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Chapter 8: Problem 14

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?

Short Answer

Expert verified
a. The point estimate of \(\mu\) is the sample mean. b. The 98% confidence interval for \(\mu\) is given by \(\bar{x} \pm z \times \text{Standard Error}\). c. The margin of error for the estimate is \(z \times \text{Standard Error}\). Replace \(\bar{x}\), \(z\), and Standard Error with the values obtained from the steps above.

Step by step solution

01

Calculate the sample mean

First we will calculate the sample mean, which is the sum of all values divided by the number of values in the set. For the given sample, sum all the observations and divide by 35.
02

Find the point estimate

The point estimate of the population mean \(\mu\) is the sample mean calculated in Step 1.
03

Calculate the standard error of the mean

The standard error is given by \(\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the standard deviation and \(n\) is the number of observations. Use the given \(\sigma\) of 2.65 and \(n\) of 35.
04

Find z-value for 98% confidence

A 98% confidence interval requires a z-value of approximately 2.33. This value is obtained from a z-table or z-calculator which gives the number of standard deviations away from the mean that corresponds to the desired level of confidence.
05

Construct Confidence Interval

A confidence interval for the population mean is given by the formula \(\bar{x} \pm z \times \text{Standard Error}\), where \(\bar{x}\) is the sample mean, \(z\) is the z value corresponding to the level of confidence, and Standard Error is calculated in Step 3.
06

Calculate the Margin of Error

The margin of error is the amount by which the estimate could be off which is given by \(z \times \text{Standard Error}\), the same as in Step 5.

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Key Concepts

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

Point Estimate
When we talk about the point estimate in statistics, we're referring to a single value that provides the best guess for an unknown population parameter. In this case, it's used to estimate the population mean, denoted as \( \mu \). The point estimate is typically the sample mean when dealing with a sample data set.
To find the point estimate:
  • Add all the sample values together.
  • Divide the total by the number of observations.
The result is your sample mean, which serves as the point estimate for the population mean. By focusing on this one number, you get a simple yet effective representation of the entire data set.
Standard Error
The standard error measures the variability or dispersion of the sample mean from the true population mean. It's crucial for building confidence intervals.
The formula for standard error is:\[ SE = \frac{\sigma}{\sqrt{n}} \]where \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
This formula calculates how much the sample mean is expected to vary from the population mean. A smaller standard error indicates that the sample mean is a more precise estimate of the population mean. Understanding the standard error helps in assessing the reliability of the point estimate.
Margin of Error
The margin of error reflects the range within which we expect the true population parameter to lie, based on the sample data. It shows how much the point estimate could be off.
The margin of error is calculated as:\[ \text{Margin of Error} = z \times SE \]where \( z \) is the z-value corresponding to the desired confidence level, and \( SE \) is the standard error.
A 98% confidence interval, for example, uses a z-value of 2.33. This means that if we constructed confidence intervals from many samples, 98% of them would contain the true population mean. The margin of error ensures us how wide this interval should be, providing a buffer around the point estimate.
Sample Mean
The sample mean is a fundamental concept in statistics, often used to summarize data by giving an average value. It's central to calculating other statistical measures like the point estimate and standard error.
To compute the sample mean:
  • Sum all the values in the sample.
  • Divide this total by the number of observations.
The sample mean gives a quick, easy snapshot of the data set's central tendency. It's the first step in many statistical calculations and offers a straightforward way to compare different samples. Its simplicity and reliability make it a staple in statistical analyses.

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Most popular questions from this chapter

The standard deviation for a population is \(\sigma=14.8\). A random sample of 25 observations selected from this population gave a mean equal to \(143.72\). The population is known to have a normal distribution. a. Make a \(99 \%\) confidence interval for \(\mu\). b. Construct a \(95 \%\) confidence interval for \(\mu\). c. Determine a \(90 \%\) confidence interval for \(\mu\). d. Does the width of the confidence intervals constructed in parts a through \(\mathrm{c}\) decrease as the confidence level decreases? Explain your answer.

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a. A random sample of 400 observations taken from a population produced a sample mean equal to \(92.45\) and a standard deviation equal to \(12.20\). Make a \(98 \%\) confidence interval for \(\mu\). b. Another sample of 400 observations taken from the same population produced a sample mean equal to \(91.75\) and a standard deviation equal to \(14.50 .\) Make a \(98 \%\) confidence interval for \(\mu\). c. A third sample of 400 observations taken from the same population produced a sample mean equal to \(89.63\) and a standard deviation equal to \(13.40 .\) Make a \(98 \%\) confidence interval for \(\mu\). d. The true population mean for this population is \(90.65\). Which of the confidence intervals constructed in parts a through \(\mathrm{c}\) cover this population mean and which do not?

A large city with chronic economic problems is considering legalizing casino gambling. The city council wants to estimate the proportion of all adults in the city who favor legalized casino gambling. What is the most conservative estimate of the minimum sample size that would limit the margin of error to be within \(.05\) of the population proportion for a \(95 \%\) confidence interval?
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