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Chapter 8: Problem 13

The following sample data are from a normal population: 10,8,12,15,13,11,6,5 a. What is the point estimate of the population mean? b. What is the point estimate of the population standard deviation? c. With \(95 \%\) confidence, what is the margin of error for the estimation of the population mean? d. What is the \(95 \%\) confidence interval for the population mean?

Short Answer

Expert verified
a) 10, b) \(\approx 3.464\), c) \(\approx 2.399\), d) \((7.601, 12.399)\).

Step by step solution

01

Calculate the Point Estimate of the Population Mean

The point estimate of the population mean is the sample mean. We calculate the sample mean (\( \bar{x} \)) by summing up all the values and dividing by the number of observations. \[ \bar{x} = \frac{10+8+12+15+13+11+6+5}{8} = \frac{80}{8} = 10 \]
02

Calculate the Point Estimate of the Population Standard Deviation

Calculate the sample standard deviation to estimate the population standard deviation. First, find the deviations from the mean, square them, sum them up, divide by \( n-1 \) (because it's a sample), and then take the square root. Deviations from mean: \((10-10), (8-10), (12-10), (15-10), (13-10), (11-10), (6-10), (5-10)\) \((0, -2, 2, 5, 3, 1, -4, -5)\) Squares: \(0^2, (-2)^2, 2^2, 5^2, 3^2, 1^2, (-4)^2, (-5)^2\)Squares: \(0, 4, 4, 25, 9, 1, 16, 25\)Sum of squares: \(0+4+4+25+9+1+16+25 = 84\)Sample variance: \(\frac{84}{7} = 12\)Sample standard deviation: \(\sqrt{12} \approx 3.464\)
03

Calculate the Margin of Error for Estimation of the Population Mean

Use the formula for margin of error with a 95% confidence interval: \( ME = z \times \frac{s}{\sqrt{n}} \). Here, \(z\) is the z-score for 95% confidence which is 1.96, \(s\) is the standard deviation, and \(n\) is the number of observations.\[ ME = 1.96 \times \frac{3.464}{\sqrt{8}} \approx 1.96 \times 1.224 \approx 2.399 \]
04

Calculate the 95% Confidence Interval for the Population Mean

We use the margin of error to calculate the confidence interval:\( CI = \bar{x} \pm ME = 10 \pm 2.399\)Lower limit = \(10 - 2.399 = 7.601\)Upper limit = \(10 + 2.399 = 12.399\)Thus, the 95% confidence interval is \((7.601, 12.399)\).

