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Chapter 9 Estimating the Value of a Parameter

Chapter 9.1 Estimating a Population Proportion

Objective A :Point Estimate

A point estimate is the value of a statistic that estimates the value of a parameter.

The best point estimate of the population proportion is a sample proportion ( n

xp ˆ ).

The best point estimate of the population mean is a sample mean (x

xn

).

Since p̂ varies from sample to sample, we use an interval based on p̂ to capture the unknown

population proportion with a level of confidence.

Objective B :Confidence Interval

A confidence interval for an unknown parameter consists of an interval of numbers based on

a point estimate.

The level of confidence represents the expected proportion of intervals that will contain the

parameter if a large number of different samples is obtained.

The level of confidence is denoted as 1 100% . The level of confidence controls the width

of the interval.

Confidence interval estimates for a parameter are of the form:

Point estimate margin of error.

Confidence interval for p :

pZp ˆ2/ˆ where

n

ppp

)ˆ1(ˆˆ

provided that 10)ˆ1(ˆ ppn .

The value of /2Z is called the critical value of the distribution.

The margin of error, E , in a 1 100% confidence interval for a population proportion is

given by n

ppZE

)ˆ1(ˆ2/

. The width of the interval is determined by the margin of error.

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Example 1:Use StatCrunch to determine the critical value /2Z that corresponds to the given level

of confidence.

(a) 90% (b) 95% (c) 98% (d) 92%

(a) 90% level of confidence represents the middle area under the standard normal distribution.

The total two tail areas is called α which is 10% (100% 90%). Thus each tail area is α

2 = 5% due to symmetry.

Open StatCrunch → Select Stat → Calculator → Normal → Standard, µ=0, = 1 →

Input P (X ≤ _____) = 0.95→ Compute

𝑍𝛼/2 = 1.645

(b) 95% level of confidence

The total tail area is α = 5%.

Each tail area is α

2 = 0.25% due to symmetry.

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Open StatCrunch → Select Stat → Calculator → Normal → Standard, µ=0, = 1 →

Input P (X ≤ _____) = 0.975 → Compute

𝑍𝛼/2 = 1.96

(c) 98% level of confidence

The total tail area is α = 2%.

Each tail area is α

2 = 0.01% due to symmetry.

Similarly --> Open StatCrunch → Select Stat → Calculator → Normal → Standard µ=0, = 1 →

Input P (X ≤ _____) = 0.99 → Compute

𝑍𝛼/2 = 2.33

(d) 92% level of confidence

The total tail area is α = 8%.

Each tail area is α

2 = 0.04% due to symmetry.

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Open StatCrunch → Select Stat → Calculator → Normal → Standard →

Input P (X ≤ _____) = 0.96 → Compute

𝑍𝛼/2 = 1.75

Example 2: Determine the margin of error for p with 540x and 900n at a 99% level of confidence.

𝑝 = 𝑥

𝑛 =

540

900 = 0.60

𝐸 = 𝑍𝛼/2 * 𝑝 (1−𝑝 )

𝑛

Open StatCrunch → Select Stat → Calculator → Normal → Standard →

Input P (X ≤ _____) = 0.99 → Compute

𝑍𝛼/2 = 2.58

E = 2.58 * √ 0.60 (0.40)

900 = 0.0421

Example 3: A Rasmussen Reports national survey of 1000 adult Americans found that 18% dreaded

Valentine's Day. Construct a 95% confidence interval for the population proportion of adult

Americans who dread Valentine's Day. Explain what does the interval mean?

n = 1000 p = 0.18 95% confidence

Method I --> Use the C. I. formula --> 𝑝 ± 𝐸

𝐸 = 𝑍𝛼/2 * 𝑝 (1−𝑝 )

𝑛

Open StatCrunch → Select Stat → Calculator → Normal → Standard →

Input P (X ≤ _____) = 0.95 → Compute

𝑍𝛼/2 = 1.96

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𝑍𝛼/2 * √ 0.18 (0.82)

1000 = 1.96*√0.1476

1000= ≈ 0.023716

The confidence interval is 𝑝 ± 𝐸 = 0.18 ± 0.0237 = (0.1563, 0.2037).

