Ch. 8-2 Estimating a Population Proportionbrinkhusecr.weebly.com/.../ch._8-2_powerpoint_pdf.pdf · Ch. 8-2 Estimating a Population Proportion . 150 We’ll have 10 students come up

Ch. 8-2 Estimating a Population Proportion

150

We’ll have 10 students come up and take a random sample of 15

beads to make a total of 150. Each person will count how many

purple beads they drew and we’ll get our point estimator.

𝑝 = _________ = _________

Work with a partner. Take

about 5 minutes to try to

come up with a 90%

confidence interval.

Estimate ± Margin of Error

Estimate ± (critical value)(std dev)

how? how? 𝑝

As always, inference is based on the sampling distribution of a

statistic. We went over sampling distributions of a sample proportion

𝑝 in Section 7-2. Here is a brief review of its important properties:

Sampling Dist of 𝑝

Approximately normal if

𝑛𝑝 ≥ 10 & 𝑛 1 − 𝑝 ≥ 10

The mean is 𝑝 if 𝑝 is an unbiased estimator

If 10% cond is satisfied,

then 𝜎𝑝 =𝑝(1 − 𝑝)

𝑛

N 𝑝,𝑝 1−𝑝

𝑛

𝑝

In practice, of course, we don’t know the value of 𝑝. If we did, we wouldn’t need to construct a confidence interval for it!

So we have to modify the way we construct a confidence interval.

The class took an SRS of 150 beads from the chest.

We cannot check whether 𝑛𝑝 or 𝑛 1 − 𝑝 are ≥ 10. However, in large samples, 𝑝 will be close to 𝑝. Therefore, we replace 𝑝 with 𝑝 in checking the Normal condition.

𝑛𝑝 = _______________ ≥ 10 𝑛 1 − 𝑝 = ____________________ ≥ 10

Since we sampled without replacement, we have to check the 10%

condition.

At least 10 150 = 1500 beads need to be in the population

There are over 2000 beads in the chest, so this condition is satisfied.

Estimate ± (critical value)(std dev)

Since we don’t know the

value of 𝑝, we use 𝑝 for 𝜎𝑝 𝑝 ± (critical value)

𝑝 ± (critical value) 𝜎 𝑝

𝑝 ± (critical value)

𝑝 1 − 𝑝

𝑛

𝑝 1 − 𝑝

𝑛

Standard Error

When the standard deviation of a

statistic is estimated from data.

Now what about the critical value? In 8-1, we were using the 68-95-

99.7 rule for the critical values. You will learn how to calculate more

accurate critical values using a calculator or z-table.

Write this every time!!

Since we will be using z-scores, critical values will now be denoted by:

𝑧∗

One-Sample 𝑧-interval for a Population Proportion

Estimate ± (critical value)(std dev of statistic)

𝑝 ± 𝑧∗ 𝑝 1 − 𝑝

𝑛


1) For a 90% confidence level, we

need to catch the central 90% of the

standard Normal distribution.

0.90

Standard

Normal curve

2) In catching the central 90%, we

leave out 10%, or 5% in each tail.

0.05

3) The desired critical value 𝑧∗ is the point with area 0.05 to the right. Use the 𝑧-table to find the point −𝑧∗ with area 0.05 to its left.

4) The point is between 𝑧 = −1.64 and 𝑧 = −1.65. invNorm(0.05,0,1) gives 𝑧 = −1.645, so we use 𝒛∗ = 𝟏. 𝟔𝟒𝟓 as our critical value.

0.05

−𝑧∗ 𝑧∗ = −1.645 = 1.645

N(0,1)

Standard

Normal curve

C%

−𝑧∗ 𝑧∗

1 − C%

2

1 − C%

2

Find 𝑧∗ and use that as the critical value.

N(0,1)

Estimate ± (critical value)(std dev of statistic)

𝑝 ± 𝑧∗ 𝑝 1 − 𝑝

𝑛

______ ± 1.645 ______ 1 − ______

𝑛

(________, ________)

______ ± ______

We are 90% confident that the interval from ________ to ________

captures the actual proportion of purple beads in Mr. Brinkhus’

treasure chest.

Alcohol abuse has been described by college presidents as the number

one problem on campus, and it is an important cause of death in

young adults. How common is it? A survey of 10,904 randomly selected

U.S. college students collected information on drinking behavior and

alcohol-related problems. The researchers defined “frequent binge

drinking” as having five or more drinks in a row three or more times in

the past two weeks. According to this definition, 2486 students were

classified as frequent binge drinkers.

1) Identify the population and the parameter of interest.

