sampling distribution & confidence...

Sampling Distribution & Confidence Interval

CI - 1

1

A Normal Distribution

Example: Consider the distribution of serum

cholesterol levels for 40- to 70-year-old

males living in community A has a mean of

211 mg/100 ml, and the standard deviation

of 46 mg/100 ml. If an individual is selected

from this population, what is the probability

that his/her serum cholesterol level is higher

than 225?

2

P(X > 225) = ?

225

0

z

x

211

X ~ N (m = 211, s = 46)

.30

225 - 211

46 = .30 z =

.382

.382

Statistical Inference

• Estimation

• Testing Hypothesis

3 4

Statistics Used to Estimate Population Parameters

Sample Mean,

Sample Variance, s2

Sample Proportion,

…

Estimators

p̂

x m population mean

s 2 population variance

p population proportion

Parameters Statistics

5

Sampling Distribution

Sampling distribution is probability distribution of the sample Statistic.

What is the sampling distribution of mean?

• Shape: Normal

• Parameters: Mean, Standard Deviation

In many situations, mean and standard

deviation can completely determine the

distribution of a specific shape.

6

Sampling Distribution of Mean (Parameters)

nx

ss mm x


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7

Sampling Distribution

s = 2

m = 8

x Population

Distribution

m = 8

x

4.025

2xsSampling

Distribution of Mean (Sample size n=25)

8

Standard Error of Mean

1. Formula

2. Standard Deviation of the sampling distribution of the Sample Means,X

3. Less Than Pop. Standard Deviation

n

s

nx

ss

ss

n

9

Sampling Distribution of Mean (Distribution shape)

Normal distribution theorem: If a random

sample is taken from a normally distributed population, then the sampling distribution of mean would be normal.

Central Limit Theorem: When a relative

large random sample is taken from any

population, regardless of the distribution of

the population, the sampling distribution of

mean would be approximately normal.

10

X

Central Limit Theorem

As

sample

size gets

large

enough

(n 30) ...

sampling

distribution

becomes

almost

normal.

ss

xn

m mx

11

A Random Sample from Population

Population mean = 19.9, standard deviation = 12.6

Random Sample of Size 400 from Population

110.0

100.090.0

80.070.0

60.050.0

40.030.0

20.010.0

0.0

120

100

80

60

40

20

0

Std. Dev = 12.92

Mean = 20.7

N = 400.00

12

Simulated Sampling Distribution of Means

SIZE2

77.073.0

69.065.0

61.057.0

53.049.0

45.041.0

37.033.0

29.025.0

21.017.0

13.09.0

5.01.0

70

60

50

40

30

20

10

0

Std. Dev = 8.88

Mean = 20.3

N = 400.00

n=2 SIZE4

77.073.0

69.065.0

61.057.0

53.049.0

45.041.0

37.033.0

29.025.0

21.017.0

13.09.0

5.01.0

70

60

50

40

30

20

10

0

Std. Dev = 5.40

Mean = 19.4

N = 400.00

n=4 SIZE10

77.073.0

69.065.0

61.057.0

53.049.0

45.041.0

37.033.0

29.025.0

21.017.0

13.09.0

5.01.0

100

80

60

40

20

0

Std. Dev = 4.32

Mean = 19.9

N = 400.00

n=10

SIZE25

77.00

73.00

69.00

65.00

61.00

57.00

53.00

49.00

45.00

41.00

37.00

33.00

29.00

25.00

21.00

17.00

13.009.00

5.001.00

200

100

0

Std. Dev = 2.23

Mean = 19.84

N = 400.00

n=25 SIZE50

77.00

73.00

69.00

65.00

61.00

57.00

53.00

49.00

45.00

41.00

37.00

33.00

29.00

25.00

21.00

17.00

13.009.00

5.001.00

200

100

0

Std. Dev = 1.64

Mean = 19.75

N = 400.00

n=50 SIZE100

77.00

73.00

69.00

65.00

61.00

57.00

53.00

49.00

45.00

41.00

37.00

33.00

29.00

25.00

21.00

17.00

13.009.00

5.001.00

300

200

100

0

Std. Dev = 1.20

Mean = 19.81

N = 400.00

n=100


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13

Probability Related to Mean

Example: Consider the distribution of serum

cholesterol levels for 40- to 70-year-old

males living in community A has a mean of

211 mg/100 ml, and the standard deviation

of 46 mg/100 ml. If a random sample of

100 individuals is taken from this population,

what is the probability that the average

serum cholesterol level of these 100

individuals is higher than 225?