This interval means that we are 95% confident that 15.63% to 20.37% of adult Americans

dreaded Valentine’s Day.

Method II --> Use StatCrunch to find the C. I.

Open StatCrunch → Stat → Proportion Stats → One Sample → with summary → Input number

of successes --> 180 (18% of 1000) and observations --> 1000, select confidence interval and

input 95% → compute and record results

Example 4: Construct a confidence interval of the population proportion at the given level of confidence.

80, 200, 96%x n confidence

𝑝 = 𝑥

𝑛 =

80

200= 0.4

Method I--> Use the C. I. formula --> 𝑝 ± 𝐸

Open StatCrunch → Select Stat → Calculator → Normal → Standard →

Input P (X ≤ _____) = 0.98 → Compute

𝑍𝛼/2 ≈ 2.05.

𝐸 = 𝑍𝛼/2 * 𝑝 (1−𝑝 )

𝑛

𝑍𝛼/2 * √ 0.4 (0.6)

200 = 2.05*√0.24

200= = 2.05*0.0346 = 0.071

The 96% confidence interval is 𝑝 ± 𝐸 = 0.40 ± 0.071 = (0.329, 0.471).

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Method II --> Use StatCrunch to find the C. I.

Open StatCrunch → Stat → Proportion Stats → One Sample → with summary → Input number

of successes --> 80 and observations --> 200, select confidence interval and input 96% →

compute and record results

Example 5: In a study of 1228 randomly selected medical malpractice lawsuits, it is found that 856

of them were later dropped or dismissed.

(a) What is the best point of estimate of the proportion of medical malpractice lawsuits that

are dropped or dismissed?

The best point estimator for 𝑝 is 𝑝 .

𝑝 = 856

1228 ≈ 0.6971

(b) Use StatCrunch to construct a 99% confidence interval for the population proportion of

medical malpractice lawsuits that are dropped or dismissed?

Open StatCrunch → Stat → Proportion Stats → One Sample → with summary → Input number

of successes --> 856 and observations --> 1228, select confidence interval and input 99% →

compute and record results

Proportion Count Total Sample Prop. Std. Err. L. Limit U. Limit

p 856 1228 0.6970684 0.013113264 0.66329087 0.73084593

(c) Interpret the interval.

The 99% level of confidence interval for p is between 0.6633 and 0.7308. We are 99% confident

that the percentage of malpractice lawsuits that are dropped or dismissed is between 66.33% and

73.08%.

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Objective C : Sample Size Needed for Estimating the Population Proportion p

The sample size required to obtain a 1 100% confidence interval for p with a

margin of error E is given by

2

2/)ˆ1(ˆ

E

Zppn Round up to the next integer

p̂ is a prior estimate of p

If a prior estimate of p is unavailable, the sample size required is

2

/20.25Z

nE

Round up to the next integer

Example 1 : An urban economist wishes to estimate the proportion of Americans who own

their homes. What size sample should be obtained if he wishes the estimate to be

within 0.02 with 90% confidence if

(a) he uses a 2010 estimate of 0.669 obtained from the U.S Census Bureau?

𝑝 = 0.669 E = 0.02 90% confidence

Use StatCrunch to find 𝑍𝛼/2 for a 90% level of confidence → 𝑍𝛼/2= 1.645

2

2/)ˆ1(ˆ

E

Zppn = ( 0.669)(1-0.669)

1.645

0.02

2

= (0.669) (0.331) 1.645

0.02

2

≈ 1498.4867

Round up to the next whole number -->1499.

(b) he does not use any prior estimates?

n = 0.25 𝑍𝛼/2

𝐸

2

= 0.25 1.645

0.02

2

= 1691.265625

Round up to the next whole number1692.