2) Check conditions for constructing a confidence interval for the

parameter.

3) Find the critical value for a 99% confidence interval. Show your

method. Then calculate the interval.

4) Interpret the interval in context.









1) Identify the population and the parameter of interest.

2) Check conditions for constructing a confidence interval for the

parameter.

Population: U.S. college students.

Parameter: true proportion who are classified as binge drinkers.

Random: students were chosen randomly.

Normal: 2486 successes and 8414 failures, both are at least 10.

Independent: 10,904 is clearly less than 10% of all U.S. college

students.

All conditions are met.









3) Find the critical value for a 99% confidence interval. Show your

method. Then calculate the interval.

4) Interpret the interval in context.

99% confidence level → 𝑧∗ = 2.576

0.218, 0.238

We are 99% confident that the interval from 0.218 to 0.238

captures the true proportion for U.S. college students who are

binge drinkers.

All Hershey’s kisses tosses

50 kisses tosses

We are 95% confident that the interval from ______ to ______

captures the actual proportion of a Hershey’s kiss landing on

its base.

We want to estimate the actual proportion 𝑝 of ECRCHS students that have seen this movie at a 95% confidence level.

One-sample 𝑧 interval for proportion

Random:

Normal:

Independent:

The sample is an SRS.

𝑛𝑝 ≥ 10

𝑛 1 − 𝑝 ≥ 10

→ 100 .23 = 23 ≥ 10

→ 100 .77 = 77 ≥ 10

So the sampling distribution of 𝑝 is approximately normal.

Sampling without replacement so check 10% condition

There are more than 10(100) = 1000 students at ECRCHS.

𝑝 =23

100= .23

𝑝 ± 𝑧∗ 𝑝 1 − 𝑝

𝑛


.23 ± 1.96 (.23) .77

100

.95 .025 .025

invNorm(.975, 0, 1) = 1.96

𝑧∗ = 1.96

.23 ± 0.0824

0.1475, 0.3124

With calculator:

x:

n:

C-Level:

23 100

0.95

0.14752, 0.31248

We are 95% confident that the interval from 0.1475 to

0.3124 captures the true proportion of the population at

ECRCHS that have seen the particular movie.

STAT TESTS 1-PropZInt (A)

If you’re ever confused about what to put for the context part, take

it from the original question.

𝑝 =23

100= .23

4 volunteers will come up to write each step on the board.

We want to estimate the actual proportion 𝑝 of the U.S. population that classify themselves as Democrat at a 99%

confidence level.

One-sample 𝑧 interval for proportion

Random:

Normal:

Independent:

The sample is an SRS.

𝑛𝑝 ≥ 10

𝑛 1 − 𝑝 ≥ 10

→ 500 .42 = 210 ≥ 10

→ 500 .58 = 290 ≥ 10

So the sampling distribution of 𝑝 is approximately normal.

There are more than 10(500) = 5000 U.S. citizens.

𝑝 ± 𝑧∗ 𝑝 1 − 𝑝

𝑛


.42 ± 2.576 (.42) .58

500

.42 ± 0.0569

0.3631, 0.4769

We are 99% confident that the interval from 0.3631 to

0.4769 captures the true proportion of the U.S.

population that classifies themselves as Democrat.

𝑧∗ = invNorm(.995, 0, 1) = 2.576

With calculator:

x:

n:

C-Level:

210 500

0.99

0.36314, 0.47686

STAT TESTS 1-PropZInt (A)

𝑝 =210

500= .42

𝑧∗𝑝 1 − 𝑝

𝑛

𝑝

This will maximize the

margin of error.

a pilot study or past experiences

Margin of Error = 𝑧∗𝑝 1 − 𝑝

𝑛

0.03 = 1.96(0.5) 0.5

𝑛

Use conservative approach: 𝑝 = 0.5

95% confidence level → 𝑧∗ = 1.96

0.0153 =0.25

𝑛

0.000234 =0.25

𝑛

0.000234𝑛 = 0.25

𝑛 = 1067.11… = 1068

We round up to ensure

that the margin of error is

no more than 3%.

ALWAYS round up!

higher sample size

higher sample size

1) 0.03 = 1.96(0.8)(0.2)

𝑛 → 𝑛 = 683

2)

0.03 = 2.576(0.8)(0.2)

𝑛 → 𝑛 = 1179

This will require a larger sample size.

Documents

Ch. 8-2 Estimating a Population Proportionbrinkhusecr.weebly.com/.../ch._8-2_powerpoint_pdf.pdf · Ch. 8-2 Estimating a Population Proportion . 150 We’ll have 10 students come up