),(n

NX xx

ssmm

14

P(X > 225) = ? Cholesterol Level has a mean 211, sd. 46.

The sampling distribution of the mean is normally distributed.

6.4100

46

211

nx

x

ss

mmParameters of the sampling distribution of the mean:

211

x100

)6.4,211(~

n

NX xx sm

15

P(X > 225) = ?

225

3.04

225 - 211

4.6 = 3.04

.001

Cholesterol Level has a mean 211, s.d. 46.

001.0

)04.3()225(

ZPXP

211

n = 100

x

)6.4,211( xxNX sm

0

z

16

Introduction to Estimation

Confidence Intervals

&

Sample Size

HT - 17

Sampling Distribution of Sample Proportion

Parameter: Population Proportion p (or p)

(Percentage of people has no health insurance)

Statistic: Sample Proportion n

xp ˆ

x is number of successes

n is sample size

Data: 1, 0, 1, 0, 0 4.5

2ˆ p

4.5

00101

x

xp ˆ

HT - 18

For a large random sample of size n from a population with proportion of successes p , and

therefore proportion of failures 1 – p , the sampling distribution of sample proportion,

= x/n, where x is the number of successes in the sample, is approximately normal with

a mean = p

and standard deviation =

under the following two conditions.

p̂

n

pp )1( -

Sampling Distribution of Sample Proportion, 𝑝


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19

Sample Size Condition: The sampling distribution of sample proportion is approximately normal under the assumptions that

• np and n(1-p) > 10, i.e., at least 10

failures and 10 successes in the sample [Central Limit Theorem], and

• p is not too close to 0 or 1.

20

Population Size Condition: The standard deviation of the sampling distribution is when either • population is infinitely large, or • the sample is from a finite population and

the size of the sample is no more than 10% of this population.

n

pp )1( -

21

Disadvantage of Point Estimation

1. Provides Single Value

Based on Observations from 1 Sample.

* Sample Proportion = .32 or 32% is a

Point Estimate of Unknown Population proportion.

2. Gives No Information about How Close Value Is to the Unknown Population Parameter

Which of the following statistics do you prefer? a. 32% b. 32% with a margin of error 3%

p̂

22

Estimation

You’re interested in finding the percentage of people in favor of candidate A?

How can we estimate this average with a measure of reliability?

32% 30% 32% 15 % 32% 3%

23

Interval Estimation for Proportion

Margin of Error Gives Information about How Close the Estimated Value Is to the Unknown Population Proportion.

24

Sampling Error

Sample statistic

(point estimate)

p p̂

Sampling Error = | p – | p̂


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25

Key Elements of Interval Estimation

Sample statistic

(point estimate)

Confidence

limit (lower)

Confidence

limit (upper)

Confidence

interval

Confidence Level: A probability that the

population parameter falls somewhere

within the interval.

32% 3%

Margin of Error p̂26

Za/2 Notation

z

0

a/2 1 - a

a/2

za /2 - za /2

27

Za /2 Notation

z

0

.025

.95

.025

z.025 - z.025

28

A Special Notation

Z .05 .06 .07

1.8 .032 .031 .031

1.9 .026 .025 .024

2.0 .020 .020 .019

2.1 .016 .015 .015

za = the z score that the proportion of

the standard normal distribution to the

right of it is a.

z.025 = ?

0 z.025

1.96

.025

29 30

Sampling Distribution of Proportion

s

p

Within how many standard deviations of

the population proportion, p, will have

95% of the sampling distribution?

.025

.025

p - ?s

.95

p̂

p + ?s p̂p̂


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31

The Confidence Interval

95% Sample

Proportions

p

1- a = .95

Confidence Level

a/2 a/2 = .025

1.96 = z.025 s p̂

p + 1.96s p̂p - 1.96s p̂

p̂Confidence Interval => p - 1.96s p + 1.96s p̂

p̂p̂

n

pp )1( -

n

pp )ˆ1(ˆ -

32

Confidence Interval

Proportion

1. Assumptions

Normal Approximation Can Be Used If np and n(1 – p) are both greater than 10.

