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Chapter 6 Estimates and Sample Sizes. 6-1 Estimating a Population Mean: Large Samples / σ Known 6-2 Estimating a Population Mean: Small Samples / σ Unknown 6-3Estimating a Population Proportion 6-4 Estimating a Population Variance: Will cover with chapter 8. Overview.

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Chapter 6 estimates and sample sizes

Chapter 6Estimates and Sample Sizes

  • 6-1 Estimating a Population Mean: Large Samples / σ Known

  • 6-2 Estimating a Population Mean: Small Samples / σ Unknown

    6-3Estimating a Population Proportion

    6-4 Estimating a Population Variance: Will cover with chapter 8


Chapter 6 estimates and sample sizes

Overview

This chapter presents:

  •   Methods for estimating population means and proportions

  •   Methods for determining sample sizes


Chapter 6 estimates and sample sizes

6-1

Estimating a Population Mean:

Large Samples / σ Known


Chapter 6 estimates and sample sizes

Assumptions

  • Large Sample is defined as samples with n > 30 and σ known.

  • Data collected carelessly can be absolutely worthless, even if the sample is quite large.


Chapter 6 estimates and sample sizes

Definitions

  • Estimator

    a formula or process for using sample data to estimate a population parameter

  • Estimate

    a specific value or range of values used to approximate some population parameter

  • Point Estimate

    a single value (or point) used to approximate a population parameter

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


Chapter 6 estimates and sample sizes

Definition

Confidence Interval

(or Interval Estimate)

a range (or an interval) of values used to estimate the true value of the population parameter

Lower # < population parameter < Upper #

As an example

Lower # <  < Upper #


Chapter 6 estimates and sample sizes

Definition

Why Confidence Intervals

A couple of points

  • Even though x is the best estimate for  and s is the best estimate for  they do not give us an indication of how good they are.

  • A confidence interval gives us a range of values based on

    • variation of the sample data

    • How accurate we want to be

  • The width of the range of values gives us an indication of how good the estimate is.

  • The width is called the Margin of Error (E). We will discuss how to calculate this later.


Chapter 6 estimates and sample sizes

Definition

Degree of Confidence

(level of confidence or confidence coefficient)

  • Proportion of times that the confidence interval actual contains the population parameter

  • Degree of Confidence = 1 - 

    • often expressed as a percentage value usually 90%, 95%, or 99%

      So ( = 10%), ( = 5%), ( = 1%)


Chapter 6 estimates and sample sizes

Interpreting a Confidence Interval

98.08o <µ < 98.32o

Let: 1 -  = .95

Correct: we are 95% confident that the interval from 98.08 to 98.32 actually does contain the true value of .

This means that if we were to select many different samples of sufficient size and construct the confidence intervals, 95% of them would actually contain the value of the population mean .

Wrong: There is a 95% chance that the true value of  will fall between 98.08 and 98.32. (there is no way to calculate the probability for a population parameter only a sample statistic)


Chapter 6 estimates and sample sizes

Confidence Intervals from 20 Different Samples

Simulations

http://www.ruf.rice.edu/~lane/stat_sim/conf_interval/index.html


Chapter 6 estimates and sample sizes

Definition

Critical Value

The number on the borderline separating sample statistics that are likely to occur from those that are unlikely to occur. The number z/2is a critical value that is a zscore with the property that it separates an area /2in the right tail of the standard normal distribution.

ENGLISH PLEASE!!!!


Chapter 6 estimates and sample sizes

The Critical Value

z2

2

2

z2

-z2

z=0

Found from calculator


Chapter 6 estimates and sample sizes

Finding z2for 95% Degree of Confidence

95%

 = 5%

2 = 2.5% = .025

.95

.025

.025

z2

-z2

Critical Values


Chapter 6 estimates and sample sizes

Finding z2for 95% Degree of Confidence

 = 0.05

 = 0.025

.025

Use calculator

to find a z score of 1.96

z2 = 1.96

.025

.025

- 1.96 1.96


Chapter 6 estimates and sample sizes

Finding z2for other Degrees of Confidence

Examples:

1 - 

1 - 

1 - 

1 - 

1 - (will use on test for ease of calculation)

Find critical value and sketch


Chapter 6 estimates and sample sizes

Definition

Margin of Error

is the maximum likely difference observed

between sample mean x and true population mean μ.

denoted by E

μ

upper limit

lower limit


Chapter 6 estimates and sample sizes

E = z/2 •

n

Confidence Interval (or Interval Estimate) for Population Mean µ(Based on Large Samples: n >30)

x - E < µ < x + E

Where


Chapter 6 estimates and sample sizes

When can we use zα/2?

