Use of Chebyshev’s Theorem to Determine Confidence Intervals

1

Use of Chebyshev’s Theorem to Determine Confidence Intervals

• Use when a sample comes from a relatively small sample n>.05N• Non normally distributed• (1 – 1/k2) = C%

k(SE)X

2

Find an 80% and a 90% confidence interval for a population of 500 with a sample size n = 100, X = 45 and a standard deviation s = 12.

Use Chebyshev’s Theorem 100>.05(500)

1 – 1/k2 = .8, k = 2.23 K(SE)X

100

122.2345

[42.324, 47.676]

1 – 1/k2 = .9 k = 3.16 K(SE)X

100

123.1645

[41.208, 48.792]

3

Assumptions

1. The sample is a simple random sample.

2. The conditions for the binomial distribution are satisfied.

3. The normal distribution can be used to approximate the distribution of sample proportions because np 5 and n(1 – p) 5 are both satisfied.

Confidence Intervals for Population Proportions

4

Notation for Proportions

5

p = xn sample proportion

π = population proportion

of x successes in a sample of size n

Notation for Proportions

6

DefinitionPoint Estimate

7

DefinitionPoint Estimate

The sample proportion p is the best point estimate of the population

proportion π.

8

Standard Error

SE =

n

PP

n

)1()1(

9

Confidence Interval for Population Proportion

P + Z(P) * SE

10

Confidence Interval for Population Proportion

P + Z(P) * SE

A poll of 100 students found that 60 prefer Mrs. Peloquinas a teacher compared to Mr. Roesler. Find a 95% and 85%confidence interval that prefer Mrs. Peloquin.

p = .6 (need at least 80 see pg 340)1 - p = .4

.048100

(.6)(.4)

n

p)p(1SE

.6 + 1.96 * .048

.6 + 1.44 * .048

[.505,.694]

[.531,.669]

11

Round-Off Rule for Confidence Interval Estimates of p

Round the confidence interval limits to

three significant digits.

12

Determining Sample Size

13


zME =

p(1-p)n

14


zME =

p(1-p)n

(solve for n by algebra)

15

( )2 p (1-p)


zME =

p(1-p)n

(solve for n by algebra)

zn =

ME2

16

Sample Size for Estimating Proportion p

When an estimate of p is known:

( )2 p (1-p)n =

ME2z

When no estimate of p is known:

( )2 0.25n =

ME2z

17

Example: We want to determine, with a margin of error of four percentage points, the current percentage of U.S. households using e-mail. Assuming that we want 90% confidence in our results, how many households must we survey? A 1997 study indicates 16.9% of U.S. households used e-mail.

18

n = [z/2 ]2 p (1-p)

ME2


19

n = [z/2 ]2 p(1-p)

ME2

= [1.645]2 (0.169)(0.831)

0.042


20

= [1.645]2 (0.169)(0.831)

n = [z/2 ]2 p (1-p)

ME2

= 237.51965= 238 households


To be 90% confident that our sample percentage is within four percentage points of the true percentage for all households, we should randomly select and survey 238 households.

0.042

21

Example: We want to determine, with a margin of error of four percentage points, the current percentage of U.S. households using e-mail. Assuming that we want 90% confidence in our results, how many households must we survey? There is no prior information suggesting a possible value for the sample percentage.

22

n = [z/2 ]2 (0.25)

ME2


23

n = [z/2 ]2 (0.25)

ME2

= (1.645)2 (0.25)

0.042


= 422.81641 = 423 households

24

n = [z/2 ]2 (0.25)

ME2

= (1.645)2 (0.25)

= 422.81641 = 423 households

With no prior information, we need a larger sample to achieve the same results with 90% confidence and an error of no more than 4%.


0.042

26

If 1) n 302) The sample is a simple random sample.3) The sample is from a normally distributed

population.

Case 1 ( is known): Largely unrealistic; Use z-scores

Case 2 (is unknown): Use Student t distribution

Small SamplesAssumptions

27

Student t Distribution

If the distribution of a population is essentially normal, then the distribution of

t =x - µ

sn

28

Student t Distribution

If the distribution of a population is essentially normal, then the distribution of

is essentially a Student t Distribution for all samples of size n.

is used to find critical values denoted by

t =x - µ

sn

t/ 2

29

DefinitionDegrees of Freedom (df )

corresponds to the number of sample values that can vary after certain restrictions have imposed on all data values

30

DefinitionDegrees of Freedom (df )


df = n - 1in this section

31

DefinitionDegrees of Freedom (df ) = n - 1


Any

#Specific

#

so that x = 80

Any

#

n = 10 df = 10 - 1 = 9

Any

#Any

#Any

#Any

#Any

#Any

#Any

#

32

Degreesof

freedom

1234567891011121314151617181920212223242526272829

Large (z)

