chapter 7 sampling distributions...the practice of statistics, 5th edition starnes, tabor, yates,...

The Practice of Statistics, 5th Edition

Starnes, Tabor, Yates, Moore

Bedford Freeman Worth Publishers

CHAPTER 7Sampling Distributions

7.3

Sample Means

The Practice of Statistics, 5th Edition 2

7.2 Multiple choice

Page 449: 43-46

43. A

44. B

45. B

46. B


Homework

38. What sample size would be required to reduce the standard deviation of

the sampling distribution to one-third the value you found in Exercise 36?

Because the standard deviation is found by dividing by 𝑛, using 9n for the

sample size reduces the standard deviation of the sampling distribution to one-

third of the previous value ( 9𝑛 = 3 𝑛); we would need to sample 9(1785)

=16,065 adults.


Homework

40. Harley-Davidson motorcycles make up 14% of all the motorcycles

registered in the United States. You plan to interview an SRS of 500

motorcycle owners. How likely is your sample to contain 20% or more

who own Harleys? Show your work.

We want to find the probability that more than 20% of those sampled own a

Harley with n = 500 and p = 0.14.

𝜇 ො𝑝 = 𝑝 = 0.14

Since it is reasonable that there are more than 10(500) = 5000 motorcycle

owners in the US, 𝜎 ො𝑝 =(0.14)(0.86)

500= 0.0155

We can use the Normal approximation because 500 0.14 = 70 ≥ 10 and

500 0.86 = 430 ≥ 10.

Step 1: 𝑃 Ƹ𝑝 ≥ 0.20 with 𝑁 0.14, 0.0155

Step 2: 𝑧 =0.2−0.14

0.155= 3.87; 𝑃 Ƹ𝑝 ≥ 20 = 1 − 𝑃 𝑧 ≤ 3.87 ≈ 0;

𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 𝑙𝑜𝑤𝑒𝑟: 0.20, 𝑢𝑝𝑝𝑒𝑟: 1000, 𝜇: 0.14, 𝜎: 0.0155 = 0.00005Step 3: There is a 0.0001 probability that 20% or more of the motorcycle

owners in the sample own Harleys.


Homework42. The Harvard College Alcohol Study finds that 67% of college students support

efforts to “crack down on underage drinking.” Does this result hold at a large local

college? To find out, college administrators survey an SRS of 100 students and

find that 62 support a crackdown on underage drinking.

a) Suppose that the proportion of all students attending this college who support

a crackdown is 67%, the same as the national proportion. What is the

probability that the proportion in an SRS of 100 students is 0.62 or less?

Show your work.

We want to find the probability that the proportion of students sampled who favor a “crackdown on

underage drinking” is 0.62 or less with n = 100 and p = 0.67.

𝜇 ො𝑝 = 𝑝 = 0.67

Since it is reasonable that there are more than 10(100) = 1000 students at a “large local college”,

𝜎 ො𝑝 =(0.67)(0.33)

100= 0.0470

We can use the Normal approximation because 100(0.67) = 67 ≥ 10 and 100 0.33 = 33 ≥ 10.

Step 1: 𝑃 Ƹ𝑝 ≤ 0.62 with 𝑁(0.67, 0.0470)

Step 2: 𝑧 =0.62−0.67

0.0470= −1.06; 𝑃 Ƹ𝑝 ≤ 0.62 = 𝑃 𝑧 ≤ −1.06 = 0.1446;

𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 𝑙𝑜𝑤𝑒𝑟:−1000, 𝑢𝑝𝑝𝑒𝑟: 0.62, 𝜇: 0.67, 𝜎: 0.0470 = 0.1437Step 3: There is a 0.1437 probability that 62% or fewer students in an SRS of 100 support a

crackdown on underage drinking.