  •   If n> 30 and we know 

  •   If n 30, the population must have a  normal distribution and we must know .

  • Knowing  is largely unrealistic.


Chapter 6 estimates and sample sizes

Round-Off Rule for Confidence Intervals Used to Estimate µ

1. When using the original set of data, round the confidence interval limits to one more decimal place than used in original set of data.

2. When the original set of data is unknown and only the summary statistics (n, x, s)are used, round the confidence interval limits to the same number of decimal places used for the sample mean.


Chapter 6 estimates and sample sizes

Example: A study found the starting salaries of 100 college graduates who have taken a statistics course. The sample mean was $43,704 and the sample standard deviation was $9,879. Find the margin of error E and the 95% confidence interval.

n = 100

x = 43704

σ = 9879

 = 0.95

 = 0.05

/2 = 0.025

z/ 2= 1.96

E = z/ 2•  = 1.96 • 9879 = 1936.3

n

100

x - E <  < x + E

43704 - 1936.3 <<43704 + 1936.3

$41,768 << $45,640

Based on the sample provided, we are 95% confident the population (true) mean of starting salaries is between 41,768 & 45,640.


Chapter 6 estimates and sample sizes

TI-83 Calculator

Finding Confidence intervals using z

  • Press STAT

  • Cursor to TESTS

  • Choose ZInterval

  • Choose Input: STATS*

  • Enter σ and x and confidence level

  • Cursor to calculate

*If your input is raw data, then input your raw data in L1 then use DATA


Chapter 6 estimates and sample sizes

Width of Confidence Intervals

Test Question

What happens to the width of confidence intervals with changing confidence levels?


Chapter 6 estimates and sample sizes

Finding the Point Estimate and E from a Confidence Interval

Point estimate of x:

x =(upper confidence interval limit) + (lower confidence interval limit)

2

Margin of Error:

E = (upper confidence interval limit) - x


Chapter 6 estimates and sample sizes

Example

Find x and E

  • 26 < µ < 40

    x = (40 + 26) / 2 = 33

    E = 40 - 33 = 7

Use for #4 on hw


Chapter 6 estimates and sample sizes

z

E =

/ 2 •

n

Sample Size for Estimating Mean 

(solve for n by algebra)

2

z

/ 2

n

=

E

z/2 = critical z score based on the desired degree of confidence

E = desired margin of error

 = population standard deviation


Chapter 6 estimates and sample sizes

Example:If we want to estimate the mean weight of plastic discarded by households in one week, how many households must be randomly selected to be 99% confident that the sample mean is within 0.25 lb of the true population mean? (A previous study indicates the standard deviation is 1.065 lb.)

2

2

n = z = (2.575)(1.065)

 = 0.01

z = 2.575

E = 0.25

σ = 1.065

E

0.25

= 120.3 = 121 households

If n is not a whole number, round it upto the next higher whole number.


Chapter 6 estimates and sample sizes

Example:If we want to estimate the mean weight of plastic discarded by households in one week, how many households must be randomly selected to be 99% confident that the sample mean is within 0.25 lb of the true population mean? (A previous study indicates the standard deviation is 1.065 lb.)

2

2

n = z = (2.575)(1.065)

 = 0.01

z = 2.575

E = 0.25

σ = 1.065

E

0.25

= 120.3 = 121 households

We would need to randomly select 121 households to be 99% confident that this mean is within 1/4 lb of the population mean.


Chapter 6 estimates and sample sizes

Example:How large will the sample have to be if we want to decrease the margin of error from 0.25 to 0.2? Would you expect it to be larger or smaller?

2

2

n = z = (2.575)(1.065)

 = 0.01

z = 2.575

E = 0.20

σ = 1.065

E

0.2

= 188.01 = 189 households

We would need to randomly select a larger sample because we require a smaller margin of error.


Chapter 6 estimates and sample sizes

What happens when E is doubled ?

