STA 291Fall 2009

Lecture 15Dustin Lueker

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Confidence Intervals To calculate the confidence interval, we

use the Central Limit Theorem (np and nq ≥ 5)

Also, we need a that is determined by the confidence level

Formula for 100(1-α)% confidence interval for μ

/ 2z

STA 291 Fall 2009 Lecture 15

nppZp )ˆ1(ˆˆ 2/

Incorrect statement◦ With 95% probability, the population mean will fall

in the interval from 3.5 to 5.2

To avoid the misleading word “probability” we say◦ We are 95% confident that the true population

mean will fall between 3.5 and 5.2

Interpreting Confidence Intervals

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Confidence Interval Changing our confidence level will change

our confidence interval◦ Increasing our confidence level will increase the

length of the confidence interval A confidence level of 100% would require a

confidence interval of infinite length Not informative

There is a tradeoff between length and accuracy◦ Ideally we would like a short interval with high

accuracy (high confidence level)

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The width of a confidence interval◦ as the confidence level increases◦ as the error probability decreases◦ as the standard error increases◦ as the sample size n decreases

Why?

Facts about Confidence Intervals

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Start with the confidence interval formula for a population proportion p

ME denotes the margin of error Mathematically we need to solve the above

equation for n

Choice of Sample Size

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MEpnppZp

ˆ)ˆ1(ˆˆ 2/

MEZppn 2/)ˆ1(ˆ

This formula requires guessing before taking the sample, or taking the safe but conservative approach of letting = .5◦ Why is this the worst case scenario? (conservative

approach)

Choice of Sample Size

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MEZppn 2/)ˆ1(ˆ

p̂

p̂

Example ABC/Washington Post poll (December 2006)

◦ Sample size of 1005◦ Question

Do you approve or disapprove of the way George W. Bush is handling his job as president? 362 people approved

Construct a 95% confidence interval for p What is the margin of error?

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Example If we wanted B=2%, using the sample

proportion from the Washington Post poll, recall that the sample proportion was .36

◦ n=2212.7, so we need a sample of 2213 What do we get if we use the conservative

approach?

9

21.960.36 (1 0.36)0.02

n

STA 291 Fall 2009 Lecture 15

To account for the extra variability of using a sample size of less than 30 the student’s t-distribution is used instead of the normal distribution

Confidence Interval for Unknown σ

10

nstx 2/

STA 291 Fall 2009 Lecture 15

t-distributions are bell-shaped and symmetric around zero

The smaller the degrees of freedom the more spread out the distribution is

t-distribution look much like normal distributions

In face, the limit of the t-distribution is a normal distribution as n gets larger

t-distribution

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Need to know α and degrees of freedom (df)◦ df = n-1

α=.05, n=23◦ tα/2=

α=.01, n=17◦ tα/2=

α=.1, n=20◦ tα/2=

Finding tα/2

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Compute a 95% confidence interval for μ if we know that s=12 and the sample of size 36 yielded a mean of 7

Example

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