Homework42. The Harvard College Alcohol Study finds that 67% of college students support

efforts to “crack down on underage drinking.” Does this result hold at a large local

college? To find out, college administrators survey an SRS of 100 students and

find that 62 support a crackdown on underage drinking.

b) A writer in the college’s student paper says that “support for a crackdown is

lower at our school than nationally.” Write a short letter to the editor

explaining why the survey does not support this conclusion.

Because the probability isn’t very small, it is

plausible that the 67% claim is correct and that

the lower than expected percentage is due to

chance alone.

Learning Objectives

After this section, you should be able to:


✓ FIND the mean and standard deviation of the sampling distribution

of a sample mean. CHECK the 10% condition before calculating the

standard deviation of a sample mean.

✓ EXPLAIN how the shape of the sampling distribution of a sample

mean is affected by the shape of the population distribution and the

sample size.

✓ If appropriate, use a Normal distribution to CALCULATE

probabilities involving sample means.

Sample Means


The Sampling Distribution of

When we record quantitative variables we are interested in other

statistics such as the median or mean or standard deviation. Sample

means are among the most common statistics.

Consider the mean household earnings for samples of size 100.

Compare the population distribution on the left with the sampling

distribution on the right. What do you notice about the shape, center,

and spread of each?

x



When we choose many SRSs from a population, the sampling

distribution of the sample mean is centered at the population mean µ

and is less spread out than the population distribution. Here are the

facts.

Sampling Distribution of a Sample Mean

as long as the 10% condition is satisfied: n ≤ (1/10)N.

x

The standard deviation of the sampling distribution of x is

s x =s

n

The mean of the sampling distribution of x is mx = m

Suppose that x is the mean of an SRS of size n drawn from a large population

with mean m and standard deviation s. Then :

Note: These facts about the mean and standard deviation of x are true

no matter what shape the population distribution has.


Sampling From a Normal Population

We have described the mean and standard deviation of the sampling

distribution of the sample mean x but not its shape. That's because the

shape of the distribution of x depends on the shape of the population

distribution.

In one important case, there is a simple relationship between the two

distributions. If the population distribution is Normal, then so is the

sampling distribution of x . This is true no matter what the sample size is.

Sampling Distribution of a Sample Mean from a Normal Population

Suppose that a population is Normally distributed with mean m

and standard deviation s . Then the sampling distribution of x

has the Normal distribution with mean m and standard

deviation s / n, provided that the 10% condition is met.


The Central Limit Theorem

Most population distributions are not Normal. What is the shape of the

sampling distribution of sample means when the population distribution

isn’t Normal?

It is a remarkable fact that as the sample size increases, the distribution

of sample means changes its shape: it looks less like that of the

population and more like a Normal distribution!

When the sample is large enough, the distribution of sample means is

very close to Normal, no matter what shape the population distribution

has, as long as the population has a finite standard deviation.

Draw an SRS of size n from any population with mean m and finite

standard deviation s. The central limit theorem (CLT) says that when n

is large, the sampling distribution of the sample mean x is approximately

Normal.



Consider the strange population distribution

from the Rice University sampling distribution

applet.

Describe the shape of the sampling

distributions as n increases. What do you

notice?



Normal/Large Condition for Sample Means

If the population distribution is Normal, then so is the

sampling distribution of x . This is true no matter what

the sample size n is.

If the population distribution is not Normal, the central

limit theorem tells us that the sampling distribution

of x will be approximately Normal in most cases if

n ³ 30.

The central limit theorem allows us to use Normal probability

calculations to answer questions about sample means from many

observations even when the population distribution is not Normal.

As the previous example illustrates, even when the population

distribution is very non-Normal, the sampling distribution of the sample

mean often looks approximately Normal with sample sizes as small as

n = 25.



x


Mean texts

Suppose that the number of texts sent during a typical day by students

at a particular high school follows a right-skewed distribution with a

mean of 45 and a standard deviation of 35.

Problem: How likely is it that a random sample of 50 students will have

sent more than a total of 2500 texts in the last day?

