AP Statistics Chap 10-1

Confidence Intervals

AP Statistics Chap 10-2

Confidence Intervals

Population Mean

σ Unknown(Lock 6.5)

ConfidenceIntervals

PopulationProportion(Lock 6.1)

σ Known(Lock 5.2)

Central Limit Theorem

AP Statistics Chap 10-3

Normal if:Asymmetric: 30+Symmetric: 15+Normal: >1

AP Statistics Chap 10-4

Confidence Interval for μ(σ Known)

Assumptions Population standard deviation σ is known Population is normally distributed If population is not normal, use large sample

Confidence interval estimate

n

σzx α/2

AP Statistics Chap 10-5

Common Levels of Confidence

Commonly used confidence levels are 90%, 95%, and 99%

Confidence Level

Confidence Coefficient,

z value,

1.28

1.645

1.96

2.33

2.58

3.08

3.27

.80

.90

.95

.98

.99

.998

.999

80%

90%

95%

98%

99%

99.8%

99.9%

1 /2z

AP Statistics Chap 10-6

Finding the Critical Value

Consider a 95% confidence interval:

z.025= -1.96 z.025= 1.96

.951

.0252

α .025

2

α

Point EstimateLower Confidence Limit

UpperConfidence Limit

z units:

x units: Point Estimate

0

1.96zα/2

Confidence Intervals

90% Samples

95% Samples

99% Samples

+1.65x +2.58x

x_

X

+1.96x

-2.58x -1.65x

-1.96x

X= ± Zx

AP Statistics Chap 10-8

μμx

Interval and Level of Confidence

Confidence Intervals

Intervals extend from

to

100(1-)%of intervals constructed contain μ;

100% do not.

Sampling Distribution of the Mean

n

σzx /2

n

σzx /2

x

x1

x2

/2 /21

AP Statistics Chap 10-9

Margin of Error(MoE)

Margin of Error (e): the amount added and subtracted to the point estimate to form the confidence interval

n

σzx /2

n

σze /2

Example: Margin of error for estimating μ, σ known:

AP Statistics Chap 10-10

Factors Affecting MoE

Data variation, σ : e as σ

Sample size, n : e as n

Level of confidence, 1 - : e if 1 -

n

σze /2

AP Statistics Chap 10-11

Example 1

A sample of 11 circuits from a large normal population has a mean resistance of 2.20 ohms. We know from past testing that the population standard deviation is .35 ohms.

Determine a 95% confidence interval for the true mean resistance of the population.

AP Statistics Chap 10-12

Example 1

A sample of 11 circuits from a large normal population has a mean resistance of 2.20 ohms. We know from past testing that the population standard deviation is .35 ohms.

(continued)

n

σ /2zx Solution:

2.20±1.96(.35/sqrt(11)

2.20±.2068

[1.99, 2.4068]

Z-Interval w/ Calculator

AP Statistics Chap 10-13

AP Statistics Chap 10-14

Interpretation

We are 95% confident that the true mean resistance is between 1.9932 and 2.4068 ohms

Although the true mean may or may not be in this interval, 95% of intervals formed in this manner will contain the true mean

An incorrect interpretation is that there is 95% probability that this

interval contains the true population mean.

(This interval either does or does not contain the true mean, there is no probability for a single interval)

Thinking Challenge

You’re a Q/C inspector for Gallo. The for 2-liter bottles is .05 liters. A random sample of 100 bottles showsX = 1.99 liters. What is the 90% confidence interval estimate of the true mean amount in 2-liter bottles?

2 liter

© 1984-1994 T/Maker Co.

2 liter

Confidence Interval Solution

AP Statistics Chap 10-17

Computer Simulation

AP Statistics Chap 10-18

Example 2

Tim Kelley weighs himself once a week for several years. Last month he weighed himself 4 times with an average of 190.5. Examination of Tim’s past data reveals that over relatively short periods of time, his weight measurements are approximately normal with a standard deviation of about 3. Find a 90% confidence interval for his mean weight for last month. Then, find a 99% confidence interval.

97.192,03.1884

3645.15.190

CI

37.194,63.1864

358.25.190

CI

AP Statistics Chap 10-19

Example 2

Suppose Tim had only weighed himself once last month and that his one observation was x=190.5 (the same as the mean before). Estimate µ with 90% confidence.

(continued)

24.198,76.1821

3645.15.190

CI

AP Statistics Chap 10-20

Example 2

Tim wants to have a margin of error of only 2 pounds with 95% confidence. How many times must he weigh himself to achieve this goal?

(continued)

93

96.12 nn

AP Statistics Chap 10-21

Confidence Intervals

with

Unknown Population

Standard Deviation

AP Statistics Chap 10-22

Confidence Intervals

Population Mean

σ Unknown

ConfidenceIntervals

PopulationProportion

σ Known

AP Statistics Chap 10-23

If the population standard deviation σ is unknown, we can substitute the sample standard deviation, s

This introduces extra uncertainty, since s is variable from sample to sample

So we use the t distribution instead of the normal distribution

Confidence Interval for μ(σ Unknown)

AP Statistics Chap 10-24

Assumptions Population standard deviation is unknown Population is normally distributed If population is not normal, use large sample

Use Student’s t Distribution Confidence Interval Estimate

Confidence Interval for μ(σ Unknown)

n

stx /2

(continued)

AP Statistics Chap 10-27

Student’s t Distribution

t0

t (df = 5)

t (df = 13)t-distributions are bell-shaped and symmetric, but have ‘fatter’ tails than the normal

Standard Normal

(t with df = )

