Sampling Distributions

Chapter 9

Sampling Distributions

1

9.1 Introduction  In real life calculating parameters of populations is prohibitive because populations are very large.  Rather than investigating the whole population, we take a sample, calculate a statistic related to the parameter of interest, and make an inference.  The sampling distribution of the statistic is the tool that tells us how close is the statistic to the parameter. 2

9.2 Sampling Distribution of the Mean  An example 



A die is thrown infinitely many times. Let X represent the number of spots showing on any throw. The probability distribution of X is x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6

E(X) = 1(1/6) + 2(1/6) + 3(1/6)+ ………………….= 3.5 V(X) = (1-3.5)2(1/6) + (2-3.5)2(1/6) + …………. …= 2.92

3

Throwing a die twice – sample mean

 Suppose we want to estimate m from the mean x of a sample of size n = 2.  What is the distribution of x ?

4

Throwing a die twice – sample mean

Sample 1 2 3 4 5 6 7 8 9 10 11 12

1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6

Mean Sample Mean 1 13 3,1 2 1.5 14 3,2 2.5 2 15 3,3 3 2.5 16 3,4 3.5 3 17 3,5 4 3.5 18 3,6 4.5 1.5 19 4,1 2.5 2 20 4,2 3 2.5 21 4,3 3.5 3 22 4,4 4 3.5 23 4,5 4.5 4 24 4,6 5

Sample 25 26 27 28 29 30 31 32 33 34 35 36

Mean 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6

3 3.5 4 4.5 5 5.5 3.5 4 4.5 5 5.5 6 5

Sample 1 2 3 4 5 6 7 8 9 10 11 12

1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6

Mean Sample Mean 1 13 3,1 2 1.5 14 3,2 2.5 2 15 3,3 3 2.5 16 3,4 3.5 3 17 3,5 4 3.5 18 3,6 4.5 1.5 x 19 x4,1 2.5 2 20 4,2 3 2.5 21 4,3 3.5 3 22 4,4 4 3.5 23 4,5 4.5 4 24 4,6 5

Sample 25 26 27 28 29 30 2 31 x 32 33 34 35 36

Mean 5,1 5,2 5,3 5,4 5,5 2 5,6 x 6,1 6,2 6,3 6,4 6,5 6,6

3 3.5 4 4.5 5 5.5 3.5 4 4.5 5 5.5 6

The distribution of x when n = 2 Note : m

m

and







2

E( x ) =1.0(1/36)+ 1.5(2/36)+….=3.5

6/36 5/36

V(X) = (1.0-3.5)2(1/36)+ (1.5-3.5)2(2/36)... = 1.46

4/36 3/36 2/36 1/36

1

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5 6.0

x

6

Sampling Distribution of the Mean n5

n  10

n  25

m x  3 .5

m x  3 .5

m x  3 .5

2

2 x

  . 5833 ( 

x 5

2

2

)

2

6

 x  . 2917 ( 

x 10

)

2 x

  . 1167 ( 

x 25

)

7

Sampling Distribution of the Mean n5

n  10

n  25

m x  3 .5

m x  3 .5

m x  3 .5 2

2

2

 x  . 5833 ( 

x 5

)

2 x

  . 2917 ( 

x 10

2

)

2 x

  . 1167 ( 

x 25

)

Notice that  x2 is smaller than .x. The larger the sample size the 2 smaller  x . Therefore, x tends to fall closer to m, as the sample size increases. 2

8

Sampling Distribution of the Mean Demonstration: The variance of the sample mean is smaller than the variance of the population. Mean = 1.5 Mean = 2. Mean = 2.5

1.5 2.5 22 3 1.5 2.5 22 1.5 2.5 1.5 2 2.5 1.5 2.5 2 Compare the variability of the population 1.5 2.5 1.5 22 of the2.5 to the variability sample mean. 1.5 2.5 1.5 2.5 2 1.5 2.5 1.5 2 2.5 1.5 2 2.5 Let us take samples 1.5 2 2.5 Population

1

of two observations

9

Sampling Distribution of the Mean Also, Expected value of the population = (1 + 2 + 3)/3 = 2

Expected value of the sample mean = (1.5 + 2 + 2.5)/3 = 2

10

The Central Limit Theorem  If a random sample is drawn from any population, the sampling distribution of the sample mean is approximately normal for a sufficiently large sample size.  The larger the sample size, the more closely the sampling distribution of x will resemble a normal distribution.

