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Chapter 6 Estimates and Sample SizesPowerPoint Presentation

Chapter 6 Estimates and Sample Sizes

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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
6-3 Estimating a Population Proportion

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

This chapter presents:

- Methods for estimating population means and proportions
- Methods for determining sample sizes

- 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.

- 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.

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 #

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.

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%)

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

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)

Confidence Intervals from 20 Different Samples

Simulations

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

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!!!!

Finding z2for 95% Degree of Confidence

95%

= 5%

2 = 2.5% = .025

.95

.025

.025

z2

-z2

Critical Values

Finding z2for 95% Degree of Confidence

= 0.05

= 0.025

.025

Use calculator

to find a z score of 1.96

z2 = 1.96

.025

.025

- 1.96 1.96

Finding z2for other Degrees of Confidence

Examples:

1 -

1 -

1 -

1 -

1 - (will use on test for ease of calculation)

Find critical value and sketch

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

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

- If n> 30 and we know
- If n 30, the population must have a normal distribution and we must know .
- Knowing is largely unrealistic.

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.

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.

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

Test Question

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

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

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

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.

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.

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.

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.

- 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

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

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

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.

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

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

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?

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

Table A-3 / Calculators / Excel

- Table from website
- TI – 84 (only)
- Excel function (tinv)

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

Critical z Value vs Critical t Values

See “t distribution pdf.xls”

Finding t2for 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

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

Using the Normal and t Distribution

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

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 = t2 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.

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

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

Estimating a population proportion

- 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.

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

DefinitionPoint Estimate

ˆ

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

Round-Off Rule for Confidence Interval Estimates of p

- Round the confidence interval limits to
three significant digits.

ˆ

p q

z

E =

n

Determining Sample Size

(solve for n by algebra)

z

ˆ

ˆ

()2

p q

n=

E2

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

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

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)

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

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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