Note: t z as n increases

AP Statistics Chap 10-28

Student’s t Table

Upper Tail Area

df .25 .10 .05

1 1.000 3.078 6.314

2 0.817 1.886 2.920

3 0.765 1.638 2.353

t0 2.920The body of the table contains t values, not probabilities

Let: n = 3 df = n - 1 = 2 = .10 /2 =.05

/2 = .05

AP Statistics Chap 10-29

t distribution values

With comparison to the z value

Confidence t t t z Level (10 d.f.) (20 d.f.) (30 d.f.) ____

.80 1.372 1.325 1.310 1.28

.90 1.812 1.725 1.697 1.64

.95 2.228 2.086 2.042 1.96

.99 3.169 2.845 2.750 2.58

Note: t z as n increases

AP Statistics Chap 10-30

Example

A random sample of n = 25 has x = 50 and s = 8. Form a 95% confidence interval for μ

2.0639t so, 24, 1– n d.f. .025,241n,/2 t

25

8(2.0639)50

n

stx /2

CI = 46.698, 53.302

The confidence interval is:

Thinking Challenge

You’re a time study analyst in manufacturing. You’ve recorded the following task times (min.): 3.6, 4.2, 4.0, 3.5, 3.8, 3.1.What is the 90% confidence interval estimate of the population mean task time?

Confidence Interval Solution*

X = 3.7 S = 3.8987 n = 6, df = n - 1 = 6 - 1 = 5 S / n = 3.99 / 6 = 1.592

t.05,5 = 2.0150

3.7 - (2.015)(1.592) 3.7 + (2.015)(1.592)

0.492 6.908

AP Statistics Chap 10-33

Approximation for Large Samples

Since t approaches z as the sample size increases, an approximation is sometimes used when n 30:

n

stx /2

n

szx /2

Correct formula

Approximation for large n

AP Statistics Chap 10-36

If σ is unknown

If unknown, σ can be estimated when using the required sample size formula

Use a value for σ that is expected to be at least as large as the true σ

Select a pilot sample and estimate σ with the sample standard deviation, s

AP Statistics Chap 10-37

Chapter 10

Confidence Intervals

for

Population Proportions

AP Statistics Chap 10-38

Confidence Intervals

Population Mean

σ Unknown

ConfidenceIntervals

PopulationProportion

σ Known

AP Statistics Chap 10-39

Confidence Intervals for the Population Proportion, p

An interval estimate for the population proportion ( p ) can be calculated by adding an allowance for uncertainty to the sample proportion ( ) p̂

AP Statistics Chap 10-40

Confidence Intervals for the

Population Proportion, p

Recall that the distribution of the sample proportion is approximately normal if the sample size is large, with standard deviation

We will estimate this with sample data:

(continued)

n

)p(1ps

p

n

p)p(1σp

AP Statistics Chap 10-41

Confidence interval endpoints

Upper and lower confidence limits for the population proportion are calculated with the formula

where z is the standard normal value for the level of confidence desired p is the sample proportion n is the sample size

n

)p(pzp /2

1

AP Statistics Chap 10-42

Example

A random sample of 100 people

shows that 25 are left-handed.

Form a 95% confidence interval for

the true proportion of left-handers

AP Statistics Chap 10-43

Example A random sample of 100 people shows

that 25 are left-handed. Form a 95% confidence interval for the true proportion of left-handers.

1.

2.

3.

.0433 .25(.75)/n)/np(1pS

.2525/100 p

p

0.3349 . . . . . 0.1651

(.0433) 1.96 .25

(continued)

AP Statistics Chap 10-44

Interpretation

We are 95% confident that the true percentage of left-handers in the population is between

16.51% and 33.49%.

Although this range may or may not contain the true proportion, 95% of intervals formed from samples of size 100 in this manner will contain the true proportion.

AP Statistics Chap 10-45

Changing the sample size

Increases in the sample size reduce the width of the confidence interval.

Example: If the sample size in the above example is

doubled to 200, and if 50 are left-handed in the sample, then the interval is still centered at .25, but the width shrinks to

.19 …… .31

Estimation Example Proportion

A random sample of 400 graduates showed 32 went to grad school. Set up a 95% confidence interval estimate for p.

Estimation Example Proportion

A random sample of 400 graduates showed 32 went to grad school. Set up a 95% confidence interval estimate for p.

ˆ ˆ ˆ ˆˆ ˆ

α/2 α/2

p×( 1-p) p×( 1-p)p-Z × ≤p≤p+Z ×

n n

.08×( 1-.08) .08×( 1-.08).08-1.96× ≤p≤.08+1.96×

400 400

.053 p .107

Thinking Challenge

You’re a production manager for a newspaper. You want to find the % defective. Of 200 newspapers, 35 had defects. What is the 90% confidence interval estimate of the population proportion defective?

Confidence Interval Solution*

ˆ ˆ ˆ ˆˆ ˆ

a/2 a/2

p×( 1-p) p×( 1-p)p-z × ≤p≤p+z ×

n n

.175×( .825) .175×( .825).175-1.645× ≤p≤.175+1.645×

200 200.1308≤p≤.2192

AP Statistics Chap 10-50

Finding the Required Sample Size

for proportion problems

n

)p(pze /2

1

Solve for n:

Define the margin of error:

2/2

e

)p(pzn

12

p can be estimated with a pilot sample, if necessary (or conservatively use p = .50)

AP Statistics Chap 10-51

What sample size...?

How large a sample would be necessary to estimate the true proportion defective in a large population within 3%, with 95% confidence?

(Assume a pilot sample yields p = .12)

AP Statistics Chap 10-52

What sample size...?

Solution:For 95% confidence, use Z = 1.96E = .03p = .12, so use this to estimate p

So use n = 451

450.74(.03)

.12)(.12)(1(1.96)

e

)p(pzn

2

2/2

2

2 1

(continued)

AP Statistics Chap 10-54

Confidence Intervals

End

of

Basic Confidence Intervals