63.6579.9255.8414.6044.0323.7073.5003.3553.2503.1693.1063.0543.0122.9772.9472.9212.8982.8782.8612.8452.8312.8192.8072.7972.7872.7792.7712.7632.7562.575

.005(one tail)

.01(two tails)

31.8216.9654.5413.7473.3653.1432.9982.8962.8212.7642.7182.6812.6502.6252.6022.5842.5672.5522.5402.5282.5182.5082.5002.4922.4852.4792.4732.4672.4622.327

12.7064.3033.1822.7762.5712.4472.3652.3062.2622.2282.2012.1792.1602.1452.1322.1202.1102.1012.0932.0862.0802.0742.0692.0642.0602.0562.0522.0482.0451.960

6.3142.9202.3532.1322.0151.9431.8951.8601.8331.8121.7961.7821.7711.7611.7531.7461.7401.7341.7291.7251.7211.7171.7141.7111.7081.7061.7031.7011.6991.645

3.0781.8861.6381.5331.4761.4401.4151.3971.3831.3721.3631.3561.3501.3451.3411.3371.3331.3301.3281.3251.3231.3211.3201.3181.3161.3151.3141.3131.3111.282

1.000.816.765.741.727.718.711.706.703.700.697.696.694.692.691.690.689.688.688.687.686.686.685.685.684.684.684.683.683.675

.01(one tail)

.02(two tails)

.025(one tail)

.05(two tails)

.05(one tail)

.10(two tails)

.10(one tail)

.20(two tails)

.25(one tail)

.50(two tails)

Table A-3 t Distribution

33

Given a variable that has a t-distribution with the specified degrees of freedom, what percentage of the time will it be in the indicated region?

a. 10 df, between -1 .81 and 1 .81.

b. 10 df, between -2.23 and 2.23.

c. 24 df, between -2.06 and 2.06.

d. 24 df, between -2.80 and 2.80.

e. 24 df, outside the interval from -2.80 and 2.80.

f. 24 df, to the right of 2.80.

g. 10 df, to the left of -1.81.

90%

95%

95 %

99 %

1 %

.5 %

5 %

34

What are the appropriate t critical values for each of the confidence intervals?

a. 95 % confidence, n = 17 b. 90 % confidence, n = 12c. 99 % confidence, n = 14 d. 90 % confidence, n = 25e. 90 % confidence, n = 13f. 95 % confidence, n = 10

2.12

1.803.01

1.711.78

2.262

35

Margin of Error E for Estimate of Based on an Unknown and a Small Simple Random

Sample from a Normally Distributed Population

where t/ 2 has n - 1 degrees of freedom

nME = t

s2

36

Confidence Interval for the Estimate of ME

Based on an Unknown and a Small Simple Random Sample from a Normally Distributed Population

x - ME < µ < x + ME

where ME = t/2 ns

37

Degreesof

freedom

1234567891011121314151617181920212223242526272829

Large (z)

63.6579.9255.8414.6044.0323.7073.5003.3553.2503.1693.1063.0543.0122.9772.9472.9212.8982.8782.8612.8452.8312.8192.8072.7972.7872.7792.7712.7632.7562.575

.005(one tail)

.01(two tails)

31.8216.9654.5413.7473.3653.1432.9982.8962.8212.7642.7182.6812.6502.6252.6022.5842.5672.5522.5402.5282.5182.5082.5002.4922.4852.4792.4732.4672.4622.327

12.7064.3033.1822.7762.5712.4472.3652.3062.2622.2282.2012.1792.1602.1452.1322.1202.1102.1012.0932.0862.0802.0742.0692.0642.0602.0562.0522.0482.0451.960

6.3142.9202.3532.1322.0151.9431.8951.8601.8331.8121.7961.7821.7711.7611.7531.7461.7401.7341.7291.7251.7211.7171.7141.7111.7081.7061.7031.7011.6991.645

3.0781.8861.6381.5331.4761.4401.4151.3971.3831.3721.3631.3561.3501.3451.3411.3371.3331.3301.3281.3251.3231.3211.3201.3181.3161.3151.3141.3131.3111.282

1.000.816.765.741.727.718.711.706.703.700.697.696.694.692.691.690.689.688.688.687.686.686.685.685.684.684.684.683.683.675

.01(one tail)

.02(two tails)

.025(one tail)

.05(two tails)

.05(one tail)

.10(two tails)

.10(one tail)

.20(two tails)

.25(one tail)

.50(two tails)


38

Important Properties of the Student t Distribution1. The Student t distribution is different for different sample sizes (see Figure

6-5 for the cases n = 3 and n = 12).

