AP Statistics Chap 10-1 Confidence Intervals. AP Statistics Chap 10-2 Confidence Intervals...

AP Statistics Chap 10-1

Confidence Intervals



Population Mean

σ Unknown(Lock 6.5)

ConfidenceIntervals

PopulationProportion(Lock 6.1)

σ Known(Lock 5.2)

Central Limit Theorem


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


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


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


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


90% Samples

95% Samples

99% Samples

+1.65x +2.58x

x_

X

+1.96x

-2.58x -1.65x

-1.96x

X= ± Zx


μμx

Interval and Level of Confidence


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


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:


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


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.


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



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


Computer Simulation


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


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


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



with

Unknown Population

Standard Deviation



Population Mean

σ Unknown

ConfidenceIntervals

PopulationProportion

σ Known


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)


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)


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


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


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


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


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


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


Chapter 10


for

Population Proportions



Population Mean

σ Unknown

ConfidenceIntervals

PopulationProportion

σ Known


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


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


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


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


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)


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.


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


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)


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)


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)



End

of

Basic Confidence Intervals

AP Statistics Chap 10-1 Confidence Intervals. AP Statistics Chap 10-2 Confidence Intervals...

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