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Example 2: In a Gallup poll conducted in October 2010, 64% of the people polled answered

"more strict" to the following question: "Do you feel that the laws covering the sale

of firearms should be made more strict as they are now?" Suppose the margin of

error in the poll was 3.5% and the estimate was made with 95% confidence. At

least how many people were surveyed?

E = 0.035 95% confidence

95% level of confidence → 𝑍𝛼/2 = 1.96

2

2/)ˆ1(ˆ

E

Zppn

= (0.64)(1-0.64) 1.96

0.035

2

= (0.64)(0.36) 1.96

0.035

2

= 722.5344

Round up to 723 people.

Example 3: A Gallup poll conducted in November 2010 found that 493 of 1050 adult Americans

believe it is the responsibility of the federal government to make sure all Americans

have healthcare coverage.

(a) Obtain a point estimate for the proportion of adult Americans who believe it is the

responsibility of the federal government to make sure all Americans have health care coverage.

The best point estimate for p is 𝑝 .

(b) Verify the requirements for constructing a confidence interval for p are satisfied.

(1050)(0.470)(1-0.470) ≥ 10

(1050)(0.470)(0.530) ≥ 10

≈ 262 ≥ 10

Yes, 𝑝 is normally distributed.

64.0ˆ p

470.01050

493ˆ p

10)ˆ1(ˆ ppn

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(c) Use StatCrunch to construct a 95% confidence interval for the proportion of adult

Americans who believe it is the responsibility of the federal government to make sure

all Americans healthcare coverage. Interpret the interval.

Open StatCrunch → Stat → Proportion Stats → One Sample → with summary →Input number

of successes - 493 and observations – 1050, select confidence interval and input 95% → compute

and record results

Proportion

Count Total Sample Prop. Std. Err. L. Limit U. Limit

p 493 1050 0.46952381 0.015401645 0.43933714 0.49971048

The 95% level of confidence interval for p is between 0.4393 and 0.4997. We are 95% confident

that the percentage of Americans that believe it is the responsibility of the federal government to

make sure that all Americans have healthcare coverage is between 43.93% and 49.97% .

(d) You wish to conduct your own study for the proportion of adult Americans who believe it is

the responsibility of the federal government to make sure all Americans have healthcare

coverage. What sample size would be needed for the estimate to be within 3 percentage

points with 90% confidence if you use the estimate obtained in part (a).

Use StatCrunch to estimate 𝑛 for 𝑝.

Open StatCrunch → Stat → Proportion Stats → One Sample → Power/Sample size

Confidence Interval Input confidence level: 0.90, Target Proportion: 0.470 ( 𝑝 =495

1050≈

0.470) and Width: .06 (𝐸 = 0.03., Total Width = 2𝐸) compute and record results.

The sample size needed to be within 3 percentage points is 749.

(e) You wish to conduct your own study for the proportion of adult Americans who believe it

is the responsibility of the federal government to make sure all Americans have healthcare

coverage. What sample size would be needed for the estimate to be within 3 percentage

points with 90% confidence if you do not have a prior estimate?

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Open StatCrunch → Stat → Proportion Stats → One Sample → Power/Sample Size →

Confidence Interval→ Input confidence level: 0.90, Target Proportion: 0.50 (no prior estimate

of 𝑝, we use 𝑝 = 0.5) and Width: 0.06 (𝐸 = 0.03., Total Width = 2𝐸) → compute and record

results.

The sample size needed to be within 3 percentage points is 752.

Chapter 9.2 Estimating a Population Mean

Objective A :Point Estimate

The best point estimate of the population mean, , is the sample mean, x .

Objective B :Student's t - distribution

Properties of the t - distribution

1. The t - distribution is different for different degrees of freedom ( 1df n ).

2. The t - distribution has the same general symmetric bell shape as the standard normal

distribution but its area in the tails is a little greater than the area in the tails of the standard

normal distribution due to the greater variability that is expected with small samples.

3. The t - distribution has a mean of 0t at the center of the distribution.

4. As the sample size n gets larger, the t - distribution gets closer to the standard normal

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distribution.