2

(z )

2

z

/ 2

/ 2

n = =

1

1

2

(z )

2

z

/ 2

/ 2

n = =

4

2

E = 1 :

E = 2 :

  • Sample size nis decreased to 1/4 of its original value if E is doubled.

  • Larger errors allow smaller samples.

  • Smaller errors require larger samples.


Chapter 6 estimates and sample sizes

Class Assignment

  • Use OLDFAITHFUL Data in Datasets File

  • Construct a 95% and 90% confidence interval for the mean eruption duration. Write a conclusion for the 95% interval. Assume σ to be 58 seconds

  • Compare the 2 confidence intervals. What can you conclude?

  • How large a sample must you choose to be 99% confident the sample mean eruption duration is within 10 seconds of the true mean

    Guidelines:

  • Choose a partner

  • Suggest having one person working the calculator and one writing

  • Due at the end of class (5 HW points)

  • Each person must turn in a paper


Chapter 6 estimates and sample sizes

6-2

Estimating a Population Mean:

Small Samples / σ Unknown


Chapter 6 estimates and sample sizes

Small SamplesAssumptions

  • n 30

  • The sample is a random sample.

  • The sample is from a normally distributed population.

    Case 1 ( is known): Largely unrealistic;

    Case 2 (is unknown): Use Student t distribution if normal ; if n is very large use z


Chapter 6 estimates and sample sizes

Determining which distribution to use

Case 1 ( is known):

Case 2 (is unknown):


Chapter 6 estimates and sample sizes

Determining which distribution to use

  • n = 150 ; x = 100 ; s = 15 skewed distribution

  • n = 8 ; x = 100 ; s = 15 normal distribution

  • n = 8 ; x = 100 ; s = 15 skewed distribution

  • n = 150 ; x = 100 ; σ = 15 skewed distribution

  • n = 8 ; x = 100 ; σ = 15 skewed distribution


Chapter 6 estimates and sample sizes

Important Facts about the Student t Distribution

  • Developed by William S. Gosset in 1908

  • Density function is complex

  • Shape is determined by “n”

  • Has the same general symmetric bell shape as the normal distribution but it reflects the greater variability (with wider distributions) that is expected with small samples.

  • The Student t distribution has a mean of t = 0, but the standard deviation varies with the sample size and is always greater than 1

  • Is essentially the normal distribution for large n. For values of n > 30, the differences are so small that we can use the critical z or t value.


Chapter 6 estimates and sample sizes

Student t Distributions for n = 3 and n = 12

Student t

distribution

with n = 12

Standard

normal

distribution

Student t

distribution

with n = 3

0

Greater variability than standard normal due to small sample size


Chapter 6 estimates and sample sizes

Student t Distribution

If the distribution of a population is essentially normal, then the distribution of

x - µ

t =

s

n

t/ 2

  • critical values denoted by


Chapter 6 estimates and sample sizes

Book Definition

Degrees of Freedom (df )

Corresponds to the number of sample values that can vary after certain restrictions have imposed on all data values.

This doesn’t help me, how about you?


Chapter 6 estimates and sample sizes

Definition

Degrees of Freedom (df )

In general, the degrees of freedom of an estimate is equal to the number of independent scores (n) that go into the estimate minus the number of parameters estimated.

In this section

df = n - 1

because we are estimating  with x


Chapter 6 estimates and sample sizes

Table A-3 / Calculators / Excel

  • Table from website

  • TI – 84 (only)

  • Excel function (tinv)


Chapter 6 estimates and sample sizes

Table A-3 t Distribution

.005

(one tail)

.01

(two tails)

.01

(one tail)

.02

(two tails)

.025

(one tail)

.05

(two tails)

.05

(one tail)

.10

(two tails)

.10

(one tail)

.20

(two tails)

.25

(one tail)

.50

(two tails)

Degrees

of

freedom

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

Large (z)

