# Distributions of the Sample Mean

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Distributions of the Sample Mean
Chapter 8 Section 1 Distributions of the Sample Mean

Chapter 8 – Section 1 Learning objectives
Understand the concept of a sampling distribution Describe the distribution of the sample mean for samples obtained from normal populations Describe the distribution of the sample mean for samples obtained from a population that is not normal 1 2 3

Chapter 8 – Section 1 Learning objectives
Understand the concept of a sampling distribution Describe the distribution of the sample mean for samples obtained from normal populations Describe the distribution of the sample mean for samples obtained from a population that is not normal 1 2 3

Chapter 8 – Section 1 Lets look at a small population. 2,4,6 8,4,3,7
Find the mean. Is this mu or x bar? Now lets take all the possible samples of 2 from this population.

2,4,6 8,4,3,7 2,2 4,4 6,6 8,8 4,4 2,4 4,6 6,8 8,4 4,3 2,6 4,8 6,4 8,3 4,7 2,8 4,4 6,3 8,7 2,4 4,3 6, 2,3 4, ,7 2,7 7,7

2,4,6 8,4,3,7 2,2 4,4 6,6 8,8 4,4 2,4 4,6 6,8 8,4 4,3 2,6 4,8 6,4 8,3 4,7 2,8 4,4 6,3 8,7 2,4 4,3 6, 2,3 4, ,7 2,7 7,7

Chapter 8 – Section 1 Often the population is too large to perform a census … so we take a sample How do the results of the sample apply to the population? What’s the relationship between the sample mean and the population mean? What’s the relationship between the sample standard deviation and the population standard deviation? This is statistical inference

Chapter 8 – Section 1 We want to use the sample mean x to estimate the population mean μ If we want to estimate the heights of eight year old girls, we can proceed as follows Randomly select 100 eight year old girls Compute the sample mean of the 100 heights Use that as our estimate This is using the sample mean to estimate the population mean

Chapter 8 – Section 1 However, if we take a series of different random samples Sample 1 – we compute sample mean x1 Sample 2 – we compute sample mean x2 Sample 3 – we compute sample mean x3 Etc. Each time we sample, we may get a different result The sample mean x is a random variable!

Chapter 8 – Section 1 Because the sample mean is a random variable
The sample mean has a mean The sample mean has a standard deviation The sample mean has a probability distribution This is called the sampling distribution of the sample mean

CENTRAL LIMIT THEOREM The Central Limit Theorem is about the distribution of sample means. x The Central Limit Theorem is what we will base most or the statistical concepts on for the rest of this course.

Definition Sampling Distribution of the mean
is the probability distribution of sample means, with all samples having the same sample size n. page 256 of text

Central Limit Theorem Given:
1. The random variable x has a distribution (which may or may not be normal) with mean µ and standard deviation . 2. Samples all of the same size n are randomly selected from the population of x values. page 257 of text

Central Limit Theorem Conclusions:

Central Limit Theorem Conclusions:
1. The distribution of sample x will, as the sample size increases, approaches a normal distribution.

Central Limit Theorem Conclusions:
1. The distribution of sample x will, as the sample size increases, approach a normal distribution. 2. The mean of the sample means will be the population mean µ.

Central Limit Theorem Conclusions: n
1. The distribution of sample x will, as the sample size increases, approach a normal distribution. 2. The mean of the sample means will be the population mean µ. 3. The standard deviation of the sample means will approach  n

Practical Rules Commonly Used:
1. For samples of size n larger than 30, the distribution of the sample means can be approximated reasonably well by a normal distribution. The approximation gets better as the sample size n becomes larger. 2. If the original population is itself normally distributed, then the sample means will be normally distributed for any sample size n (not just the values of n larger than 30).

Notation

the mean of the sample means
Notation the mean of the sample means  µx = µ

the standard deviation of sample mean
Notation the mean of the sample means the standard deviation of sample mean  µx = µ x = n

the standard deviation of sample mean
Notation the mean of the sample means the standard deviation of sample mean  (often called standard error of the mean) µx = µ x = n

As the sample size increases, the sampling distribution of sample means approaches a normal distribution. page 259 of text

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability that her weight is greater than 150 lb. b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb. Example on page of text

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability that her weight is greater than 150 lb. z = = 0.24 29 P(x>150) = P(z>.24) = = P(z>.24) = = This problem is one that could have appeared in section 5-3. 0.5948  = 143 150 = 29 0.24

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb. The (b) problem exemplifies one that uses the central limit theorem to compute. Note the different wording to that of the (a) problem.

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb. The different standard deviation will have to be computed for this distribution. Note that the standard deviation is smaller than that of the population. x = 143 150 x= 29 = 36

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb. z = = 1.45 29 36 With a different standard deviation, there will be a different z score computation. 0.9265 x = 143 150 x= 1.45

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb. z = = 1.45 29 36 = 0.9265 x = 143 150 x= 1.45

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, the probability that their mean weight is greater than 150 lb is z = = 1.45 29 36 = Interpretation of numerical answer. 0.4265 x = 143 150 x= 1.45

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, Comparison of two different problems and answers.

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability that her weight is greater than 150 lb. P(x > 150) =

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability that her weight is greater than 150 lb. P(x > 150) = b.) if 36 different women are randomly selected, their mean weight is greater than 150 lb. P(x> 150) = The answers are significantly different. Discussion of why they are different should be held.

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability that her weight is greater than 150 lb. P(x > 150) = b.) if 36 different women are randomly selected, their mean weight is greater than 150 lb. P(x36 > 150) = It is much easier for an individual to deviate from the mean than it is for a group of 36 to deviate from the mean. The last statement on this slide should be discussed in detail.

Sampling Without Replacement
If n > 0.05 N page 262 of text This topic can be omitted without loss of understanding of future topics.

Sampling Without Replacement
If n > 0.05 N N - n x = n N - 1

Sampling Without Replacement
If n > 0.05 N N - n x = n N - 1 finite population correction factor

Distribution of 50 Sample Means for 50 Students
15 Frequency 10 5 Students should note that the standard deviation of the means would be a smaller number than that of the population. Also discuss position of the mean of the means versus the mean of the population and the shape of the distribution of means versus the distribution shape of the population. Figure 5-20

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