Example 1: Use StatCrunch to determine the t -value.

(a) Find the t -value such that the area in the right tail is 0.05 with 19 degreesof freedom.

Open StatCrunch → Select Stat → Calculator → T → Input DF = 19 and P (X ≥___) = 0.95 →

compute and record results.

t-value = 1.7291328

(b) Find the t -value such that the area left of the t -value is 0.02 with 6 degrees of freedom.

Open StatCrunch → Select Stat → Calculator → T → Input DF = 6 and P (X ≤___) = 0.028 →

compute and record results.

t-value = -2.3637046

(c) Find the critical t -value that corresponds to 90% confidence. Assume 12 degrees of

freedom.

Open StatCrunch → Select Stat → Calculator → T → Input DF and P (X ≤___) = 0.95

(90% is the middle area and 5% is each tail area ) → compute and record results.

t-value = ±1.7822876

In general, the population standard deviation is unknown for estimating a population mean based

on a sample mean. The t -distribution is used to off-set the additional variability introduced by using

s in place of .

Objective C :Confidence Interval for a Population Mean

Constructing a 1 100% Confidence Interval for

Point estimate margin of error

/2

sx t

n where

n

stE 2/ .

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provided the data come from a population that is normally distributed, or the sample size is

large.

Example 1: A simple random sample of size n < 30has been obtained. From the normal probability

plot and boxplot, judge whether a t -interval should be constructed.

(a)

All the data lie within the bounds of the normality probability plot, indicating

the data could come from a population that is normally distributed.

The box plot looks like an almost bell-shaped curve without outliers. The

requirements for constructing a t- interval are satisfied.

(b)

One data point lies outside the bounds of the normality probability plot, indicating the data may

not come from a population that is normally distributed.

The box plot is skewed to the left. t-interval should not be used.

Example 2: A simple random sample of size n is drawn from a population that is normally distributed.

The sample mean, x , is found to be 50, and the sample standard deviation, s , is found to be 8.

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(a) Use StatCrunch to construct a 98% confidence interval for if the sample size, n , is 20.

Open StatCrunch→ Select Stat → T Stats → One sample → with summary → Input Sample

mean 50, Sample std. dev. 8, Sample size 20

98% confidence interval results:

μ : Mean of population Mean Sample Mean Std. Err. DF L. Limit U. Limit

μ 50 1.7888544 19 45.457234 54.542766

𝐸 = (𝑈. 𝐿𝑖𝑚𝑖𝑡 − 𝐿. 𝐿𝑖𝑚𝑖𝑡) ÷ 2 = (54.542766 - 45.457234)/2 = 4.545

(b) Use StatCrunch to construct a 98% confidence interval for if the sample size, n , is 15.

How does decreasing the sample size affect the margin of error, E ?

98% confidence interval results:

μ : Mean of population Mean Sample Mean Std. Err. DF L. Limit U. Limit

μ 50 2.0655911 14 44.578868 55.421132

𝐸 = (𝑈. 𝐿𝑖𝑚𝑖𝑡 − 𝐿. 𝐿𝑖𝑚𝑖𝑡) ÷ 2 = (55.421132 - 44.578868)/2 = 5.421132

Decreasing the sample size increases the margin of error.

(c) Construct a 95% confidence interval for if the sample size, n , is 20.

95% confidence interval results:

μ : Mean of population

Mean Sample Mean Std. Err. DF L. Limit U. Limit

μ 50 1.7888544 19 46.255885 53.744115

Compare the results to those obtained in part (a).

How does decreasing the level of confidence affect the margin of error, E ?

𝐸 = (𝑈. 𝐿𝑖𝑚𝑖𝑡 − 𝐿. 𝐿𝑖𝑚𝑖𝑡) ÷ 2 = (53.744115 - 46.255885)/2 = 3.744115

Decreasing the level of confidence decreases the margin of error.

(d) Could we have computed the confidence intervals in parts (a) to (c) if the population

had not been normally distributed? Why?

