modular 13 ch 8.1 to 8.2. ch 8.1 distribution of the sample mean objective a : shape, center, and...

Modular 13Ch 8.1 to 8.2

Ch 8.1 Distribution of the Sample Mean

Objective A : Shape, Center, and Spread of the Distribution of x

Objective B : Finding Probability of that is Normally Distributed

x

Ch 8.2 Distribution of the Sample Proportion

Objective A : Shape, Center, and Spread of the Distribution of p̂


p̂

A1. Sampling Distributions of Mean

Assume equal chances for each number to be selected.

This population distribution of is normally distributed.

}3,2,1{S

,3

1)1( P


3

1

1 x

)(xP

2 3

,3

1)2( P

3

1)3( P


x

List out all possible combinations (sample space) and for each combination.

x

1

123

)1,1( 12

11

x )2,1( 5.1

2

21

x )3,1( 2

2

31

x

2123

)1,2( 5.12

12

x )2,2( 2

2

22

x )3,2( 5.2

2

32

x

3123

)1,3( 22

13

x )2,3( 5.2

2

23

x )3,3( 3

2

33

x

Sampling Probability distribution of mean .x

Let’s say we select two elements ( ) from with replacement. (Independent case)

}3,2,1{S2n

9

1

1

)(xP

5.1 2 5.2 3x

9

1

9

2

9

3

9

2

Probability distribution of is summarized in the table shown below.x

Probability histogram for .x

91

1

)(xP

5.1 5.22 3 x

92

93

Let’s compare the distribution shape of and .x x

Is this by chance that is normally distributed?x

91

1

)(xP

5.1 5.22 3 x

92

93

The population distribution of is uniformly distributed .3

1

1 x

)(xP

2 3

The sampling distribution of is normally distributed.

x

x

The mean of the sampling distribution of is .

A2. Central Limit TheoremA. If the population distribution of is normally distributed, the sampling distribution of is normally distributed regardless of the sample size .

xn

If the population distribution is not normally distributed, the sampling distribution of is guaranteed to be normally distributed if . 30n

x

B. Mean/standard deviation of a sampling distribution of vs mean/standard deviation of a population distribution of .

x

x x

The standard deviation of the sampling distribution of is .

nx

x

x

x

The mean and standard deviation of population distribution are and respectively.

x

x x

15,6,27 n

27x

549.115

6

nx

Example 1 : Determine and from the given parameters of the population and the sample size.

x x

Example 2 : A simple random sample of is obtained from a population with and . 64 18

(a) If the population distribution is skewed to the right, what condition must be applied in order to guarantee the sampling distribution of is normally distributed?x

Since the population distribution is not normally distributed, the selected sample size must be greater than or equal to 30 (i.e ).30n

(b) If the sample size is , what must be true regarding the distribution of the population in order to guarantee the sampling distribution of to be normally distributed?

x

9n

For small sample size, the population distribution must be normally distributed in order to guarantee the sampling distribution of to be normally distributed.

x




x




p̂

Standardize to x Z

x

ZRecall : Standardize to : x Z

Now : Standardize to : x Zx

xxZ

Ch 8.1 Distribution of the Sample MeanObjective B : Finding Probability of that is Normally Distributed

x

(a) Describe the sampling distribution .x

336

18

nx

64x

(b) What is ?

Example 1 : A simple random sample of size is obtained from a population mean and population standard deviation .

36n64

18

Since , is normally distributed. 30n x

)6.62( xP

47.03

4.1

3

646.62

x

xxZ

)47.0()6.62( ZPxP

)47.0(ZP

047.0Z

From Table V07.0

3192.04.0

3192.0

3192.0

Example 2: The upper leg of 20 to 29 year old males is normally distributed with a mean length of 43.7cm and a standard deviation of 4.2cm.

(a) What is the probability that a random sample of 12 males who are

20 to 29 years old results in a mean upper leg length that is between 42cm and 48cm?Population is normally distributed.

Since the population distribution is normally distributed, is normally distributed for any sample size.

x

7.43)4842( xP 2.4 12n

2124.112

2.4

nx

7.43x

Z0 55.340.1

x

xxZ

42x

40.12124.1

7.1

2124.1

7.4342

48x

x

xxZ

55.32124.1

3.4

2124.1

7.4348

)55.340.1( ZP

0808.09998.0

From Table V

)areagreenwholethe(919.0

0808.040.1 9998.055.3

9998.0

0808.0

Z0 12.2

(b) A random sample of 15 males who are 20 to 29 years old results in

a mean upper leg length greater than 46 cm. Do you find the result unusual? Why?In order to know if it is unusual or not, we need to find the

probability.

0844.115

2.4

nx

7.43x

15n)46( xP 7.43 2.4

12.20844.1

7.4346

x

xxZ

)12.2( ZP

From Table V

9830.019830.012.2

017.0

9830.0

Since 0.017 is smaller than 0.05, this result is unusual.

017.0




x




p̂

Distribution of the Sample Proportions


A. Sampling distribution of sample proportion , where .p̂

The shape of the sampling distribution of is approximately normally provided by,

p̂

10)1( pnp

,10npqor where pq 1

B. Finding the mean and standard deviation of p̂

pp ˆn

ppp

)1(ˆ

n

xp ˆ





x




p̂

p

ppZ

ˆ

ˆˆ

Standardize to p̂ Z


p̂


where andpp ˆn

ppp

)1(ˆ

provided is approximately normally distributed.p̂

Example 1: A nationwide study indicated that 80% of college students who use a cell phone, send and receive text messages on their phone. A simple random sample of college students using a cell phone is obtained.(a) Describe sampling distribution of .

10)1( pnp

p̂

10)8.01)(8.0)(200(

10)2.0)(8.0)(200(

1032

Since , is normally distributed.10npq p̂

200n

200,8.0 np

Check to see if

006.1Z

(b) What is the probability that 154 or fewer college students in the sample send and receive text messages on the cell phone? Is this unusual?

Sample proportion : 77.0200

154ˆ n

xp

)77.0ˆ( pPLooking for probability :

Standardize to p̂ Z

06.1028284.0

03.0

028284.0

8.077.0ˆ

ˆ

ˆ

p

ppZ

8.0ˆ pp 028284.0200

)8.01)(8.0()1(ˆ

n

ppp

)06.1( ZPFrom Table V

1446.006.1 1446.0

Since 0.1446 is bigger than 0.05, this result is usual.

1446.0

modular 13 ch 8.1 to 8.2. ch 8.1 distribution of the sample mean objective a : shape, center, and...

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