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What is the point estimator of the population mean, \(\mu ?\) How would you calculate the margin of error for an estimate of \(\mu\) ?

Short Answer

Expert verified
The point estimator for the population mean \(\mu\) is the sample mean \(\bar{x}\). The margin of error for an estimate of \(\mu\) is calculated using the formula \(E = Z \times \frac{\sigma}{\sqrt{n}}\). If we have a sample mean of \(\bar{x}\) and a margin of error of \(E\), then our estimate of the population mean (\(\mu\)) is \(\bar{x} \pm E\).

Step by step solution

01

Identifying the Point Estimator

The sample mean (\(\bar{x}\)) is often used as the point estimator for the population mean (\(\mu\)). The formula for the sample mean is: \(\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_{i}\). Here, \(x_{i}\) represents each value from the sample and \(n\) is the number of values in the sample.
02

Calculating the Margin of Error

The margin of error for an estimate of \(\mu\) is calculated using the formula: \(E = Z \times \frac{\sigma}{\sqrt{n}}\). Here, \(E\) is the margin of error, \(Z\) is the z-score (which depends on your desired confidence level), \(\sigma\) is the population standard deviation, and \(n\) is the sample size. If you do not know \(\sigma\), you can use the sample standard deviation, \(s\), instead.
03

Interpretation

Once the margin of error is calculated it needs to be interpreted. If we have a sample mean of \(\bar{x}\) and a margin of error of \(E\), then our estimate of the population mean (\(\mu\)) with \(95%\) confidence is \(\bar{x} \pm E\). Which means we are \(95%\) confident the population mean (\(\mu\)) lies within \(\bar{x} \pm E\).

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Key Concepts

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

Sample Mean
The sample mean, denoted by \( \bar{x} \), is a critical concept in statistics. It serves as a point estimator for the population mean, \( \mu \). The purpose of using the sample mean is to estimate the average value of a larger population based on a smaller sample size. To calculate the sample mean, you take the sum of all observations in the sample, represented as \( x_i \), and divide it by the number of observations, \( n \). The formula is:
  • \( \bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_{i} \)
This formula provides a straightforward method for estimating the center of a dataset, allowing for a practical approximation of the population mean. Since collecting data from an entire population is often impractical or impossible, the sample mean is crucial for making statistical inferences.
Margin of Error
The margin of error indicates the amount of error that can be expected in the results obtained from a sample survey. It quantifies the uncertainty inherent in using a sample to estimate a population parameter, such as the population mean.The formula to calculate the margin of error, \( E \), uses several critical components:
  • Z-score (\( Z \)): This value is determined by the desired confidence level for the estimation. It's derived from the standard normal distribution.
  • Population standard deviation (\( \sigma \)) or sample standard deviation (\( s \)): If \( \sigma \) is unknown, \( s \) can be used as a substitute.
  • Sample size (\( n \)): The number of observations in the sample.
The formula for margin of error is:
  • \( E = Z \times \frac{\sigma}{\sqrt{n}} \)
This formula shows that the margin of error decreases with increasing sample size, offering more precise estimates of the population mean. It plays a vital role in the construction of confidence intervals.
Population Mean
The population mean, denoted as \( \mu \), is a central concept in statistics and represents the average of all data points in an entire population. Unlike the sample mean, which is derived from a subset, the population mean pertains to the whole data set hence providing an ideal measure of central tendency for the population.Finding the true population mean can be challenging, as it requires access to data from the entire population, which is often impractical. This is why statisticians often rely on sample means as estimates of \( \mu \).Another important point is that while individual samples provide only estimates of \( \mu \), the philosophy of point estimation involves using the sample mean \( \bar{x} \) as the best single estimate of \( \mu \). Thus, the population mean is a theoretical value around which sample means are expected to center.
Confidence Interval
A confidence interval is an essential part of inferential statistics, offering a range of values that is likely to contain the population mean, \( \mu \). By accounting for variability in the sample data and the sample mean, confidence intervals provide a measure of precision for the estimate of \( \mu \).The construction of a confidence interval involves the sample mean \( \bar{x} \) and a computed margin of error \( E \). The formula for a confidence interval is:
  • \( \bar{x} \pm E \)
This means that the true population mean is expected to lie within the interval from \( \bar{x} - E \) to \( \bar{x} + E \).The level of confidence, typically expressed as a percentage (e.g., 95%), specifies how certain we are that the interval contains the population mean. Higher confidence levels increase the margin of error, making the interval wider. This relationship highlights the balance between confidence and precision in statistical analysis and decision making.

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Most popular questions from this chapter

The mean time taken to design a house plan by 40 architects was found to be 23 hours with a standard deviation of \(3.75\) hours. a. Construct a \(98 \%\) confidence interval for the population mean \(\mu\). b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Describe all possible alternatives. Which alternative is the best and why?

A casino player has grown suspicious about a specific roulette wheel. Specifically, this player believes that the slots for the numbers 0 and 00 , which can lead to larger payoffs, are slightly smaller than the rest of 36 slots, which means that the ball would land in these two slots less often than it would if all of the slots were of the same size. This player watched 430 spins on this roulette wheel, and found that the ball landed in 0 or 00 slot 14 times. a. What is the value of the point estimate of the proportion of all roulette spins on this wheel in which the ball would land in 0 or 00 slot? b. Construct a \(95 \%\) confidence interval for the proportion of all roulette spins on this wheel in which the ball would land in 0 or 00 slot. c. If all of the slots on this wheel are of the same size, the ball should land in 0 or 00 slot \(5.26 \%\) of the time. Based on the confidence interval you calculated in part b, does the player's suspicion seem reasonable?

The express check-out lanes at Wally's Supermarket are limited to customers purchasing 12 or fewer items. Cashiers at this supermarket have complained that many customers who use the express lanes have more than 12 items. A recently taken random sample of 200 customers entering express lanes at this supermarket found that 74 of them had more than 12 items. a. Construct a \(98 \%\) confidence interval for the percentage of all customers at this supermarket who enter express lanes with more than 12 items. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=2.467\) and \(d f=28\) b. \(t=-1.672\) and \(d f=58\) c. \(t=-2.670\) and \(n=55\) d. \(t=2.383\) and \(n=23\)

Check if the sample size is large enough to use the normal distribution to make a confidence interval for \(p\) for each of the following cases. a. \(n=80\) and \(\hat{p}=.85\) b. \(n=110\) and \(\hat{p}=.98\) c. \(n=35\) and \(\hat{p}=.40\) d. \(n=200\) and \(\hat{p}=.08\)

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