statisticssite.iugaza.edu.ps/mriffi/files/2018/02/ch08.pdf · copyright © 2017, 2013, 2010 pearson...

STATISTICSINFORMED DECISIONS USING DATAFifth Edition

Chapter 8

Sampling Distributions

Copyright © 2017, 2013, 2010 Pearson Education, Inc. All Rights Reserved


8.1 Distribution of the Sample MeanLearning Objectives

1. Describe the distribution of the sample mean: normal population

2. Describe the distribution of the sample mean: nonnormal population


8.1 Distribution of the Sample MeanIntroduction (1 of 2)


8.1 Distribution of the Sample MeanIntroduction (2 of 2)

The sampling distribution of a statistic is a probability distribution for all possible values of the statistic computed from a sample of size n.


8.1 Distribution of the Sample MeanIllustrating Sampling Distributions (1 of 3)

Step 1: Obtain a simple random sample of size n.



Step 2: Compute the sample mean.



Step 3: Assuming that we are sampling from a finite population, repeat Steps 1 and 2 until all distinct simple random samples of size n have been obtained.


8.1 Distribution of the Sample Mean8.1.1 Describe the Distribution of the Sample Mean: Normal Population (1 of 14)

Parallel Example 1: Sampling Distribution of the SampleMean-Normal Population

The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams.

Approximate the sampling distribution of the sample mean by obtaining 200 simple random samples of size n = 5 from this population.



The data on the following slide represent the sample means for the 200 simple random samples of size n = 5.

For example, the first sample of n = 5 had the following data:

2.433 2.466 2.423 2.442 2.456



Sample Means for Samples of Size n = 5

2.444 2.483 2.452 2.484 2.468 2.41 2.432 2.46 2.471 2.466

2.441 2.469 2.472 2.444 2.46 2.422 2.424 2.451 2.434 2.483

2.496 2.481 2.454 2.416 2.456 2.463 2.464 2.464 2.488 2.497

2.436 2.464 2.417 2.425 2.454 2.449 2.462 2.453 2.502 2.472

2.509 2.462 2.481 2.432 2.466 2.484 2.444 2.447 2.459 2.465

2.438 2.461 2.461 2.444 2.415 2.475 2.452 2.454 2.45 2.439

2.474 2.451 2.427 2.461 2.433 2.419 2.48 2.486 2.432 2.447

2.419 2.478 2.418 2.474 2.442 2.479 2.472 2.445 2.442 2.475

2.461 2.48 2.453 2.477 2.472 2.479 2.472 2.445 2.442 2.475

2.435 2.466 2.43 2.461 2.446 2.451 2.475 2.454 2.471 2.483



Sample Means for Samples of Size n = 5

2.457 2.445 2.439 2.485 2.472 2.467 2.459 2.421 2.462 2.467

2.456 2.462 2.447 2.48 2.47 2.5 2.491 2.448 2.445 2.468

2.472 2.469 2.496 2.445 2.449 2.492 2.481 2.457 2.466 2.465

2.441 2.44 2.473 2.483 2.444 2.464 2.465 2.447 2.473 2.437

2.464 2.466 2.472 2.47 2.454 2.438 2.443 2.492 2.439 2.465

2.483 2.465 2.472 2.478 2.456 2.427 2.492 2.48 2.492 2.451

2.458 2.415 2.466 2.432 2.472 2.46 2.483 2.485 2.462 2.472

2.471 2.47 2.498 2.444 2.456 2.452 2.425 2.47 2.463 2.485

2.467 2.468 2.451 2.46 2.429 2.479 2.471 2.455 2.483 2.446

2.459 2.449 2.497 2.499 2.432 2.498 2.443 2.457 2.493 2.461



The mean of the 200 sample means is 2.46, the same as the mean of the population.

The standard deviation of the sample means is 0.0086, which is smaller than the standard deviation of the population.

The next slide shows the histogram of the sample means.



What role does n, the sample size, play in the standard deviation of the distribution of the sample mean?



As the size of the sample increases, the standard deviation of the distribution of the sample mean decreases.



Parallel Example 2: The Impact of Sample Size on Sampling Variability

Approximate the sampling distribution of the sample mean by obtaining 200 simple random samples of size n = 20 from the population of weights of pennies minted after 1982 (μ = 2.46 grams and σ = 0.02 grams)



The mean of the 200 sample means for n = 20 is still 2.46, but the standard deviation is now 0.0045 (0.0086 for n = 5). As expected, there is less variability in the distribution of the sample mean withn = 20 than with n = 5.