11

Sampling Distribution of the Sample Mean 1. m x  m x 2. 

2 x





2 x

n

3 . If x is normal, x is normal. x is approximat

ely normally

If x is nonnormal distribute d for

sufficient ly large sample size.

12

Sampling Distribution of the Sample Mean  Example 9.1 





The amount of soda pop in each bottle is normally distributed with a mean of 32.2 ounces and a standard deviation of .3 ounces. Find the probability that a bottle bought by a customer will contain more than 32 ounces. Solution 0.7486  The random variable X is the amount of soda in a bottle. P ( x  32 )  P (

xm x



32  32 . 2

)

.3

 P ( z   . 67 )  0 . 7486

x = 32 m = 32.2 13

Sampling Distribution of the Sample Mean  Find the probability that a carton of four bottles will have a mean of more than 32 ounces of soda per bottle.  Solution 

Define the random variable as the mean amount of soda per bottle.

P ( x  32 )  P (

xm x



32  32 . 2 .3

0.9082

)

4

 P ( z   1 . 33 )  0 . 9082

0.7486 x = 32 x  32

m = 32.2

m x  32 . 2

14

Sampling Distribution of the Sample Mean  Example 9.2 





Dean’s claim: The average weekly income of B.B.A graduates one year after graduation is $600. Suppose the distribution of weekly income has a standard deviation of $100. What is the probability that 25 randomly selected graduates have an average weekly income of less than $550? Solution x  m 550  600 P ( x  550 )  P (

x



100

25

 P ( z   2 . 5 )  0 . 0062

)

15

Sampling Distribution of the Sample Mean  Example 9.2– continued 



If a random sample of 25 graduates actually had an average weekly income of $550, what would you conclude about the validity of the claim that the average weekly income is 600? Solution 



With m = 600 the probability of observing a sample mean as low as 550 is very small (0.0062). The claim that the mean weekly income is $600 is probably unjustified. It will be more reasonable to assume that m is smaller than $600, because then a sample mean of $550 becomes more probable. 16

Using Sampling Distributions for Inference  To make inference about population parameters we use sampling distributions (as in Example 9.2).  The symmetry of the normal distribution along with the sample distribution of the mean lead to: P (  1 . 96  z  1 . 96 )  . 95 , or P (  1 . 96 

- Z.025

Z.025

This can be written P (  1 . 96





 1 . 96 )  . 95

n

as

 x  m  1 . 96

n which

x m



)  . 95

n

become P ( m  1 . 96

 n

 x  m  1 . 96

 n

)  . 95 17

Using Sampling Distributions for Inference Standard normal distribution Z P ( 600  1 . 96

Normal distribution of x 100

 x  600  1 . 96

25

.95

-1.96

0

Z -1.96

m 11. 96 P ( 600 . 96

)  . 95

25

.95

.025 .025

.025

100

 100 25 n

m m600

.025  100

Pm ( 600 1 . 96  1 . 96

x

n 25 18

Using Sampling Distributions for Inference P ( 600  1 . 96

100 25

Which

reduces

 x  600  1 . 96

100

)  . 95

25 to P ( 560 . 8  x  639 . 2 )  . 95

 Conclusion  There is 95% chance that the sample mean falls within the interval [560.8, 639.2] if the population mean is 600.  Since the sample mean was 550, the population mean is probably not 600. 19

9.3 Sampling Distribution of a Proportion  The parameter of interest for nominal data is the proportion of times a particular outcome (success) occurs.  To estimate the population proportion ‘p’ we use the sample proportion. The number of successes

^ = The estimate of p = p

X n 20

9.3 Sampling Distribution of a Proportion ^ p  Since X is binomial, probabilities about can be calculated from the binomial distribution. ^ we prefer to use  Yet, for inference about p normal approximation to the binomial.

21

Normal approximation to the Binomial 

Normal approximation to the binomial works best when 





the number of experiments (sample size) is large, and the probability of success, p, is close to 0.5.

For the approximation to provide good results two conditions should be met: np



5; n(1 - p)



5

22

Normal approximation to the Binomial Example Approximate the binomial probability P(x=10) when n = 20 and p = .5 The parameters of the normal distribution used to approximate the binomial are: m = np; 2 = np(1 - p)

23

Normal approximation to the Binomial Let us build a normal distribution to approximate the binomial P(X = 10).

m = np = 20(.5) = 10; 2 = np(1 - p) = 20(.5)(1 - .5) = 5  = 51/2 = 2.24

P(9.5