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The sampling distribution of a mean tends toward one like this-a Normal model centered at \(\mu\) with standard deviation \(\sigma / \sqrt{n}\). We know that our sample mean \(\bar{y}\) is somewhere in this picture. But where? \(95 \%\) of samples will have means within 1.96 standard deviations of \(\mu .\) So, if we made \(\mu\) traps for each sample going out \(1.96 \times \frac{\sigma}{\sqrt{n}}\) on either side of \(\bar{y}, 95 \%\) of those traps would capture \(\mu .\)

Short Answer

Expert verified
The sample mean \( \bar{y} \) is within a theoretical distribution of sample means that is comparable to a normal distribution. By extending 1.96 standard deviations (calculated as \(1.96 \times \frac{\sigma}{\sqrt{n}}\)) on either side of \( \bar{y} \), it captures the population mean \( \mu \) 95% of the time. This represents our confidence level that the sample mean is a good estimate of the population mean.

Step by step solution

01

Understand the Sampling Distribution

The sampling distribution of a mean is a theoretical probability distribution of sample means for all possible samples of a certain size from the population. It tends towards a normal model, which is centered at the population mean, \( \mu \) with standard deviation equal to \( \sigma / \sqrt{n} \), where \( \sigma \) stands for population standard deviation and \( n \) is the size of the sample.
02

Locate the Sample Mean

In this scenario, the sample mean, \( \bar{y} \), is said to be located within this distribution. The exact location is not specified, but it lies somewhere within the distribution.
03

Understand the 95% Rule

By rule, 95% of samples will have means within 1.96 standard deviations of \( \mu \). This means if we created intervals in the distribution for each sample that extend 1.96 standard deviations on either side of the mean (calculated as \(1.96 \times \frac{\sigma}{\sqrt{n}}\)), 95% of those intervals would contain the population mean \( \mu \).
04

Interpret the Conclusion

Thus, positioning 'traps' (or intervals in the context of the problem) that extend \(1.96 \times \frac{\sigma}{\sqrt{n}}\) on either side of the sample mean \( \bar{y} \) will capture the population mean \( \mu \) 95% of the time. This concept illustrates our confidence that the estimated parameter (mean) from a sample will be close to the actual population parameter.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is an essential concept in statistics. It states that the sampling distribution of the sample mean approaches a normal distribution as the sample size becomes larger, regardless of the population's original distribution. This is incredibly helpful because it allows us to make inferences about the population mean using the normal distribution even when the population is not normally distributed. By using the CLT, statisticians can apply powerful tools like z-scores and other statistical measures that rely on normality.

One of the key requirements for the CLT to hold true is having a sufficiently large sample size. A rule of thumb is that a sample size of at least 30 is considered adequate to apply the CLT, but this can vary depending on the shape of the population distribution. The theorem provides a framework to estimate how a mean from sample data represents the entire population.
Confidence Intervals
Confidence intervals are a range of values used to estimate a population parameter, like a mean. They are constructed around the sample mean and are often expressed with a confidence level, such as 95%. This means that if the same population is sampled many times, and a confidence interval is computed each time, we would expect the population parameter to lie within 95% of those intervals.

The interval provides a range of plausible values for the parameter, reflecting both variability in the sample data and certain statistical assumptions. In the example from the exercise, the 95% confidence interval of a sample mean extends 1.96 standard deviations on either side of the sample mean. This is derived from the properties of a normal distribution, where roughly 95% of data falls within 1.96 standard deviations of the mean.
  • The concept of confidence intervals is vital because it provides a measure of reliability for an estimate.
  • They allow researchers to make probabilistic statements about how well the sample statistic approximates the population parameter.
However, they do not account for all possible errors or biases in sampling, such as systematic errors.
Standard Deviation
Standard deviation is a measure of the amount of variability or spread in a set of data values. In the context of a sampling distribution, the standard deviation assumes a slightly different perspective.

For a single sample drawn from a population, the standard deviation of the sampling distribution, known as the standard error, is calculated as \( \frac{\sigma}{\sqrt{n}} \). Here, \( \sigma \) is the population standard deviation, and \( n \) is the sample size. This standard error measures how much the sample mean \( \bar{y} \) is expected to vary from the true population mean \( \mu \).

The smaller the standard error, the more precise the estimate of the population mean will be. This is why increasing the sample size (\( n \)) tends to decrease the standard error, providing a more reliable estimation of the population characteristic. A small standard deviation indicates that the data points are very close to the means, while a large standard deviation indicates that the data points are spread out over a wider range of values.

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Most popular questions from this chapter

Golf balls The United States Golf Association (USGA) sets performance standards for golf balls. For example, the mean initial velocity of the ball may not exceed 250 feet per second when measured by an apparatus approved by the USGA. Suppose a manufacturer introduces a new kind of ball and provides a sample for testing. Based on these data, the USGA comes up with a \(95 \%\) confidence interval for the mean initial velocity from 240.8 to 259.9 feet. What does this say about the performance of the new ball?

Groceries A grocery store's receipts show that Sunday customer purchases have a skewed distribution with a mean of \(\$ 32\) and a standard deviation of \(\$ 20\). a. Explain why you cannot determine the probability that the next Sunday customer will spend at least \(\$ 40\). b. Can you estimate the probability that the next 10 Sunday customers will spend an average of at least \(\$ 40 ?\) Explain. c. Is it likely that the next 50 Sunday customers will spend an average of at least \(\$ 40 ?\) Explain.

Ski wax Bjork Larsen was trying to decide whether to use a new racing wax for cross-country skis. He decided that the wax would be worth the price if he could average less than 55 seconds on a course he knew well, so he planned to study the wax by racing on the course 8 times. His 8 race times were \(56.3,65.9,50.5,52.4,46.5,57.8,52.2,\) and 43.2 seconds. Should he buy the wax? Explain by using a confidence interval

Teachers Software analysis of the salaries of a random sample of 288 Nevada teachers produced the confidence interval shown below. Which conclusion is correct? What's wrong with the others? with \(90.00 \%\) Confidence, \(t\) -interval for \(\mu: 43454<\mu(\) TchPay \()<45398\) a. If we took many random samples of 288 Nevada teachers, about 9 out of 10 of them would produce this confidence interval. b. If we took many random samples of Nevada teachers, about 9 out of 10 of them would produce a confidence interval that contained the mean salary of all Nevada teachers. c. About 9 out of 10 Nevada teachers earn between \(\$ 43,454\) and \(\$ 45,398 .\) d. About 9 out of 10 of the teachers surveyed earn between \(\$ 43,454\) and \(\$ 45,398\). e. We are \(90 \%\) confident that the average teacher salary in the United States is between \(\$ 43,454\) and \(\$ 45,398\).

Parking Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period (44 weekdays), daily fees collected averaged \(\$ 126,\) with a standard deviation of \(\$ 15 .\) a. What assumptions must you make in order to use these statistics for inference? b. Write a \(90 \%\) confidence interval for the mean daily income this parking garage will generate. c. Interpret this confidence interval in context. d. Explain what "90\% confidence" means in this context. e. The consultant who advised the city on this project predicted that parking revenues would average \(\$ 130\) per day. Based on your confidence interval, do you think the consultant was correct? Why?

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