2. Confidence Interval Estimate

(for large sample)

) )ˆ1(ˆ

ˆ , )ˆ1(ˆ

ˆ ( 22n

ppzp

n

ppzp

-

-- aa

n

ppp

)ˆ1(ˆzˆ

2

- a

33

95% Samples

s

p

2.5% 2.5%

95 % of

intervals

contain p.

5% do not.


34

Factors Affecting Interval Width

• Data Dispersion (Affects standard error)

• Sample Size (Affects standard error)

• Level of Confidence, 1 - a, (Affects Za/2)

)ˆ1(ˆ

n

pp -

) )ˆ1(ˆ

ˆ , )ˆ1(ˆ

ˆ ( 22n

ppzp

n

ppzp

-

-- aa

Standard Error:

35

90% Samples

95% Samples

99% Samples

m + 1.65s m + 2.58s

s

m+1.96s

m - 2.58s m - 1.65s

m-1.96s

m

Size of Interval

p̂

p̂p̂ p̂

p̂

p̂

p̂

p̂

36

Estimation Example Proportion

A random sample of 400 from a large community showed that 32 have diabetes. Set up a 95% confidence interval estimate for p, the percentage of people that have diabetes.

96.1

08.0400

32ˆ

400

025.2

zz

p

n

a

)ˆ1(ˆ

ˆ2/

n

ppZp

- a

400

)08.1(08.96.108.

-

) %7.10 , %3.5 (

%7.2%8 .027 .08


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37

Thinking Challenge

A member of a health department wish to see what percentage of people in a community will support an environmental policy. Of 200 survey forms sent and received, 35 responded that they support the policy and the rest of them do not support the policy.

Find a 90% confidence interval estimate of the percentage of the population in this community that support the policy?

38

Confidence Interval Solution*

645.1 ,200 175.200

35ˆ

2/ azn,p

)ˆ1(ˆ

ˆ 2/n

ppzp

- a

200

)825(.175.645.1175.

) %92.21 , %08.13 (

4.42%17.5%0442. .175

39

Example:

Researchers wish to estimate the percentage of hospital employees infected by SARS in a certain country. Out of 500 randomly chosen hospital employees, 14 were infected. Find the 95% confidence interval estimate for percentage of hospital employees infected by SARS in this country.

40

Sample Size

to get the largest sample to

achieve the goal.

if pilot study is done.

n

ppp

)ˆ1(ˆzˆ :C.I.

2

- a

n

ppZB

)ˆ1(ˆError ofMargin

2

- a

)ˆ1(ˆ2

2

2 ppB

zn -

a

25.0

or

2

2

2 B

zn

a

41

Sample Size (No prior information on p)

Sample Size Example: If one wishes to do a survey to estimate the population proportion with 95% confidence and a margin of error of 3%, how large a sample is needed?

Za/2 = 1.96; B = .03

n = (1.962/.032) x .25 = 1067.11

A sample of size 1068 is needed.

42

Sample Size (With prior information on p)

Sample Size Example: If one wishes to to estimate the percentage of people infected with West Nile in a population with 95% confidence and a margin of error of 3%, how large a sample is needed? (A pilot study has been done, and the sample proportion was 6%.)

Za/2 = 1.96; B = .03

n = (1.962/.032) x .06 x (1 – .06) = 240.7

A sample of size 241 is needed.

How large a sample was used for pilot study?


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43

Interval Estimation for Mean

Margin of Error Gives Information about How Close the Estimated Value Is to the Unknown Population Mean.

44

Sampling Error in Estimating Mean

Sample statistic

(point estimate)

x m

Sampling Error = | m – x |

45

Key Elements of Interval Estimation

Sample statistic

(point estimate)

Confidence

limit (lower)

Confidence

limit (upper)

Confidence

interval

Confidence Level: A probability that the

population parameter falls somewhere

within the interval.

x Margin of Error

98 1 F

46

(1-a)·100% Confidence Interval Estimate for mean of a normal population

or

) , ( 2/2/n

zxn

zxss

aa -

2/n

zxs

a Margin of Error

Confidence Interval for Mean

(s Known)

“s Known” may mean that we have very good estimate of s.

