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When a researcher selects all possible pairs of samples from a population in order to find the difference between the means of each pair, what will be the shape of the distribution of the differences when the original distributions are normally distributed? What will be the mean of the distribution? What will be the standard deviation of the distribution?

Short Answer

Expert verified
The distribution is normal with mean 0 and standard deviation \( \sigma \sqrt{\frac{2}{n}} \).

Step by step solution

01

Understanding the Task

We need to determine the shape, mean, and standard deviation of the distribution of differences between means of pairs of samples, given that the original population distribution is normal.
02

Shape of the Distribution

When samples are taken from a normally distributed population, the differences between the means of these samples will also be normally distributed. This is because the normal distribution characteristic is preserved in the difference of normal random variables.
03

Mean of the Distribution

For any pair of samples taken from a population, the expected (mean) difference between the means of these samples is zero. This is because each sample mean is an unbiased estimator of the population mean.
04

Standard Deviation of the Distribution

To find the standard deviation of the distribution of differences, if each sample has size n from a population with standard deviation , we use the formula for the standard deviation of the difference between two samples: \( \sigma_{\bar{X}_1 - \bar{X}_2} = \sigma \sqrt{\frac{2}{n}} \). This formula arises from the standard deviation of the normal distribution and properties of variance.

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Key Concepts

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

Sampling Distribution
Sampling distribution refers to the probability distribution of a statistic obtained from many samples drawn from a specific population. When we talk about a sampling distribution of the mean, we are referring to the distribution of sample means gathered from multiple sample groups within the same population. This concept helps statisticians understand how the sample mean behaves compared to the population mean.
  • This distribution provides valuable insights because it allows calculations about the probability of obtaining certain sample results.
  • Understanding the shape of the sampling distribution helps in making predictions and inferences about the population.
In the given exercise, the researcher is interested in the distribution of differences between means for pairs of samples. When the original population is normally distributed, this sampling distribution also tends to display a normal shape.
This characteristic stems from the consistency of normal distribution properties when dealing with means.
Mean Difference
The mean difference is an important concept in statistics, especially when comparing two groups. It represents the average of all differences between the pairs of sample means. In the exercise, one must calculate the mean difference when the samples are drawn from a normally distributed population. Due to the properties of normal distribution and unbiased estimators:
  • The mean difference is expected to be zero.
  • Each sample mean serves as a reliable representation of the population mean.
Since any two samples are being compared, the law of large numbers suggests that the average outcome of repeated experiments gets closer to the expected value.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. When applied to a distribution's mean differences, it tells us how scattered these differences are. For the setup discussed in the exercise, the standard deviation of the distribution of differences between two sample means is calculated with the formula:\[\sigma_{\bar{X}_1 - \bar{X}_2} = \sigma \sqrt{\frac{2}{n}} \]where \( \sigma \) is the standard deviation of the population, and \( n \) is the size of each sample.
  • This reveals how spread out the differences are, despite the expectation of a mean difference of zero.
  • The larger the sample size, \( n \), the less variability there is among the mean differences.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics asserting that, given a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normally distributed, no matter the shape of the original population distribution.
In simpler terms, regardless of whether the population distribution is normal, skewed, or of any other shape, the distribution of the sample means tends to be normal. However, our exercise begins with a normally distributed population. Thus, when you draw pairs of samples and calculate their mean differences:
  • The distribution of those differences shares the normal shape due to the inheritance of properties from the original population.
  • This makes it easier to perform statistical tests and draw conclusions.
The CLT aids in the reliability of sample-based inferences about the population, highlighting the power of sample size in achieving a normal distribution.

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Most popular questions from this chapter

Teacher Salaries A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers in a large school district. The mean of the salaries of a random sample of 26 elementary school teachers is dollar 48,256, and the sample standard deviation is dollar 3,912.40 . The mean of the salaries of a random sample of 24 secondary school teachers is dollar 45,633 . The sample standard deviation is dollar 5533 . At \(\alpha=0.05,\) can it be concluded that the mean of the salaries of the elementary school teachers is greater than the mean of the salaries of the secondary school teachers? Use the \(P\) -value method.

Weights of Running Shoes The weights in ounces of a sample of running shoes for men and women are shown. Test the claim that the means are different. Use the \(P\) -value method with \(\alpha=0.05 .\) $$ \begin{array}{ll|ccc}{\text { Men }} & {} & {\text { Women }} & {} \\ \hline 10.4 & {12.6} & {10.6} & {10.2} & {8.8} \\ {1.1} & {14.7} & {9.6} & {9.5} & {9.5} \\ {10.8} & {12.9} & {10.1} & {11.2} & {9.3} \\ {11.7} & {13.3} & {9.4} & {10.3} & {9.5} \\ {12.8} & {14.5} & {9.8} & {10.3} & {11.0} \\\ \hline\end{array} $$

Explain the difference between testing a single mean and testing the difference between two means.

ACT Scores A random survey of 1000 students nationwide showed a mean ACT score of 21.4 . Ohio was not used. A survey of 500 randomly selected Ohio scores showed a mean of 20.8 . If the population standard deviation is \(3,\) can we conclude that Ohio is below the national average? Use \(\alpha=0.05 .\)

Age Differences In a large hospital, a nursing director selected a random sample of 30 registered nurses and found that the mean of their ages was \(30.2 .\) The population standard deviation for the ages is \(5.6 .\) She selected a random sample of 40 nursing assistants and found the mean of their ages was \(31.7 .\) The population standard deviation of the ages for the assistants is \(4.3 .\) Find the \(99 \%\) confidence interval of the differences in the ages.

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