The requirement of using a t-interval of 𝑛 < 30 is the sample data must be

obtained from a population that is normally distributed in order to guarantee the

sample data are approximately normally distributed.

(Recall: The t - distribution has the same general symmetric bell shape)

If the population had not been normally distributed, there was no guaranteed that

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𝑥 would have been normally distributed with 𝑛 = 15 or 20 (small sample size).

Thus we could not compute the t-interval.

Example 3: Determine the point estimate of the population mean and margin of error for the

following confidence interval.

Lower bound: 5

Upper bound: 23

The point estimate of the population mean is (5+23)/2 = 28/2 = 14 <-- Similar to finding

the midpoint between two endpoints.

The margin of error is (23-5)/2 = 18/2 = 9

Example 4 : How much time do Americans spend eating or drinking? Suppose for a random

sample of 1001 Americans age 15 or older, the mean amount of time spent eating

or drinking per day is 1.22 hours with a standard deviation of 0.65 hour.

(a) A histogram of time spent eating and drinking each day is skewed right. Use this

result to explain why a large sample size is needed to construct a confidence interval

for the mean time spent eating and drinking each day.

Since the population distribution is not normally distributed, 𝑥 is guaranteed to be

normally distributed when 𝑛 ≥ 30. Thus a large sample size is necessary to guarantee

that the mean amount of time spent eating or drinking per day is normally distributed.

(b) Use StatCrunch to determine and a 95% confidence interval for the mean amount of

time Americans age 15 or older spend eating and drinking each day.

Interpret the interval.

95% confidence interval results:

μ : Mean of population Standard deviation = 0.65 Mean n Sample Mean Std. Err. L. Limit U. Limit

μ 1001 1.22 0.020544535 1.1797335 1.2602665

The 95% level of confidence interval for µ is between 1.1797 and 1.2603. We are 95%

confident that the percentage of Americans, age 15 or older, spend between 1.18 hours

and 1.26 hours eating and drinking each day.

(c) Could the interval be used to estimate the mean amount of time a 9-year-old American

spends eating and drinking each day? Explain

No, our sample data statistics is pertaining to 15 year old Americans, only.

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Objective D : Determining the Sample Size n The sample size required to estimate the population mean, , with a level of confidence

1 100% within a specified margin of error, E , is given by

2

/2Z sn

E

where n is rounded up to the nearest whole number.

Note: The t -distribution approaches the standard normal z - distribution as the sample size increases.

Example 1: A researcher wanted to determine the mean number of hours per week(Sunday through

Saturday) the typical person watches television. Results from the Sullivan Statistics

Survey indicate that 7.5s hours.

(a) How many people are needed to estimate the number of hours people watch television

per week within 2 hours with 95% confidence?

S = 7.5, E = 2, 95% level of confidence

Open StatCrunch → Stat → Z Stats → One Sample → Power/Sample Size → Select confidence

interval Confidence Interval: 0.95, Standard deviation: 7.5 and width of 4 → compute and

record results.

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55 people are needed to estimate the number of hours people watch television per week

within 2 hours with 95% confidence.

(b) How many people are needed to estimate the number of hours people watch television

per week within 1 hour with 95% confidence?

S = 7.5, E = 1, 95% level of confidence

Open StatCrunch → Stat → Z Stats → One Sample → Power/Sample Size → Select confidence

interval Confidence Interval: 0.95, Standard deviation: 7.5 and width of 2 → compute and

record results.

217 people are needed to estimate the number of hours people watch television per week

within 1 hour with 95% confidence.

(c) What effect does doubling the required accuracy have on the sample size?

217/55 = 3.94545...

Quadruple the sample size.

Chapter 9 Estimating a Population Standard Deviation (Supplementary Materials)

Objective A : Point Estimate

The best point estimate of the population variance, 2 , is the sample variance, 2s .

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Objective B : Chi-Square Distribution

Example 1: Use StatCrunch to find the critical values 2

2/1 and 2

2/ for the given level of confidence

and sample size.

