central limit theorem for means - hampden-sydney...

Central LimitTheorem for

Means

Robb T.Koether

HomeworkReview

The CLT forProportions

Computingthe SamplingDistribution ofx

The CentralLimit Theoremfor Means

Assignment

Central Limit Theorem for MeansLecture 27Section 8.4

Robb T. Koether

Hampden-Sydney College

Tue, Mar 3, 2009


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Outline

1 Homework Review

2 The CLT for Proportions

3 Computing the Sampling Distribution of x

4 The Central Limit Theorem for Means

5 Assignment


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Homework Review

Exercise 8.14, page 529

A random sample of 300 holiday shoppers in ashopping mall is selected and 168 are in favor of havinglonger shopping hours.Is this sufficient evidence to conclude that a majority ofall shoppers favor longer shopping hours?Using a 5% significance level, test H0 : p = 0.50 versusH1 : p > 0.50.


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Homework Review

Exercise 8.14, page 529

(a) If the true proportion of shoppers who favor longerhours is 0.50, what is the sampling distribution for thesample proportion p̂?

(b) Compute the observed sample proportion p̂. Using thesampling distribution in part (a), report thecorresponding p-value for this test.

(c) Give your decision and a written conclusion in thecontext of the problem.


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Homework Review

Solution(a) Assuming that p = 0.50, the CLT says that the sampling

distribution of p̂ is normal with mean 0.50 and standard

deviation√

(0.50)(0.50)300 = 0.02887. That is, p̂ is

N(0.50, 0.02887).


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Homework Review

Solution

(b) The observed proportion is p̂ = 168300 = 0.56. The p-value

of 0.56 isnormalcdf(0.56,E99,0.50,0.02887) = 0.0188.


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Homework Review

Solution(c) The decision is to reject H0. The conclusion is that more

than 50% of shoppers at that mall favor longer shoppinghours.


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Proportions and the CLT

Example (Proportions and the CLT)Bag A contains millions of cards, 60% of which arelabeled “yes” and 40% are labeled “no.”Bag B contains millions of cards, 61% of which arelabeled “yes” and 39% are labeled “no.”We are handed one of the two bags, but we do notknow which one.We select 3,000 cards from the bag and compute theproportion of yeses.


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Example (Proportions and the CLT)Let the hypotheses be

H0: The bag is Bag A.H1: The bag is Bag B.

Describe the sampling distribution of p̂ under H0.Describe the sampling distribution of p̂ under H1.


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Example (Proportions and the CLT)The sampling distribution of p̂ under H0:

0.58 0.59 0.60 0.61 0.62 0.63

20

40

60

80

100

120

140


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Assignment



0.58 0.59 0.60 0.61 0.62 0.63

20

40

60

80

100

120

140


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Assignment


Example (Proportions and the CLT)Increase the sample size to n = 10000.Describe the sampling distribution of p̂ under H0.Describe the sampling distribution of p̂ under H1.


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0.58 0.59 0.60 0.61 0.62 0.63

20

40

60

80

100

120

140


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Assignment


Example (Proportions and the CLT)Increase the sample size to n = 30000.Describe the sampling distribution of p̂ under H0.Describe the sampling distribution of p̂ under H1.


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Assignment



0.58 0.59 0.60 0.61 0.62 0.63

20

40

60

80

100

120

140


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Assignment


Example (n = 1000)

0.58 0.59 0.60 0.61 0.62 0.63

50

100

150

200

250

300


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Assignment


Example (n = 2000)

0.58 0.59 0.60 0.61 0.62 0.63

50

100

150

200

250

300


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Assignment


Example (n = 4000)

0.58 0.59 0.60 0.61 0.62 0.63

50

100

150

200

250

300


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Example (n = 8000)

0.58 0.59 0.60 0.61 0.62 0.63

50

100

150

200

250

300


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Example (n = 16000)

0.58 0.59 0.60 0.61 0.62 0.63

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100

150

200

250

300


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Example (n = 32000)

0.58 0.59 0.60 0.61 0.62 0.63

50

100

150

200

250

300


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Example (n = 64000)

0.58 0.59 0.60 0.61 0.62 0.63

50

100

150

200

250

300


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Example (n = 128000)

0.58 0.59 0.60 0.61 0.62 0.63

50

100

150

200

250

300


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Assignment

The Sample Mean

Now we will do the same thing with sample means.


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The Sample Mean

Definition (Sampling Distribution of x)

The sampling distribution of x is the probability distributionof x over all possible samples of a given size n.

The symbol µ represents the population mean.The symbol x represents the sample mean.So x is a random variable because its value varies fromsample to sample.


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The Population

Suppose a population consists of millions of people, 1/4of whom weigh 60 lbs, 1/4 of whom weigh 120 lbs, and1/2 of whom weigh 180 lbs.Then the mean of the population is

µ = 135

and the standard deviation is

σ = 49.75.