2. The Student t distribution has the same general symmetric bell shape as the normal distribution but it reflects the greater variability (with wider distributions) that is expected with small samples.

3. The Student t distribution has a mean of t = 0 (just as the standard normal distribution has a mean of z = 0).

4. The standard deviation of the Student t distribution varies with the sample size and is greater than 1 (unlike the standard normal distribution, which has a = 1).

5. As the sample size n gets larger, the Student t distribution gets closer to the normal distribution. For values of n > 30, the differences are so small that we can use the critical z values instead of developing a much larger table of critical t values. (The values in the bottom row of Table A-3 are equal to the corresponding critical z values from the standard normal distribution.)

39

Student tdistributionwith n = 3

Student t Distributions for n = 3 and n = 12

0

Student tdistributionwith n = 12

Standardnormaldistribution

40

Example: A study of 12 Dodge Vipers involved in collisions resulted in repairs averaging $26,227 and a standard deviation of $15,873. Find the 95% interval estimate of , the mean repair cost for all Dodge Vipers involved in collisions. (The 12 cars’ distribution appears to be bell-shaped.)

41


x = 26,227

s = 15,873

= 0.05/2 = 0.025

42

Degreesof

freedom

1234567891011121314151617181920212223242526272829

Large (z)

63.6579.9255.8414.6044.0323.7073.5003.3553.2503.1693.1063.0543.0122.9772.9472.9212.8982.8782.8612.8452.8312.8192.8072.7972.7872.7792.7712.7632.7562.575

.005(one tail)

.01(two tails)

31.8216.9654.5413.7473.3653.1432.9982.8962.8212.7642.7182.6812.6502.6252.6022.5842.5672.5522.5402.5282.5182.5082.5002.4922.4852.4792.4732.4672.4622.327

12.7064.3033.1822.7762.5712.4472.3652.3062.2622.2282.2012.1792.1602.1452.1322.1202.1102.1012.0932.0862.0802.0742.0692.0642.0602.0562.0522.0482.0451.960

6.3142.9202.3532.1322.0151.9431.8951.8601.8331.8121.7961.7821.7711.7611.7531.7461.7401.7341.7291.7251.7211.7171.7141.7111.7081.7061.7031.7011.6991.645

3.0781.8861.6381.5331.4761.4401.4151.3971.3831.3721.3631.3561.3501.3451.3411.3371.3331.3301.3281.3251.3231.3211.3201.3181.3161.3151.3141.3131.3111.282

1.000.816.765.741.727.718.711.706.703.700.697.696.694.692.691.690.689.688.688.687.686.686.685.685.684.684.684.683.683.675

.01(one tail)

.02(two tails)

.025(one tail)

.05(two tails)

.05(one tail)

.10(two tails)

.10(one tail)

.20(two tails)

.25(one tail)

.50(two tails)


43


x = 26,227

s = 15,873

= 0.05/2 = 0.025t/2 = 2.201

ME = t2 s = (2.201)(15,873) = 10,085.29

n 12

44


x = 26,227

s = 15,873

= 0.05/2 = 0.025t/2 = 2.201

ME = t2 s = (2.201)(15,873) = 10,085.3

n 12

x - ME < µ < x +ME

45


x = 26,227

s = 15,873

= 0.05/2 = 0.025t/2 = 2.201

E = t2 s = (2.201)(15,873) = 10,085.3

n 12

26,227 - 10,085.3 < µ < 26,227 + 10,085.3

x - E < µ < x + E

46


x = 26,227

s = 15,873

= 0.05/2 = 0.025t/2 = 2.201

ME = t2 s = (2.201)(15,873) = 10,085.3

n 12

26,227 - 10,085.3 < µ < 26,227 + 10,085.3

$16,141.7 < µ < $36,312.3

x - ME < µ < x + ME

47


x = 26,227

s = 15,873

= 0.05/2 = 0.025t/2 = 2.201

E = t2 s = (2.201)(15,873) = 10,085.3

n 12

26,227 - 10,085.3 < µ < 26,227 + 10,085.3

x - E < µ < x + E

We are 95% confident that this interval contains the average cost of repairing a Dodge Viper.

$16,141.7 < µ < $36,312.3

48

One Sided Confidence Intervals

A one sided confidence interval is a confidence interval thatestablishes either a likely minimum or a likely maximum valuefor µ, but not both.

Likely maximum

Likely minimum

SE)XZ(X

SE)XZ(X

Use of Chebyshev’s Theorem to Determine Confidence Intervals

Documents

Transcript of Use of Chebyshev’s Theorem to Determine Confidence Intervals