review of the binomial distribution completely determined by the number of trials (n) and the...

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Review of the Binomial 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.

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Review of the Binomial 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 Population Proportion of

Successes

n

rhatpasreadp )""(ˆ

Point Estimate for q (Population Proportion of

Failures)

phatqasreadq ˆ1)""(ˆ

For a sample of 500 airplane departures, 370 departed on time. Use this information to

estimate the probability that an airplane from the entire

population departs on time.

74.0500

370ˆ n

rp

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

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

pp ˆ

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

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

distribution

npp

zEandnr

pwhere

EppEp

c

)ˆ1(ˆˆ

ˆˆ

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

Is the use of the normal distribution justified?

74.0ˆ500 pn

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

Can we use the normal distribution?

130ˆ370ˆ qnpn

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

5ˆˆ bothareqnandpn

so the use of the normal distribution is justified.

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

confidence interval.

74.0500

370ˆ n

rp

0506.0500

)26(.74.58.2 E

99% confidence interval for the proportion of airplanes that

depart on time:

E = 0.0506

Confidence interval is:

7906.06894.0

0506.074.0506.074.

ˆˆ

p

p

EppEp

99% confidence interval for the proportion of airplanes that

depart on time

Confidence interval is

0.6894 < p < 0.7906

We are 99% confident that between 69% and 79% of the planes depart on time.

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

population.

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

statistical estimate.

Margin of Error

The margin of error is the maximal error of estimate E for a

confidence interval.

Usually, a 95% confidence interval is assumed.

Interpretation of Poll Results

The proportion responding in a certain way is

A 95% confidence interval for population proportion p is:

reportpollp

errorofinmpperrorofinmp

ˆ

argˆargˆ

Interpret the following poll 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 …”

19/20 = 0.95

A 95% confidence interval is being used.

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

84% represents the percentage of households who preferred the

new cereal. .ˆ represents %84 p

“... the results would differ by no more than 3.5 percentage

points in either direction.”

3.5% represents the margin of error, E.

The confidence interval is:

84% - 3.5% < p < 84% + 3.5%

80.5% < p < 87.5%

The poll indicates ( with 95% confidence):

between 80.5% and 87.5% of the population prefer the new

cereal.