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Chapter 8: Problem 13

The standard deviation for a population is \(\sigma=7.14\). A random sample selected from this population gave a mean equal to \(48.52\). a. Make a \(95 \%\) confidence interval for \(\mu\) assuming \(n=196\). b. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=100\). c. Determine a \(95 \%\) confidence interval for \(\mu\) assuming \(n=49\). d. Does the width of the confidence intervals constructed in parts a through c increase as the sample size decreases? Explain.

Short Answer

Expert verified
a. The 95% confidence interval for a sample size of 196 is [47.96, 49.08].\nb. The 95% confidence interval for a sample size of 100 is [47.76, 49.28].\nc. The 95% confidence interval for a sample size of 49 is [47.44, 49.60].\nd. Yes, the width of the confidence intervals increases as the sample size decreases, suggesting that the precision of the estimate decreases with smaller sample sizes.

Step by step solution

01

Find confidence intervals for different sample sizes

Plug the given mean (\(48.52\)), standard deviation (\(7.14\)), and sample sizes (\(n=196\), \(n=100\), and \(n=49\)) into the formula. This gives the following results:\na. For \(n=196\), the confidence interval is [\(48.52 - 1.96 * 7.14/\sqrt{196}, 48.52 + 1.96 * 7.14/\sqrt{196}\)].\nb. For \(n=100\), the confidence interval is [\(48.52 - 1.96 * 7.14/\sqrt{100}, 48.52 + 1.96 * 7.14/\sqrt{100}\)].\nc. For \(n=49\), the confidence interval is [\(48.52 - 1.96 * 7.14/\sqrt{49}, 48.52 + 1.96 * 7.14/\sqrt{49}\)].
02

Compute the confidence intervals

Perform the mathematical calculations to find the confidence intervals for the different sample sizes. The results are:\na. For \(n=196\), the confidence interval is [\(47.96, 49.08\)].\nb. For \(n=100\), the confidence interval is [\(47.76, 49.28\)].\nc. For \(n=49\), the confidence interval is [\(47.44, 49.60\)].
03

Analyze the widths of the confidence intervals

Observe the widths of the confidence intervals for the different sample sizes. The width of a confidence interval is the difference between its upper and lower limits. We see that as the sample size decreases, the width of the confidence interval increases. This suggests that the precision of the estimate decreases as the sample size decreases.

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Key Concepts

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

Standard Deviation
Standard deviation is a vital statistical concept that measures how spread out numbers in a data set are. Imagine you took a test alongside your classmates. If most of you scored pretty close to the class average, the standard deviation would be small. But if scores were all over the place, from very low to very high, the standard deviation would be larger. Here's why it matters:
  • It helps understand the variability of your data.
  • A lower standard deviation means data points are close to the mean, while a higher one indicates more spread.
For this exercise, the population standard deviation is given as 7.14. This means that individual measurements fall within this range away from the average.
Understanding this helps us get a sense of how "typical" or "atypical" a data point might be.
Sample Size
Sample size plays a crucial role in statistics. It refers to the number of observations or data points you have collected. With the confidence interval problem at hand, the sample sizes are varied as 196, 100, and 49. Why does it matter?
  • A larger sample size reduces the margin of error in calculations.
  • Bigger samples tend to provide more reliable and accurate estimates of the population mean.
  • Smaller samples might lead to wider confidence intervals, meaning less precision.
In this exercise, notice how the confidence intervals change with different sample sizes. A larger sample (n = 196) gives a tighter interval, [47.96, 49.08], compared to a smaller sample (n = 49) with a wider interval, [47.44, 49.60].
This contrast shows how much your results can vary, hinging on how much data you gather.
Population Mean
The population mean is a central concept in statistics, representing the average of a set of data points for an entire group. In our case, the sample mean is 48.52. What's the difference between a sample mean and a population mean?
  • Sample mean is the average of a subset drawn from the population.
  • Population mean is the hypothetical average calculated if you had all data points from the population.
The sample mean serves as an estimate of the population mean. That's because, practically, it's often impossible to measure every single individual in the population.
Calculations with the sample mean help us draft a confidence interval, giving us a range where we believe the true population mean lies, with a certain level of confidence (like 95%). It's like saying, "We're 95% sure the actual average is somewhere around here."
Statistical Precision
Statistical precision refers to how much confidence we can have in our results. It's about how "tight" or "narrow" our estimates are. When we see terms like `confidence interval`, this precision is being evaluated.
For this exercise, what we've learned is:
  • The width of a confidence interval indicates precision. The narrower, the more precise.
  • Larger sample sizes generally lead to more precise estimates, evidenced by narrower confidence intervals.
  • Precision decreases as sample size decreases, making results less reliable.
The shift in confidence intervals in the provided solutions, depending on the sample size, reflects changes in statistical precision.
Practically, when precision is high, decisions based on this data are more likely to be accurate. This understanding helps in planning experiments or surveys effectively, ensuring that conclusions are strong and dependable.

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Most popular questions from this chapter

For a population, the value of the standard deviation is \(2.65\). A random sample of 35 observations taken from this population produced the following data. \(\begin{array}{lllllll}42 & 51 & 42 & 31 & 28 & 36 & 49 \\ 29 & 46 & 37 & 32 & 27 & 33 & 41 \\ 47 & 41 & 28 & 46 & 34 & 39 & 48 \\ 26 & 35 & 37 & 38 & 46 & 48 & 39 \\ 29 & 31 & 44 & 41 & 37 & 38 & 46\end{array}\) a. What is the point estimate of \(\mu\) ? b. Make a \(98 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for part b?

What assumption(s) must hold true to use the normal distribution to make a confidence interval for the population proportion, \(p\) ?

In a January 2014 survey conducted by the Associated PressWe TV, \(68 \%\) of American adults said that owning a home is the most important thing or \(a\) very important but not the most important thing (opportunityagenda.org). Assume that this survey was based on a random sample of 900 American adults. a. Construct a \(95 \%\) confidence interval for the proportion of all American adults who will say that owning a home is the most important thing or a very important but not the most important thing. b. Explain why we need to construct a confidence interval. Why can we not simply say that \(68 \%\) of all American adults would say that owning a home is the most important thing or \(a\) very important but not the most important thing?

activities (playing games, personal communications, etc.) during this month are as follows: $$ \begin{array}{lllllllll} 7 & 12 & 9 & 8 & 11 & 4 & 14 & 1 & 6 \end{array} $$ Assuming that such times for all employees are approximately normally distributed, make a \(95 \%\) confidence interval for the corresponding population mean for all employees of this company.A company randomly selected nine office employees and secretly monitored their computers for one month. The times (in hours) spent by these employees using their computers for non- job-related

When one is attempting to determine the required sample size for estimating a population mean, and the information on the population standard deviation is not available, it may be feasible to take a small preliminary sample and use the sample standard deviation to estimate the required sample size, \(n .\) Suppose that we want to estimate \(\mu\), the mean commuting distance for students at a community college, to a margin of error within 1 mile with a confidence level of \(95 \%\). A random sample of 20 students yields a standard deviation of \(4.1\) miles. Use this value of the sample standard deviation, \(s\), to estimate the required sample size, \(n .\) Assume that the corresponding population has an approximate normal distribution.
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