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For a data set obtained from a random sample, \(n=81\) and \(\bar{x}=48.25\). It is known that \(\sigma=4.8\). a. What is the point estimate of \(\mu\) ? b. Make a \(95 \%\) 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 48.25. b. The 95% confidence interval for \(\mu\) is (47.16, 49.34). c. The margin of error of the estimate is 1.04.

Step by step solution

01

Determine Point Estimate of \(\mu\)

A point estimate of a population parameter is a single value of a statistic. In this case, the sample mean (\(\bar{x}\)) serves as the point estimate of the population mean (\(\mu\)). So, \(\bar{x}\) = 48.25 is the point estimate of \(\mu\).
02

Calculate the 95% Confidence Interval for \(\mu\)

The confidence interval is calculated using the formula: \(\bar{x} \pm Z_{\alpha/2}*\frac{\sigma}{\sqrt{n}}\). Here, \(\bar{x}\) = 48.25, \(Z_{\alpha/2}\) = 1.96 (for a 95% confidence interval), \(\sigma\) = 4.8, and \(n\) = 81. Substituting these values into the equation, we get: 48.25 \(\pm\) 1.96*(4.8/\(\sqrt{81}\)). This gives us the confidence interval (47.16, 49.34).
03

Calculate the Margin of Error

The margin of error is calculated using the formula \(Z_{\alpha/2}*\frac{\sigma}{\sqrt{n}}\). Inserting the values from our problem, we get: margin of error = 1.96*(4.8/\(\sqrt{81}\)), which simplifies to 1.04.

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Key Concepts

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

Point Estimate
When estimating the characteristics of an entire population, sometimes it is necessary to use information from a sample. One way to simplify this task is to find a point estimate. This is basically a single value that serves as an educated guess of the population parameter you are interested in.
The point estimate is especially useful when dealing with means. In our case, imagine you are trying to estimate the average value (mean) of a huge group. Instead of measuring everyone, you take a random sample and calculate its average. This average becomes your point estimate for the population mean.
For example, using the data provided in the exercise, the point estimate of the population mean \( \mu \) is the sample mean \( \bar{x} = 48.25 \). This sample mean suggests that you expect the mean of the entire population to be around this value. It’s quick, simple, and very powerful when used properly.
Margin of Error
When reporting estimates, it's crucial to acknowledge the uncertainty involved. This is where the margin of error comes into play. The margin of error tells us how far off our sample statistic might be from the true population parameter.
Essentially, it provides a range above and below the point estimate within which the actual parameter is likely to fall. It accounts for sample variability and helps in understanding the precision of our estimate.
We calculate the margin of error by multiplying the critical value (in this case, 1.96 for a 95% confidence level) by the standard deviation of the sample mean's estimate.
  • The formula is: \( Z_{\alpha/2}*\frac{\sigma}{\sqrt{n}} \)

In the provided exercise, with a \( \sigma = 4.8 \), a sample size of \( \ n = 81 \), and a critical value of 1.96, the margin of error works out to approximately 1.04.
This number tells us that we can expect our sample mean to be within 1.04 units of the true population mean with 95% confidence. It is key in constructing the confidence interval, which gives a clearer picture of our estimate's reliability.
Sample Mean
The sample mean is a fundamental concept in statistics, especially when making inferences about a population. Simply put, it is the average of the data points in a sample drawn from the overall population.
Computing the sample mean is simple: add up all sample values and divide by the number of values you have. Mathematically, it is represented as \( \bar{x} = \frac{1}{n}\sum_{i=1}^{n}x_i \), where \( \bar{x} \) is the sample mean, \( x_i \) are the sample data points, and \( n \) is the sample size.
The sample mean serves two important roles: it is itself a point estimate for the population mean, and it is used in formulas for creating confidence intervals and calculating margins of error.
  • In the exercise, the calculated sample mean is 48.25.

This number, being our point estimate, suggests a central value around which we center our analysis for the entire group or population. By applying further statistical techniques, we gain insights into the possible variance of the population mean, making our inferences more comprehensive and actionable.

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Most popular questions from this chapter

According to a Gallup poll conducted April \(3-6,2014,21 \%\) of Americans aged 18 to 29 said that college loans and/or expenses were the top financial problem facing their families. Suppose that this poll was based on a random sample of 1450 Americans aged 18 to \(29 .\) a, What is the point estimate of the corresponding population proportion? b. Construct a \(95 \%\) confidence interval for the proportion of all Americans aged 18 to 29 who will say that college loans and/or expenses were the top financial problem facing their families. What is the margin of error for this estimate?

Explain the various alternatives for decreasing the width of a confidence interval. Which is the best alternative

a. How large a sample should be selected so that the margin of error of estimate for a \(98 \%\) confidence interval for \(p\) is \(.045\) when the value of the sample proportion obtained from a preliminary sample is .53? b. Find the most conservative sample size that will produce the margin of error for a \(98 \%\) confidence interval for \(p\) equal to \(.045\).

You want to estimate the proportion of students at your college who hold off- campus (part-time or full-time) jobs. Briefly explain how you will make such an estimate. Collect data from 40 students at your college on whether or not they hold off-campus jobs. Then calculate the proportion of students in this sample who hold offcampus jobs. Using this information, estimate the population proportion. Select your own confidence level.

A random sample of 11 observations taken from a normally distributed population produced the following data: $$ \begin{array}{lllllllllll} -7.1 & 10.3 & 8.7 & -3.6 & -6.0 & -7.5 & 5.2 & 3.7 & 9.8 & -4.4 & 6.4 \end{array} $$ a. What is the point estimate of \(\mu\) ? b. Make a \(95 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for \(\mu\) in part b?

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