63.657

9.925

5.841

4.604

4.032

3.707

3.500

3.355

3.250

3.169

3.106

3.054

3.012

2.977

2.947

2.921

2.898

2.878

2.861

2.845

2.831

2.819

2.807

2.797

2.787

2.779

2.771

2.763

2.756

2.575

31.821

6.965

4.541

3.747

3.365

3.143

2.998

2.896

2.821

2.764

2.718

2.681

2.650

2.625

2.602

2.584

2.567

2.552

2.540

2.528

2.518

2.508

2.500

2.492

2.485

2.479

2.473

2.467

2.462

2.327

12.706

4.303

3.182

2.776

2.571

2.447

2.365

2.306

2.262

2.228

2.201

2.179

2.160

2.145

2.132

2.120

2.110

2.101

2.093

2.086

2.080

2.074

2.069

2.064

2.060

2.056

2.052

2.048

2.045

1.960

6.314

2.920

2.353

2.132

2.015

1.943

1.895

1.860

1.833

1.812

1.796

1.782

1.771

1.761

1.753

1.746

1.740

1.734

1.729

1.725

1.721

1.717

1.714

1.711

1.708

1.706

1.703

1.701

1.699

1.645

3.078

1.886

1.638

1.533

1.476

1.440

1.415

1.397

1.383

1.372

1.363

1.356

1.350

1.345

1.341

1.337

1.333

1.330

1.328

1.325

1.323

1.321

1.320

1.318

1.316

1.315

1.314

1.313

1.311

1.282

1.000

.816

.765

.741

.727

.718

.711

.706

.703

.700

.697

.696

.694

.692

.691

.690

.689

.688

.688

.687

.686

.686

.685

.685

.684

.684

.684

.683

.683

.675


Chapter 6 estimates and sample sizes

Critical z Value vs Critical t Values

See “t distribution pdf.xls”


Chapter 6 estimates and sample sizes

Finding t2for the following Degrees of Confidence and sample size

Examples:

1 - n = 12

1 - n = 15

1 - n = 9

1 - n = 20

Find critical value and sketch


Chapter 6 estimates and sample sizes

Confidence Interval for the Estimate of µBased on an Unknown  and a Small Simple Random Sample from a Normally Distributed Population

x - E < µ < x + E

s

E = t/2

where

n

t/2 found in Table A-3


Chapter 6 estimates and sample sizes

Using the Normal and t Distribution


Chapter 6 estimates and sample sizes

Example: Let’s do an example comparing z and t. Construct confidence interval’s for each using the following data.

n = 16

x = 50

s = 20

 = 0.05

/2 = 0.025

Now we wouldn’t use a z distribution here due to the small sample but let’s do it anyway and compare the width of the confidence interval to a confidence interval created using a t distribution


Chapter 6 estimates and sample sizes

x - E < µ < x + E

Example: A study of 12 Dodge Vipers involved in collisions resulted in repairs averaging $26,227 and a standard deviation of $15,873. Find the 95% interval estimate of , the mean repair cost for all Dodge Vipers involved in collisions. (The 12 cars’ distribution appears to be bell-shaped.)

E = t2 s =(2.201)(15,873) = 10,085.3

x = 26,227

s = 15,873

 = 0.05

/2 = 0.025

t/2 = 2.201

n

12

26,227 - 10,085.3 < µ < 26,227 + 10,085.3

$16,141.7< µ < $36,312.3

We are 95% confident that this interval contains the average cost of repairing a Dodge Viper.


Chapter 6 estimates and sample sizes

TI-83 Calculator

Finding Confidence intervals using t

  • Press STAT

  • Cursor to TESTS

  • Choose TInterval

  • Choose Input: STATS*

  • Enter s and x and confidence level

  • Cursor to calculate

    *If your input is raw data, then input your raw data in L1 then use DATA


Chapter 6 estimates and sample sizes

Table A-3 t Distribution

.005

(one tail)

.01

(two tails)

.01

(one tail)

.02

(two tails)

.025

(one tail)

.05

(two tails)

.05

(one tail)

.10

(two tails)

.10

(one tail)

.20

(two tails)

.25

(one tail)

.50

(two tails)

Degrees

of

freedom

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

Large (z)

