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Statistics Lesson 5.4.notebook
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April 27, 2016
Sampling Distributions and the Central Limit Theorem
We have been studying the relationship between the mean of a population and the values of a random variable. Now we will study the relationship between a population mean and the means of samples taken from the population.
DefinitionA sampling distribution is the probability distribution of a sample statistic that is formed when samples of size n are repeatedly taken from a population. If the sample statistic is the sample mean, then the distribution is the sampling distribution of sample means. Every sample statistic has a sampling distribution.
Page 261, look at the Venn diagram. the rectangle represents a large population and each circle represents a sample of size n. Because the sample entries can differ, so can the sample means.
Properties of Sampling Distributions of Sample Means1. The mean of the sample means is equal to the population mean μ
2. The standard deviation of the sample mean is equal to the population standard deviation σ divided by the square root of the sample size n.
The standard deviation of the sampling distribution of the sample means is called the standard error of the mean.
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You write the population values {1,3,5,7} on slips of paper and put them in a box. Then you randomly choose two slips of paper, with replacement. List all possible samples of size n =2 and calculate the mean of each of each. These means form the sampling distributions of the sample means. Find the mean, variance, and standard deviation of the sample means. Compare your results with the mean μ = 4, variance σ2 = 5, and standard deviation σ = √5 of the population.
Sample Samplesample mean, x sample mean, x
1,1 1,3 1.5 1.7 3,1 3,3 3,5 3,7
12342345
5,1 5,3 5,5 5,7 7,1 7,3 7,5 7,7
34564567
x f probability
1234567
1234321
1/16 = 0.06252/16 = 0.12503/16 = 0.18754/16 = 0.25003/16 = 0.18752/16 = 0.12501/16 = 0.0625
Make a probability distribution of the sample means, draw a histogram of the probabilities.
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Using the properties of sampling distributions of sample means, we get:
List all possible samples of size n = 3, with replacement, from the population {1,3,5}.Calculate the mean of each sample. Find the mean, variance, and standard deviation of the sample means. Compare your results with the mean μ = 3, variance σ2 = 8/3, and standard deviation σ = √8/3≈1.633 of the population.
Sample Sample mean, x1,1,11,1,31,1,51,3,11,3,31,3,51,5,11,5,31,5,53,1,13,1,33,1,53,3,13,3,33,3,5
Sample Sample mean, x
3,5,13,5,33,5,55,1,15,1,35,1,55,3,15,3,35,3,55,5,15,5,35,5,5
11.672.331.672.333
2.333
3.671.672.333
2.333
3.67
33.674.332.333
3.673
3.674.333.674.335
Probability
11.672.333
3.674.335
1367631
1/27=0.03703/27=0.11116/27=0.22227/27=0.25936/27=0.22223/27=0.11111/27=0.0370
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THE CENTRAL LIMIT THEOREM1. If samples of size n, where n≥30, are drawn from any population with a mean μ and a standards deviation σ, then the sampling distribution of the sample means approximates a normal distribution. The greater the sample size, the better the approximation. (See figures for "Any Population Distribution on next page)
2. If the population itself is normally distributed, then the sampling distribution of sample means is normally distributed for any sample size n. (See figures for "Normal Population Distribution" on next page)
In either case, the sampling distribution of sample means has a mean equal to the population mean.
The sampling distribution of the sample means has a value equal to 1/n times the variance of the population and a standard deviation equal to the population standard deviation divided by the square root of n.
mean of the sample means
Variance of the sample means
Standard deviation of the sample means
Remember that the standard deviation of the sampling distribution of the sample means, , is also called the standard error of the mean.
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Mean
Standard deviation
Any Population Distribution
Distribution of Sample means, n ≥ 30
1.
standard deviationof the sample means
mean
2. Normal Population Distribution
Mean
Standard deviation
Distribution of Sample Means (any n)
mean
standard deviationof the sample means
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EXAMPLE:
Cell phone bills for residents of a city have a mean of $47 of $9, as shown in the figure. Random samples of 100 cell phone bills are drawn from this population, and the mean of each sample is determined. Find the mean and standard deviation of the sampling distribution of the sample means. Then sketch a graph of the sampling distribution.
The mean of the sampling distribution is equal to the population mean. So,
The standard deviation of the sample means is equal to the population standard deviation divided by √n. So,
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Because the sample size is greater than 30, the sampling distribution can be approximated by a normal distribution with a mean of $47 and a standard deviation of $0.90.
Assume the training heart rates of all 20yearold athletes are normally distributes, with a mean of 135 beats per minute and a standard deviation of 18 beats per minute. Random samples of size 4 are drawn from the population, and the mean sample is determined. Find the mean and standard deviation of the sampling distribution of the sampling means. Then sketch a graph of the sampling distribution.
The sampling mean is 135 and the standard deviationof the sample means is 9. So the graph should look like:
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PROBABILITY AND THE CENTRAL LIMIT THEOREMIn section 5.2, you learned how to find the probability that a random variable x will lie in a given interval of population values. We can find the probability that a sample mean x will lie in a given interval of the x sampling distribution. To transform x to a zscore, you can use the formula
EXAMPLE 4 P. 266
The figure on P. 266 show the lengths of time people spend driving each day. You randomly select 50 drivers ages 15 to 19. What is the probability that the mean time they spend driving each day is between 24.7 and 25.5 minutes. Assume σ = 1.5.
(a) Since we need to find the probability between 24.7 and 25.5, we must find both zscores.
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We are trying to find the probability that the drivers are between 24.7 and 25.5 minutes.
This means that of the samples of 50 drivers ages 15 to 19, about 91% will have a drive time that is between 24.7 and 25.5 minutes.
You randomly select 100 drivers ages 15 to 19 from the chart on page 266. What is the probability that the mean time they spend driving each day is between 24.7 and 25.5 minutes? Use μ = 25 and σ = 1.5.
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The mean room and board expense per year at fouryear colleges is $9126. You randomly select 9 fouryear colleges. What is the probability that the mean room and board is less than $9400? Assume that the room and board expenses are normally distributed with a standard deviation of $1500.
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The average credit card debt carried by undergraduates is normally distributed, with a meanof $3173 and a standard deviation of $1120.
(a) What is the probability that a randomly selected undergraduate, who is a credit card holder, has a credit card balance less than $2700?
μ = 3173 σ = 1120
P(x < 2700) = 0.3364
(b) You randomly select 25 undergraduates who are credit card holders. What is the probability that their mean credit card balance is less than $2700.
P(x < 2700) = 0.0174
There is a 34% chance that an undergraduate will have a balance less than $2700, but there is only about a 2% chance that a sample of 25 graduates will have a balance less than $2700.
Try It Yourself 6 P. 268
P. 269 1 37 odds