It is not practical to assume that we know s.

47


95% Sample

Means

s x _

X

m + 1.96sx m - 1.96sx

m

1- a = .95

Confidence Level

a/2 a/2 = .025

1.96 = z.025

x + 1.96sx x - 1.96sx

x

Confidence Interval => 48

Confidence Interval of Mean

(s unKnown and n 30)

(1-a)·100% Confidence Interval Estimate for mean of a population when sample size is relative large

or

) , ( 2/2/n

szx

n

szx - aa

2/n

szx a


CI - 9

49


95% Samples

s x _

X

m + 1.96sx m - 1.96sx

m

x - 1.96sx x + 1.96sx

x

Confidence Interval =>

95% Confidence

Interval

50

95% Samples

s x _

X m

2.5% 2.5%

95 % of

intervals

contain m.

5% do not.


51

Factors Affecting Interval Width

1. Data Dispersion Measured by s

2. Sample Size Affects standard error:

3. Level of Confidence (1 - a) Affects Za/2

n

x

ss

) , ( 2/2/n

zxn

zxss

aa -

52

90% Samples

95% Samples

99% Samples

m + 1.65s x m + 2.58sx

s x _

X

m+1.96s x

m - 2.58s x m - 1.65sx

m-1.96s x

m

Size of Interval

53

Estimation Example Mean (s Known)

The average weight of a random sample of n = 25 subjects isX = 140. Set up a 95% confidence interval estimate for m if s = 10. (Assume Normal population.)

3.92140or ) 92.341 , 08.631 (

) 25

1096.1041 ,

25

1096.1041 (

) , (

1.96. z .025, 2

.05, ,95.1

2/2/

2

-

-

-

nZX

nZX

ss

aaa

aa

a

2/n

zxs

a

143.92) (136.08,

92.3 140 25

1096.1401

54

Interpretation

We can be 95% confident that the population mean is in (136.08, 143.92).

We can be 95% confident that the maximum sampling error using this interval estimate for estimating mean is within 3.92.


CI - 10

55

Confidence Interval of Mean

(s unKnown and n 30)

(1-a)·100% Confidence Interval Estimate for mean of a population when sample size is relative large

or

) , ( 2/2/n

szx

n

szx - aa

2/n

szx a

56

Thinking Challenge

Example: A city uses a certain noise index to monitor the noise pollution at a certain area of the city. A random sample of 100 observations from randomly selected days around noon showed an average index value of x = 1.99 and standard deviation s = 0.05. Find the 90% confidence interval estimate of the average noise index at noon.

57


) 998.1 , 982.1 (

0.008 1.99100

05.64.199.1

1.64z z

.05 /2 .1, 90.1 .90, 1

2/

.052 /

n

szx

--

a

a

aaa

58


In a survey, the BMI for a random sample of 64 individuals who lived in a community were measured. The sample mean was 26.5, and the sample standard deviation was 3.5. Find the 95% confidence interval estimate for the average BMI for people lived in this community.

59

Finding Sample Sizes

for Estimating m

B = Margin of Error or Bound

2

22

2

2

2

Error of Margin

nzx :C.I.

B

zn

nzB

s

s

s

a

a

a

60

Sample Size Example

What sample size is needed to be 90% confident of being correct within 5? A pilot study suggested that the standard deviation is 45.

2202.2195

45645.12

22

2

22

05. B

zn

s


CI - 11

61

Thinking Challenge

You plan to survey residents in your county to find the average health insurance premium that they are paying. You want to be 95% confident that the sample mean is within ± $50. A pilot study showed that s was about $400. What sample size should you use?

62

Sample Size Solution*

24686.245

50

40096.12

22

2

22

025.0

B

zn

s

63

Confidence Interval Mean (s Unknown & n < 30)

1. Assumptions

Population Standard Deviation Is Unknown

Population Must Be Normally Distributed

2. Use Student’s t Distribution

3. Confidence Interval Estimate

) , ( 1,2/1,2/n

stx

n

stx nn - -- aa

n

stx

n

-1 ,2

a64

t

Student’s t Distribution

0

t (df = 5)

Z

Standard

Normal (Z)

Bell-Shaped

Symmetric

‘Fatter’ Tails

t (df = 13)

ns

xt

m-

65

Student’s t Table

t values

t0

.05

For a 90% C.I.:

n = 3

df = n - 1 = 2

a = .10

a/2 =.05

ta/2 = ?