(a) 90% confidence, 23n

Open StatCrunch Stat Calculators Chi-Square DF 22 (n-1), P(x ≤ ____) = 0.05

compute and record. Repeat for the right-tail inserting P(x ≤ ____) = 0.95.

The critical values are 12.338015 and 33.924438.

Check the answer:

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(b) 99% confidence, 15n

Open StatCrunch Stat Calculators Chi-Square DF 14 (n-1), P(x ≤ ____) = 0.005

compute and record. Repeat for the right-tail inserting P(x ≤ ____) = 0.995.

The critical values are 4.074675 and 31.31935.

Check the answer:

Objective C : Confidence Interval for a Population Variance or Standard Deviation

%100)1( of the values of 2 will lie between 2

2/1 and 2

2/ . ( Recall: 2

2

2 )1(

sn )

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To find a %100)1( confidence interval about , take the square root of the lower bound and upper bound.

Example 1: A simple random sample of size n is drawn from a population that is known to be normally

distributed. The sample variance, 2s , is determined to be 19.8.

(a) Use StatCrunch to construct a 95% confidence interval for 2 if the sample size, n , is 10.

Open StatCrunch Stat Variance Stats One Sample with summary Sample

variance: 19.8, sample size: 10 Confidence interval for 𝜎2: 0.95 compute and record the

results.

95% confidence interval results:

σ2 : Variance of population

Variance Sample Var. DF L. Limit U. Limit

σ2 19.8 9 9.367722 65.99048

(b) Use StatCrunch to construct a 95% confidence interval for 2 if the sample size, n , is 25.

How does increasing the sample size affect the width of the interval?

Open StatCrunch Stat Variance Stats One Sample with summary Sample

variance: 19.8, sample size: 25 Confidence interval for 𝜎2: 0.95 compute and record the

results.

95% confidence interval results:

σ2 : Variance of population

Variance Sample Var. DF L. Limit U. Limit

σ2 19.8 24 12.07192 38.319026

Increasing the sample size decreases the width of the interval for 𝜎2 from

n = 10 (9.3677, 65.9905) to n = 25 (12.0719, 38.3190)

(c) Use StatCrunch to construct a 99% confidence interval for 2 if the sample size, n , is 10.

Compare the results with those obtained in part (a). How does increasing the level of

confidence affect the width of the confidence interval?

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Open StatCrunch Stat Variance Stats One Sample with summary Sample

variance: 19.8, sample size: 10 Confidence interval for 𝜎2: 0.99 compute and record the

results.

99% confidence interval results:

σ2 : Variance of population

Variance Sample Var. DF L. Limit U. Limit

σ2 19.8 9 7.5542562 102.71291

Increasing the level of confidence increases the width of the interval for 𝜎2 from

95% confidence (9.3677, 65.9905) to 99% confidence (7.5543, 102.7129)

Example 2: Travelers per taxes for flying, car rentals, and hotels. The following data represent the total

travel tax for a 3-day business trip in eight randomly selected cities. It was verified that the

data are normally distributed. Use StatCrunch to construct a 90% confidence interval for the

standard deviation travel tax for a 3-day business trip. Interpret the interval.

Open StatCrunch Stat Input given data Summary Statistics Columns Var1

Variance compute and record the results

Summary statistics: Column Variance

var1 151.87187

Open StatCrunch Stat Variance Stats One Sample with summary

Sample variance: 151.87187, sample size: 8 Confidence interval for 𝜎2: 0.90

compute and record the results.

Alternative way: Open StatCrunch Stat Input given data Variance Stats

One Sample with data Columns Var1 Confidence interval for 𝜎2: 0.90

compute and record the results.

90% confidence interval results:

σ2 : Variance of population

Variance Sample Var. DF L. Limit U. Limit

σ2 151.87187 7 75.573504 490.50829

Manually compute the square root of each limit.

75.573504= 8.69330224943 Lower Limit

490.50829= 22.1474217461 Upper Limit

We are 90% confidence that the standard deviation travel tax for a 3-day

business trip is between 8.693 and 22.147.