Chapter 7Chapter 7EstimationEstimation

Section 7.3Section 7.3

Estimating p in Estimating p in the Binomial the Binomial DistributionDistribution

Review of theReview of theBinomial DistributionBinomial Distribution

• Completely determined by the number of trials (n) and the probability of success (p) in a single trial.

• q = 1 – p• If np and nq are both > 5, the binomial

distribution can be approximated by the normal distribution.

A Point Estimate for p, the A Point Estimate for p, the Population Proportion of Population Proportion of

SuccessesSuccesses

Point Estimate for qPoint Estimate for q (Population Proportion of (Population Proportion of

Failures)Failures)

For a sample of 500 airplane For a sample of 500 airplane departures, 370 departed on departures, 370 departed on time. Use this information to time. Use this information to estimate the probability that estimate the probability that an airplane from the entire an airplane from the entire population departs on time.population departs on time.

We estimate that there is a 74% chance that any given flight will depart on time.

Error of Estimate for “p hat” Error of Estimate for “p hat” as a Point Estimate for pas a Point Estimate for p

Confidence Interval for p for Confidence Interval for p for Large Samples (np and nq > 5)Large Samples (np and nq > 5)

zc = critical value for confidence level c taken from a normal distribution

p̂−E < p< p̂+E

wherep̂=rn

and E =zcp̂(1−p̂)

n=zc

p̂q̂n

For a sample of 500 airplane For a sample of 500 airplane departures, 370 departed on departures, 370 departed on time. Find a 99% confidence time. Find a 99% confidence interval for the proportion of interval for the proportion of airplanes that depart on time.airplanes that depart on time.

Is the use of the normal distribution justified?

n =500 p̂≈0.74

For a sample of 500 airplane For a sample of 500 airplane departures, 370 departed on departures, 370 departed on time. Find a 99% confidence time. Find a 99% confidence interval for the proportion of interval for the proportion of airplanes that depart on time.airplanes that depart on time.

Can we use the normal distribution?

For a sample of 500 airplane For a sample of 500 airplane departures, 370 departed on departures, 370 departed on time. Find a 99% confidence time. Find a 99% confidence interval for the proportion of interval for the proportion of airplanes that depart on time.airplanes that depart on time.

So the use of the normal distribution

is justified.

Out of 500 departures, 370 Out of 500 departures, 370 departed on time. Find a departed on time. Find a 99% confidence interval.99% confidence interval.

r =370 p̂=rn

n=500 p̂=370500

p̂=0.74

q̂=130500

q̂=0.26

99% confidence interval for the proportion 99% confidence interval for the proportion of airplanes that depart on time:of airplanes that depart on time:

Confidence interval is:

0.74 −2.57580.74(0.26)

500< p< 0.74 + 2.5758

0.74(0.26)500

0.74 −2.57580.1924500

< p< 0.74 + 2.57580.1924500

0.74 −2.5758 0.003848 < p< 0.74 + 2.5758 0.0003848

0.74 −2.5758 0.01961( ) < p< 0.74 + 2.5758 0.01961( ).74 −0.0505 < p< .74 + 0.0505

0.6895 < p< 0.7905

We can say with 99% confidence that the population proportion of planes that depart on

time is between 0.6895 and 0.7905.

The point estimate and the The point estimate and the confidence interval do not confidence interval do not depend on the size of the depend on the size of the

population. population.

The sample size, however, The sample size, however, does affect the accuracy of does affect the accuracy of

the statistical estimate.the statistical estimate.

Margin of ErrorMargin of Error

the maximal error of estimate E for a confidence interval

E =±zcp̂q̂n

Interpretation of Poll ResultsInterpretation of Poll Results

The proportion responding in a certain way is

A 95% confidence interval A 95% confidence interval for population proportion for population proportion

p is:p is:

p̂−marginof error < p< p̂+marginof error

p̂=poll report

Interpret the following poll Interpret the following poll results:results:

“ A recent survey of 400 households indicated that 84% of the households surveyed preferred a new breakfast cereal to their previous brand. Chances are 19 out of 20 that if all households had been surveyed, the results would differ by no more than 3.5 percentage points in either direction.”

““Chances are 19 out of 20 …”Chances are 19 out of 20 …”

19/20 = 0.95A 95% confidence interval

is being used.

““... 84% of the households ... 84% of the households surveyed preferred …”surveyed preferred …”

84% represents the percentage of households who preferred

the new cereal.

84% represents p̂.

““... the results would differ by ... the results would differ by no more than 3.5 percentage no more than 3.5 percentage points in either direction.”points in either direction.”

3.5% represents the margin of error, E.

The confidence interval is:The confidence interval is:

84% - 3.5% < p < 84% + 3.5%80.5% < p < 87.5%

We cay say with 95% confidence that the population proportion of people that would prefer the new cereal is

between 80.5% and 87.5%.

Sample Size for Estimating p for the Binomial Distribution

Do you have a preliminary study?

Formula for Minimum Sample Size for Estimating p for the

Binomial Distribution If p is an estimate of

the population proportion,

n= p̂q̂zcE

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2

Formula for Minimum Sample Size for Estimating p for the

Binomial Distribution If we have no preliminary estimate for p, the

probability is at least c that the point estimate r/n for p will be in error by less than the

quantity E if n is at least:

n=14

zcE

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2

The manager of a furniture store wishes to estimate the proportion

of orders delivered by the manufacturer in less than three

weeks. She wishes to be 95% sure that her point estimate is in error

either way by less than 0.05. Assume no preliminary study is

done to estimate p.

She wishes to be 95% sure ...

z0.95 = 1.9600

... that her point estimate is in error either way by less than

0.05.

E = 0.05

... no preliminary study is done to estimate p.

The minimum required sample size would be 385 deliveries to construct a 95% confidence interval for the proportion of deliveries completed within 3 weeks with an error of no more than 0.05 and no preliminary study.

n=14

1.96000.05

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2

n=14

39.2( )2

n=14

1536.64( )

n=384.16n=385

A preliminary estimate of p indicated that p was

approximately equal to 0.75:

Determine the minimum required sample size if a preliminary study had

been conducted.

n= 0.75( ) 0.25( )1.96000.05

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2

n= 0.75( ) 0.25( ) 39.2( )2

n= 0.75( ) 0.25( ) 1536.64( )n=288.12n=289

The minimum required sample size would be 289 deliveries to construct a 95% confidence interval for the proportion of deliveries completed within 3 weeks with an error of no more than 0.05 and a preliminary study indicating that the proportion of deliveries that were completed within 3 weeks was 0.75.