63.657

9.925

5.841

4.604

4.032

3.707

3.500

3.355

3.250

3.169

3.106

3.054

3.012

2.977

2.947

2.921

2.898

2.878

2.861

2.845

2.831

2.819

2.807

2.797

2.787

2.779

2.771

2.763

2.756

2.575

31.821

6.965

4.541

3.747

3.365

3.143

2.998

2.896

2.821

2.764

2.718

2.681

2.650

2.625

2.602

2.584

2.567

2.552

2.540

2.528

2.518

2.508

2.500

2.492

2.485

2.479

2.473

2.467

2.462

2.327

12.706

4.303

3.182

2.776

2.571

2.447

2.365

2.306

2.262

2.228

2.201

2.179

2.160

2.145

2.132

2.120

2.110

2.101

2.093

2.086

2.080

2.074

2.069

2.064

2.060

2.056

2.052

2.048

2.045

1.960

6.314

2.920

2.353

2.132

2.015

1.943

1.895

1.860

1.833

1.812

1.796

1.782

1.771

1.761

1.753

1.746

1.740

1.734

1.729

1.725

1.721

1.717

1.714

1.711

1.708

1.706

1.703

1.701

1.699

1.645

3.078

1.886

1.638

1.533

1.476

1.440

1.415

1.397

1.383

1.372

1.363

1.356

1.350

1.345

1.341

1.337

1.333

1.330

1.328

1.325

1.323

1.321

1.320

1.318

1.316

1.315

1.314

1.313

1.311

1.282

1.000

.816

.765

.741

.727

.718

.711

.706

.703

.700

.697

.696

.694

.692

.691

.690

.689

.688

.688

.687

.686

.686

.685

.685

.684

.684

.684

.683

.683

.675


Chapter 6 estimates and sample sizes

6-3

Estimating a population proportion


Chapter 6 estimates and sample sizes

Assumptions

  • 1. The sample is a random sample.

  • 2. The conditions for the binomial distribution are satisfied (See Section 4-3.)

  • 3. The normal distribution can be used to approximate the distribution of sample proportions because np  5 and nq 5 are both satisfied.


Chapter 6 estimates and sample sizes

ˆ

p=

x

n

sample proportion

Notation for Proportions

p=

population proportion

of xsuccesses in a sample of size n

(pronounced

‘p-hat’)

ˆ

ˆ

q = 1 - p =sampleproportion

of xfailures in a sample size of n


Chapter 6 estimates and sample sizes

DefinitionPoint Estimate

ˆ

The sample proportion pis the best point estimate of the population proportion p.


Chapter 6 estimates and sample sizes

Confidence Interval for Population Proportion

ˆ

ˆ

p

p

p - E < < + E

where

ˆ

ˆ

z

p q

E =



n


Chapter 6 estimates and sample sizes

Round-Off Rule for Confidence Interval Estimates of p

  • Round the confidence interval limits to

    three significant digits.


Chapter 6 estimates and sample sizes

ˆ

ˆ

p q

z

E =



n

Determining Sample Size

(solve for n by algebra)

z

ˆ

ˆ

()2

p q



n=

E2


Chapter 6 estimates and sample sizes

z

ˆ

()2

p q



n=

E2

Sample Size for Estimating Proportion p

ˆ

When an estimate of p is known:

ˆ

When no estimate of p is known:

z

()2

0.25



n=

E2


Chapter 6 estimates and sample sizes

ˆ

ˆ

Example:We want to determine, with a margin of error of four percentage points, the current percentage of U.S. households using e-mail. Assuming that we want 90% confidence in our results, how many households must we survey? A 1997 study indicates 16.9% of U.S. households used e-mail.

n = [z/2 ]2p q

2

E

To be 90% confident that our sample percentage is within four percentage points of the true percentage for all households, we should randomly select and survey 238 households.

= [1.645]2(0.169)(0.831)

0.042

= 237.51965

= 238 households


Chapter 6 estimates and sample sizes

Round-Off Rule for Sample Size n

When finding the sample size n, if the result is not a whole number, always increase the value of n to the next larger whole number.

n = 237.51965 = 238 (rounded up)


Chapter 6 estimates and sample sizes

Example:We want to determine, with a margin of error of four percentage points, the current percentage of U.S. households using e-mail. Assuming that we want 90% confidence in our results, how many households must we survey? There is no prior information suggesting a possible value for the sample percentage.

n = [z/2 ]2(0.25)

E

2

= (1.645)2 (0.25)

With no prior information, we need a larger sample to achieve the same results with 90% confidence and an error of no more than 4%.

0.042

= 422.81641

= 423 households


Chapter 6 estimates and sample sizes

TI-83 Calculator

Finding Confidence intervals using z (proportions)

  • Press STAT

  • Cursor to TESTS

  • Choose 1-ProbZInt

  • Enter x and n and confidence level

  • Cursor to calculate


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