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Sample Size n = 1

Take a sample of 1 person.Find the sampling distribution of x.


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Sample Size n = 1

60

180

0.50

0.25

60

180

0.25

0.50

1200.25

120 0.25


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Sample Size n = 1

The probability distribution of x isx P(x)

60 0.25120 0.25180 0.50

µx = 135, and σx = 49.75.


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Assignment

Sample Size n = 2

Now take a sample of 2 people.Find the sampling distribution of x.


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Sample Size n = 2

60

180

0.50

0.25

60

180

0.50

0.25 60 0.0625

1200.25

1200.25

60

180

0.50

0.25

1200.25

60

180

0.50

0.25

1200.25

90 0.0625

120 0.1250

90 0.0625

120 0.0625

150 0.1250

120 0.1250

150 0.1250

180 0.2500


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Sample Size n = 2


60 0.062590 0.1250120 0.3125150 0.2500180 0.2500

µx = 135, and σx = 35.178.


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Sample Size n = 3

Now take a sample of 3 people.Find the sampling distribution of x.


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Assignment

Sample Size n = 3

60

180

0.50

0.25

60

180

0.50

0.25

60 0.0156

1200.25

1200.25

60

180

0.50

0.25

1200.25

60

180

0.50

0.25

1200.25

80 0.0156100 0.031380 0.0156

100 0.0156120 0.0313100 0.0313120 0.0313140 0.0625

80 0.0156100 0.0156120 0.0313100 0.0156120 0.0156140 0.0313120 0.0313140 0.0313160 0.0625

100 0.0313120 0.0313140 0.0625120 0.0313140 0.0313160 0.0625140 0.0625160 0.0625180 0.1250

0.50

0.250.25

0.50

0.250.25

0.50

0.250.25

0.50

0.250.25

0.50

0.250.25

0.50

0.250.25

0.50

0.250.25

0.50

0.250.25

0.50

0.250.25


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Sample Size n = 3


60 0.015680 0.0469100 0.1406120 0.2031140 0.2813160 0.1875180 0.1250

µx = 135, and σx = 28.722.


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Sample Size n = 4

When n = 4, the probability distribution of x isx P(x)

60 0.003975 0.015690 0.0547105 0.1094120 0.1914135 0.2188150 0.2188165 0.1250180 0.0625

µx = 135, and σx = 24.875.


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Sample Size n = 5

When n = 5, the probability distribution of x isx P(x)

60 0.001072 0.004984 0.019596 0.0488108 0.1025120 0.1572132 0.2051144 0.1953156 0.1563168 0.0781180 0.0313

µx = 135, and σx = 22.249.


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Sampling Distribution of x, n = 1

n = 1, µx = 135, and σx = 49.75.

18060 120


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n = 2, µx = 135, and σx = 35.178.

18060 120


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n = 3, µx = 135, and σx = 28.722.

18060 120


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n = 4, µx = 135, and σx = 24.875.

18060 120


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n = 5, µx = 135, and σx = 22.249.

18060 120


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n = 6, µx = 135, and σx = 20.310.

18060 120


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n = 8, µx = 135, and σx = 17.589.

18060 120


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n = 10, µx = 135, and σx = 15.732.

18060 120


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n = 12, µx = 135, and σx = 14.361.

18060 120


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n = 15, µx = 135, and σx = 12.845.

18060 120


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The Central Limit Theorem for Means

Theorem (The Central Limit Theorem for Means)For any population with mean µ and standard deviation σ,the sampling distribution of x has the following mean andstandard deviation:

µx = µ

σx =σ√n.

Furthermore, if the population is normal, then x is normal. Ifthe population is not normal, but large enough, then x isapproximately normal.


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The Central Limit Theorem for Means

The Central Limit Theorem for MeansFor our purposes, n is large enough if it is at least 30.


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Assignment

HomeworkRead Section 8.4, pages 531 - 545.Exercises 17 - 22, page 551.


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Assignment

Answers18. (a) N(69, 0.8333).

(b) N(69, 0.25).(c) The second one has a smaller standard

deviation.


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Assignment

Answers20.

12 13 14 15 16 17

0.1

0.2

0.3

0.4


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Assignment

Answers22. (a)

26 27 28 29 30 31

0.1

0.2

0.3

0.4


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Assignment

Answers22. (a) (i) 27 and 29.

(ii) 26 and 30.(iii) 25 and 31.


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Assignment

Answers22. (c)

26 27 28 29 30 31

0.2

0.4

0.6

0.8


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Assignment

Answers22. (c) (i) It is narrower (twice as narrow).

(ii) It will be half as wide (27.5 to 28.5).(d) No. A value of 26 mpg for x would be 4

standard deviations below average.

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