2.920 66


CI - 12

67

Estimation Example Mean (s Unknown)

A random sample of weights of 25 subjects, has a sample mean 140 and sample standard deviation 8. Set up a 95% confidence interval estimate for m.

) 31.341 , 69.631 (

3.31 140 25

8064.2041

064.2

.025, /2 .05,.951 .95, 1

025.024 , /2

--

tt dfa

aaa

1,2/n

stx n -a

68

Thinking Challenge

The numbers of community hospital beds per 1000 population that are available in each different regions of the country is normally distributed. A random sample 6 regions were selected and the rates of beds per 1000 were recorded and they are

3.6, 4.2, 4.0, 3.5, 3.8, 3.1.

Find the 90% confidence interval estimate of the mean bed-rate in the country.

69


= 3.7

s = 0.38987

x

1592.6

38987.

n

s

(use 90% confidence level)

n = 6, df = n - 1 = 6 - 1 = 5

t.05, 5 = 2.015

( 3.7 - (2.015)(0.1592), 3.7 + (2.015)(0.1592) )

3.7 ± 0.321 => ( 3.379, 4.021 )

n

stx n -1 ,2/a

70

Confidence interval with z-score:

The (1- a% confidence interval estimate for population mean:

Assumption: If sampled from normal

population with known variance, s,

Assumption: If large sample and if

unknown variance, s replaces s,

nzx

sa 2/

n

szx 2/a

71

Confidence interval with t-score:

The (1- a% confidence interval estimate for population mean:

Assumption: If sampled from normal

population with unknown variance, s,

n

stx ndf - 1 ,2/a

(If sample size is large the normality assumption is

insignificant.) t z as sample becomes large

72

Average Weight for Female Ten Year Children In US

Info. from a random sample: n = 10, x = 80 lb, s = 18.05 lb, assume weight is normally distributed, find the 95% confidence interval estimate for average weight.

Data: 73.80 50.00 101.40 67.20 102.20 97.80 81.00 93.40 63.20 70.00

How do we know whether normality assumption is OK?


CI - 13

73

Tests of Nor ma lity

.171 10 .200* .930 10 .452weight (pounds) of participantStatistic df Sig. Statistic df Sig.

Kolmogorov-Smirnov a Shapiro-Wilk

This is a lower bound of the true significance.*.

Lilliefors Significance Correctiona.

Both are greater than 0.05, normality assumption is acceptable.

74

Average Weight for Female Ten Year Children In US

Info. from a random sample: n = 10, x = 80 lb, s = 18.05 lb, assume weight is normally distributed, find the 95% confidence interval estimate for average weight.

ta/2 = t.05/2 = t0.025 , d.f. = 10 – 1 = 9, t0.025, 9 = 2.262

10

05.18262.2809,2/

n

stx dfa

)91.92 ,09.67( 91.1280

75

Descriptives

80.0000 5.70840

67.0867

92.9133

80.4333

77.4000

325.858

18.05153

50.00

102.20

52.20

32.5000

-.148 .687

-1.229 1.334

86.8600 3.96048

77.9008

95.8192

86.5222

82.8000

156.854

12.52413

73.80

106.00

32.20

25.5500

.553 .687

-1.422 1.334

Mean

Lower Bound

Upper Bound

95% Conf idence

Interv al for Mean

5% Trimmed Mean

Median

Variance

Std. Dev iation

Minimum

Maximum

Range

Interquart ile Range

Skewness

Kurtosis

Mean

Lower Bound

Upper Bound

95% Conf idence

Interv al for Mean

5% Trimmed Mean

Median

Variance

Std. Dev iation

Minimum

Maximum

Range

Interquart ile Range

Skewness

Kurtosis

What is your sex?

f emale

male

weight (pounds)

of participant

Stat ist ic Std. Error

80 12.91 Weight for Ten Year Old

76


In a survey, the BMI for a random sample of 16 individuals who lived in a community were measured. The sample mean was 26.5, and the sample standard deviation was 3.5. Find the 95% confidence interval estimate for the average BMI for people lived in this community.

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