Parallel Example 3: Describing the Distribution of the Sample Mean

The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams.

What is the probability that in a simple random sample of 10 pennies minted after 1982, we obtain a sample mean of at least 2.465 grams?


8.1 Distribution of the Sample Mean8.1.2 Describe the Distribution of the Sample Mean: Nonnormal Population (1 of 9)

The following table and histogram

give the probability distribution for

rolling a fair die:

Face on Die Relative Frequency

1 0.1667

2 0.1667

3 0.1667

4 0.1667

5 0.1667

6 0.1667



Histograms of the sampling distribution of the sample mean for each sample size are given on the next slide.



• The shape of the distribution of the sample mean becomes approximately normal as the sample size n increases, regardless of the shape of the underlying population.



The Central Limit Theorem



Parallel Example 5: Using the Central Limit Theorem

Suppose that the mean time for an oil change at a “10-minute oil change joint” is 11.4 minutes with a standard deviation of 3.2 minutes.

a) If a random sample of n = 35 oil changes is selected, describe the sampling distribution of the sample mean.



Parallel Example 5: Using the Central Limit Theorem

b) If a random sample of n = 35 oil changes is selected, what is the probability the mean oil change time is less than 11 minutes?


8.2 Distribution of the Sample ProportionLearning Objectives

1. Describe the sampling distribution of a sample proportion

2. Compute probabilities of a sample proportion


8.2 Distribution of the Sample Proportion8.2.1 Describe the Sampling Distribution of a Sample Proportion (1 of 11)

Point Estimate of a Population Proportion

Suppose that a random sample of size n is obtained from a population in which each individual either does or does not have a certain characteristic.



Parallel Example 1: Computing a Sample Proportion

In a Quinnipiac University Poll conducted in May of 2008, 1745 registered voters nationwide were asked whether they approved of the way George W. Bush is handling the economy. 349 responded “yes”. Obtain a point estimate for the proportion of registered voters who approve of the way George W. Bush is handling the economy.



Parallel Example 2: Using Simulation to Describe the Distribution of the Sample Proportion

According to a Time poll conducted in June of 2008, 42% of registered voters believed that gay and lesbian couples should be allowed to marry.

Describe the sampling distribution of the sample proportion for samples of size n = 10, 50, 100.

Note: We are using simulations to create the histograms on the following slides.



Key Points from Example 2

• Shape: As the size of the sample, n, increases, the shape of the sampling distribution of the sample proportion becomes approximately normal.

• Center: The mean of the sampling distribution of the sample proportion equals the population proportion, p.

• Spread: The standard deviation of the sampling distribution of the sample proportion decreases as the sample size, n, increases.



• The model on the previous slide requires that the sampled values are independent. When sampling from finite populations, this assumption is verified by checking that the sample size n is no more than 5% of the population size N (n ≤ 0.05N).



Parallel Example 3: Describing the Sampling Distribution of the Sample Proportion

According to a Time poll conducted in June of 2008, 42% of registered voters believed that gay and lesbian couples should be allowed to marry. Suppose that we obtain a simple random sample of 50 voters and determine which voters believe that gay and lesbian couples should be allowed to marry. Describe the sampling distribution of the sample proportion for registered voters who believe that gay and lesbian couples should be allowed to marry.



Solution

The sample of n = 50 is smaller than 5% of the population size (all registered voters in the U.S.). Also, np(1 − p) = 50(0.42)(0.58) = 12.18 ≥ 10.

The sampling distribution of the sample proportion is therefore

approximately normal with mean = 0.42 and


8.2 Distribution of the Sample Proportion8.2.2 Compute Probabilities of a Sample Proportion (1 of 3)

Parallel Example 4: Compute Probabilities of a Sample Proportion

According to the Centers for Disease Control and Prevention, 18.8% of school-aged children, aged 6−11 years, were overweight in 2004.

a) In a random sample of 90 school-aged children, aged 6−11 years, what is the probability that at least 19% are overweight?

b) Suppose a random sample of 90 school-aged children, aged 6−11 years, results in 24 overweight children. What might you conclude?



Solution

• n = 90 is less than 5% of the population size

• np(1 − p) = 90(.188)(1 − .188) ≈ 13.7 ≥ 10



We would only expect to see about 3 samples in 100 resulting in a sample proportion of 0.2667 or more. This is an unusual sample if the true population proportion is 0.188.

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