Let ҧ𝑥 = the average number of texts sent by 50 randomly selected students

𝜇 ҧ𝑥= 𝜇 = 45We must assume that this high school has at least 10(50) = 500 students, so

𝜎 ҧ𝑥 =𝜎

𝑛=

35

50= 4.95

Because of the Central Limit Theorem, we can use the normal distribution

because 𝑛 = 50 ≥ 30.Step 1: 𝑃 ҧ𝑥 > Τ2500

50 = 𝑃( ҧ𝑥 > 50) with N(45, 4.95)

Step 2: 𝑧 =50−45

4.95= 1.01; 𝑃 ҧ𝑥 > 50 = 𝑃 𝑧 > 1.01 = 1 − 𝑃 𝑧 ≤ 1.01 = 1 −

0.8438 = 0.1562; 𝑡𝑒𝑐ℎ: 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 50, 1000,45,4.95 = 0.1562Step 3: The probability that a random sample of students will have sent more

than 2500 in the last day is 0.1562.


Buy me some peanuts and sample means

Problem: At the P. Nutty Peanut Company, dry-roasted, shelled

peanuts are placed in jars by a machine. The distribution of weights in

the jars is approximately Normal, with a mean of 16.1 ounces and a

standard deviation of 0.15 ounces.

(a) Without doing any calculations, explain which outcome is more

likely: randomly selecting a single jar and finding that the contents

weigh less than 16 ounces or randomly selecting 10 jars and finding

that the average contents weigh less than 16 ounces.

Because averages are less variable than individual measurements,

you would expect the sample mean of 10 jars to be closer, on

average, to the true mean of 16.1 ounces. Thus, it is more likely that a

single jar would weigh less than 16 ounces than for the average

weight of 10 jars to be less than 16 ounces.







(b) Find the probability that a randomly selected jar contains less than

16 ounces of peanuts.

Step 1: Let X = the weight of the contents of a randomly selected jar of

peanuts. X has the approximately N(16.1, 0.15) distribution, and we

want to find 𝑃(𝑋 < 16).

Step 2: 𝑧 =16−16.1

0.15= −0.67;𝑃 𝑋 < 16 = 𝑃 𝑧 < −0.67 = 0.2514;

Tech:𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 −1000, 16, 16.1, 0.15 = 0.2525Step 3: There is about a 25% chance that a single jar will contain less

than 16 ounces of peanuts.







(c) Find the probability that 10 randomly selected jars contain less than

16 ounces of peanuts, on average.

Let ҧ𝑥 = the average weight of the contents of 10 randomly selected jars of peanuts.

𝜇 ҧ𝑥 = 𝜇 = 16.1Since it is reasonable that 10 is less than 10% of all jars of peanuts produced, 𝜎 ҧ𝑥 =𝜎

𝑛=

0.15

10= 0.047.

Since the original distribution was approximately Normal, we can use the Normal

approximation here, so…

Step 1: ҧ𝑥 has roughly the N(16.1, 0.047) distribution and we want to find 𝑃( ҧ𝑥 < 16).

Step 2: 𝑧 =16−16.1

0.047= −2.13; 𝑃 ҧ𝑥 < 16 = P z < −2.13 = 0.0166; Tech:

𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 −1000, 16, 16.1, 0.047 = 0.0167.

Step 3: There is about a 1.7% chance that a random sample of 10 jars will contain

less than 16 ounces of peanuts, on average. This confirms our answer to part (a).

Section Summary

In this section, we learned how to…


✓ FIND the mean and standard deviation of the sampling distribution of

a sample mean. CHECK the 10% condition before calculating the

standard deviation of a sample mean.

✓ EXPLAIN how the shape of the sampling distribution of a sample

mean is affected by the shape of the population distribution and the

sample size.

✓ If appropriate, use a Normal distribution to CALCULATE probabilities

involving sample means.

Sample Means


PAGE 46250-58EVEN

Homework

chapter 7 sampling distributions...the practice of statistics, 5th edition starnes, tabor, yates,...

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