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Key Concepts

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

Normal Distribution
The **normal distribution** is a fundamental concept in statistics that describes how data is spread out. It's often called a bell curve due to its shape. This curve is symmetric about the mean, with the majority of data points falling close to the mean.
  • Mean: The central point, where the highest peak of the curve is.
  • Standard deviation: Measures the spread of the data. A smaller deviation means the data is tightly packed around the mean.
  • Properties: Symmetric, bell-shaped, with mean, median, and mode all equal.
In the context of the exercise, assuming a normal distribution allows us to use confidence intervals effectively to estimate the population parameters with the given data. Understanding normal distribution helps us grasp the probability of occurrences within certain ranges in our data.
Confidence Interval
A **confidence interval** is a range of values that provides us with an estimate of an unknown population parameter. In the exercise, we're interested in the population mean. The interval is constructed so that we can be fairly sure the true population mean lies within this range. Confidence intervals have a confidence level, typically expressed as a percentage like 95%. This level tells us how confident we are that the interval contains the true parameter. So, a 95% confidence interval means there's a 95% chance that the interval has captured the population mean.
  • The size of the confidence interval depends on the sample size, variability, and desired confidence level.
  • Wider intervals indicate less precision but more certainty to capture the mean.
The process involves calculating the margin of error and using it to define the interval around the sample mean.
Margin of Error
The **margin of error** represents how much we expect our sample estimate to vary from the actual population parameter. It's crucial when constructing confidence intervals. In the exercise, the margin of error gives us a leeway around our calculated sample mean.To calculate the margin of error (ME), we leverage the sample's standard deviation and a z-score corresponding to our confidence level. For a 95% confidence level, we use a z-score of 1.96.
  • Calculation: \( ME = z \times \frac{s}{\sqrt{n}} \)
  • Implications: A smaller margin represents more precise estimates, larger suggests greater uncertainty.
The margin of error helps us determine the width of our confidence interval, impacting our overall confidence in the resulting estimate.
Population Mean
The **population mean** is a key concept in statistics, representing the average value of a given population. Since this is often unknown, we use the sample mean as a point estimate. The formula for the sample mean is simple:\[ \bar{x} = \frac{\text{sum of all sample values}}{\text{number of sample values}} \]
  • The sample mean is used as it provides a close estimate to the population mean, assuming the sample is random and unbiased.
  • Under a normal distribution, the sample mean is a very efficient and unbiased estimator of the population mean.
From the original exercise, the calculated sample mean is 10, offering a point estimate of what the actual population mean might be.
Sample Standard Deviation
The **sample standard deviation** measures the amount of variation or dispersion in a set of sample values. It's used to estimate the population standard deviation and is a critical step in statistics when assessing data reliability.The calculation involves:
  • Finding the deviation of each sample value from the mean.
  • Squaring these deviations.
  • Calculating the average of these squared deviations. (Here we divide by \(n-1\) to account for a sample rather than a whole population.)
  • Taking the square root of that average to return to the original unit of measure.
Being aware of variability through standard deviation is vital because it directly affects the margin of error and confidence intervals. In the exercise, the sample standard deviation is approximately 3.464, providing a measure of how much the sample data deviates from the mean value.

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Most popular questions from this chapter

The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .30 a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of \(.02 ?\) Use \(95 \%\) confidence. b. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population? c. What is the \(95 \%\) confidence interval for the proportion of smokers in the population?

A simple random sample of 40 items resulted in a sample mean of \(25 .\) The population standard deviation is \(\sigma=5\) a. What is the standard error of the mean, \(\sigma_{z} ?\) b. \(\quad\) At \(95 \%\) confidence, what is the margin of error?

A USA Today/CNN/Gallup poll for the presidential campaign sampled 491 potential voters in June (USA Today, June 9, 2000). A primary purpose of the poll was to obtain an estimate of the proportion of potential voters who favor each candidate. Assume a planning value of \(p^{*}=.50\) and a \(95 \%\) confidence level. a. For \(p^{*}=.50,\) what was the planned margin of error for the June poll? b. Closer to the November election, better precision and smaller margins of error are desired. Assume the following margins of error are requested for surveys to be conducted during the presidential campaign. Compute the recommended sample size for each survey

The motion picture Harry Potter and the Sorcerer's Stone shattered the box office debut record previously held by The Lost World: Jurassic Park (The Wall Street Journal, November 19,2001 ). A sample of 100 movie theaters showed that the mean three-day weekend gross was \(\$ 25,467\) per theater. The sample standard deviation was \(\$ 4980\) a. What is the margin of error for this study? Use \(95 \%\) confidence. b. What is the \(95 \%\) confidence interval estimate for the population mean weekend gross per theater? c. The Lost World took in \(\$ 72.1\) million in its first three-day weekend. Harry Potter and the Sorcerer's Stone was shown in 3672 theaters. What is an estimate of the total Harry Potter and the Sorcerer's Stone took in during its first three-day weekend? d. An Associated Press article claimed Harry Potter "shattered" the box office debut record held by The Lost World. Do your results agree with this claim?

The National Quality Research Center at the University of Michigan provides a quarterly measure of consumer opinions about products and services (The Wall Street Journal, February 18,2003 ). A survey of 10 restaurants in the Fast Food/Pizza group showed a sample mean customer satisfaction index of \(71 .\) Past data indicate that the population standard deviation of the index has been relatively stable with \(\sigma=5\) a. What assumption should the researcher be willing to make if a margin of error is desired? b. Using \(95 \%\) confidence, what is the margin of error? c. What is the margin of error if \(99 \%\) confidence is desired?
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