate the true value of a population parameter, like the mean. It provides an interval within which we expect the population parameter to lie, given a certain level of confidence. For example, a 95% confidence interval implies that we are 95% confident that the interval contains the true population mean.

The formula for constructing a confidence interval for the mean when the population standard deviation is known involves the sample mean (\(\bar{x}\)) and the Z-score associated with the desired confidence level:
  • \(\bar{x} \pm Z_{\frac{\alpha}{2}} \cdot \frac{\sigma}{\sqrt{n}}\)
Here, \(Z_{\frac{\alpha}{2}}\) is the Z-score for the confidence level, \(\sigma\) is the known population standard deviation, and \(n\) is the sample size.

Confidence intervals are a crucial tool in statistics because they provide a range of plausible values for unknown parameters, giving researchers a way to quantify the uncertainty inherent in samples.
Margin of Error
The margin of error represents the amount by which the sample estimate, like the sample mean, is expected to vary from the true population mean. It gives us an idea of how much error we can expect in our estimate due to the inherent variability in sampling.

In the context of confidence intervals, the margin of error is expressed as:
  • \(E = Z_{\frac{\alpha}{2}} \cdot \frac{\sigma}{\sqrt{n}}\)
This equation shows that the margin of error depends on the Z-score for the desired confidence level and the standard deviation of the population, divided by the square root of the sample size.

A smaller margin of error indicates a more precise estimate of the population mean, which can be achieved by increasing the sample size or by selecting a lower confidence level. However, the trade-off between confidence level and margin of error should be considered carefully in any research study.
Z-score
The Z-score is a statistical measure that helps us determine how far a point is from the mean of a standard normal distribution. In the context of confidence intervals, the Z-score corresponds to the desired confidence level. It helps to determine the critical value, which defines the boundaries of the confidence interval.

For example, when a 95% confidence interval is required, the Z-score used is approximately 1.96. This value comes from the properties of the normal distribution, where approximately 95% of the data falls within 1.96 standard deviations from the mean.

To find a Z-score for a given confidence level:
  • Use a standard normal distribution table.
  • Utilize statistical calculators or online tools.
Z-scores are fundamental components in hypothesis testing and confidence interval estimation, providing a bridge between sample data and broader population inferences.
Population Mean Estimation
Estimating the population mean involves using sample data to infer about the mean of the entire population. This is particularly useful when it's impractical or impossible to survey the whole population.

The process entails collecting a random sample, calculating its mean (\(\bar{x}\)), and then using statistical techniques to infer the population mean (\(\mu\)). In practical terms, this often involves constructing a confidence interval based on sample statistics.

Key steps in population mean estimation include:
  • Determine the confidence level desired.
  • Compute the sample mean.
  • Use the sample data and confidence level to calculate the margin of error.
  • Construct the confidence interval around the sample mean.
These steps help statisticians and researchers understand how likely it is that the true mean falls within a given range, given the statistical evidence presented by the sample. This understanding guides decisions and predictions based on the estimated population parameters.

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Most popular questions from this chapter

Multimedia Kids Do our children spend as much time enjoying the outdoors and playing with family and friends as previous generations did? Or are our children spending more and more time glued to the television, computer, and other multimedia equipment? A random sample of 250 children between the ages of 8 and 18 showed that 170 children had a TV in their bedroom and that 120 of them had a video game player in their bedroom. a. Estimate the proportion of all 8 - to 18 -year-olds who have a TV in their bedroom, and calculate the margin of error for your estimate. b. Estimate the proportion of all 8 - to 18 -year-olds who have a video game player in their bedroom, and calculate the margin of error for your estimate.

Exercise 8.19 discussed a research poll conducted for \(A B C\) News and the Washington Post that included questions about illegal immigration into the United States, and the federal and state responses to the problem. \({ }^{3}\) Suppose that you were designing a poll of this type. a. Explain how you would select your sample. What problems might you encounter in this process? b. If you wanted to estimate the percentage of the population who agree with a particular statement in your survey questionnaire correct to within \(1 \%\) with probability .95, approximately how many people would have to be polled?

Even though we know it may not be good for us, many Americans really enjoy their fast food! A survey conducted by Pew Research Center \(^{19}\) graphically illustrated our penchant for eating out, and in particular, eating fast food: a. This survey was based on "telephone interviews conducted with a nationally representative sample of 2250 adults, ages 18 years and older, living in continental U.S. telephone households." What problems might arise with this type of sampling? b. How accurate do you expect the percentages given in the survey to be in estimating the actual population percentages? (HINT: Find the margin of error. c. If you want to decrease your margin of error to be \(\pm 1 \%,\) how large a sample should you take?

Baseball Fans The first day of baseball comes in late March, ending in October with the World Series. Does fan support grow as the season goes on? Two CNN/USA Today/Gallup polls, one conducted in March and one in November, both involved random samples of 1001 adults aged 18 and older. In the March sample, \(45 \%\) of the adults claimed to be fans of professional baseball, while \(51 \%\) of the adults in the November sample claimed to be fans. \({ }^{14}\) a. Construct a \(99 \%\) confidence interval for the difference in the proportion of adults who claim to be fans in March versus November. b. Does the data indicate that the proportion of adults who claim to be fans increases in November, around the time of the World Series? Explain.

Do well-rounded people get fewer colds? A study on the Chronicle of Higher \(E d u\) cation was conducted by scientists at Carnegie Mellon University, the University of Pittsburgh, and the University of Virginia. They found that people who have only a few social outlets get more colds than those 16 who are involved in a variety of social activities. Suppose that of the 276 healthy men and women tested, \(n_{1}=96\) had only a few social outlets and \(n_{2}=105\) were busy with six or more activities. When these people were exposed to a cold virus, the following results were observed: a. Construct a \(99 \%\) confidence interval for the difference in the two population proportions. b. Does there appear to be a difference in the population proportions for the two groups? c. You might think that coming into contact with more people would lead to more colds, but the data show the opposite effect. How can you explain this unexpected finding?

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