Point Estimate

Transcript Point Estimate

Business Statistics: A Decision-Making Approach

8 th Edition

Chapter 8 Estimating Single Population Parameters

Chapter Goals

After completing this chapter, you should be able to:

 Distinguish between a point estimate and a confidence interval estimate  Construct and interpret a confidence interval estimate for a single population mean using both the z and t distributions  Determine the required sample size to estimate a single population mean or a proportion within a specified margin of error  Form and interpret a confidence interval estimate for a single population proportion Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-2

Overview of the Chapter

     Builds upon the material from Chapter 1 and 7 Introduces using sample statistics to estimate population parameters 

Because gaining access to population parameters can be expensive, time consuming and sometimes not feasible

Confidence Intervals for the Population Mean, μ   when Population Standard Deviation σ is Known when Population Standard Deviation σ is Unknown Confidence Intervals for the Population Proportion, p Determining the Required Sample Size for means and proportions Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-3

Population

(mean, μ, is unknown)

Estimation Process

Confidence Level Random Sample

( point estimate ) Mean x = 50 I am 95% confident that μ is between 40 & 60.

Sample confidence interval

Point Estimate

  Suppose a poll indicate that 62% (sample mean) of the people favor limiting property taxes to 1% of the market value of the property.   The 62% is the

point estimate

of the true population of people who favor the property-tax limitation. EPA Automobile Mileage Test Result (point estimate) A

point estimate

is a single number, used to estimate an

unknown

Confidence Interval

   The point estimate is not likely to exactly equal the population parameter because of sampling error.

 Probability of “sample

mean

= population

mean”

zero

With sample mean, it is impossible to determining how far the sample mean is from the population mean.

To overcome this problem, “confidence interval” can be used as the most common procedure.

Confidence Level

   Confidence Level (not same as critical value, α)  Describes how strongly we believe that a particular sampling method will produce a confidence interval that includes the true population parameter. A percentage (less than 100%)  Most common: 90 % (α = 0.1), 95% (α = 0.05), 99% Suppose confidence level = 95%  In the long run, 95% of all the confidence intervals will contain the unknown true parameter Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-7

General Formula



The general formula for the confidence interval is: x



z σ n

Point Estimate



(Critical Value)(Standard Error) z-value (or t value) based on the level of confidence desired

How to Find the Critical Value

 The Central Limit Theorem states that the sampling distribution of a statistic will be normal or nearly normal, if any of the following conditions apply.

  n > 30 : will give a sampling distribution that is nearly normal The sampling distribution of the mean is normally distributed because the population distribution is normally distributed. Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-9

How to Find the Critical Value

  When one of these conditions is satisfied, the critical value can be expressed as a z score or as a t score . To find the critical value, follow these steps.

  Compute alpha (α): α = 1 - (confidence level / 100) = 1 - 0.95 = 0.05

Find the critical probability (p*): p* = 1 – α/2 (because there are lower confidence limit and upper confidence limit) =1 - 0.05/2 = 0.975

 To express the critical value as a z score, find the z score having a cumulative probability equal to the critical probability (p*).

t score as the critical value

 When the population standard deviation is unknown or when the sample size is small, the t score is preferred.

 Find the degrees of freedom (DF) . When estimating a mean score or a proportion from a single sample, DF is equal to the sample size minus one. For other applications, the degrees of freedom may be calculated differently.  The critical t score (t*) is the t score having degrees of freedom equal to DF and a cumulative probability equal to the critical probability (p*).

From Chapter 7

 The standard deviation of the possible

sample means computed from all random samples

of size n is

equal to

the

population standard deviation divided by the square root

of the sample size: x  z σ n

Also called the standard error

Margin of Error

 Margin of Error (e): the amount added and subtracted to the point estimate to form the confidence interval Example: Margin of error for estimating μ, σ known: σ σ x  z n e  z n Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-13

Point and Interval Estimates

 So, a confidence interval provides additional information about variability within a range of z-values 

The interval incorporates the sampling error Lower Confidence Limit Point Estimate Width of confidence interval Upper Confidence Limit

Using Analysis ToolPak

Download “Confidence Interval Example” Excel file Confidence Interval 119.9

(mean)

± 2.59

117.31 ------ 122.49

Don’t even worry about p* = 1 α/2

Margin of Error

Using Analysis ToolPak

Factors Affecting Margin of Error

e  z σ n  Data variation, σ :  Sample size, n :  Level of confidence, 1  : e as σ e as n e if 1  

Video Lecture: Confidence Intervals

Confidence Intervals

Confidence Intervals Population Mean Population Proportion σ Known σ Unknown

Confidence Interval for μ ( σ Known)

 Assumptions  Population standard deviation σ is known  If population is not normal, use larger sample n > 30 (Central Limit Theorem)  Confidence interval estimate

x



z σ n

Finding the Critical Value

 Consider a 95% confidence interval:

0.95/2 = 0.475 find on z table Normsinv(0.5 – 0.475)=1.959963…

z   1.96

(1   ) / 2  .95/2 α 2  .025

z units: x units: -z = -1.96

Lower Confidence Limit

x α 2  .025

z = 1.96

Upper Confidence Limit

 z σ n 8-22

Common Levels of Confidence

 Commonly used confidence levels are 90%, 95%, and 99%

Confidence Level

80% 90% 95% 98% 99% 99.8% 99.9%

Critical value, z

1.28

1.645

1.96

2.33

2.58

3.08

3.27

8-23

Computing a Confidence Interval Estimate for the Mean (

known )

Select a random sample of size n Specify the confidence level Compute the sample mean Determine the standard error Determine the critical value (z) from the normal table Compute the confidence interval estimate Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-24

Example

 A sample of

circuits from a large

normal

population has a mean resistance of 2.20 ohms. We know from past testing that the population standard deviation is 0.35 ohms.  Determine a 95 % (α = 0.05) confidence interval for the true mean resistance of the population.

Example Solution Solution:

x  z σ n  2.20

 1.96

(0.35/  2.20

 .2068

11 ) 1.9932

2.4068

(continued)

Interpretation

 We are 95% confident that the true mean resistance is between 1.9932 and 2.4068 ohms  Although the true mean may or may not be in this interval, 95% of intervals formed in this manner will contain the true mean  An

incorrect

interpretation is that there is

95% probability

Confidence Interval for μ ( σ Unknown )

 In most real world situations, population mean and StdDev are NOT KNOWN.

 When the population standard deviation is unknown or when the sample size is small , the

t score

is preferred.

 When the sample size is large (n > 30), it doesn't make much difference. Both approaches yield similar results.

 So we

use the t distribution

instead of the normal distribution.

Confidence Interval for μ ( σ Unknown)

(continued)

  Assumptions  Population standard deviation is unknown  Sample size is small  If population is not normal, use large sample n > 30 Use Student’s

Student’s t Distribution

 The

is a family of distributions  The

value depends on degrees of freedom (d.f.)  For example, if n = 28, then the d.f. is 27.

 As the d.f. increase, the t distribution approaches the normal distribution ( see the

t distribution simulation

one the class website )  d.f. = n – 1 Only n-1 independent pieces of data information left in the sample because the sample mean has already been obtained Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-30

Student’s t Distribution

Note: t compared to z as n increases Standard Normal (t with df =  )

As n the estimate of

becomes better so t converges to z

(

= 13) t-distributions are bell shaped and symmetric, but have ‘fatter’ tails than the normal

(

= 5) 0

t

Student’s t Table

Confidence Level df

0.50

0.80

0.90

1 1.000 3.078 6.314

0.817 1.886

2.920

3 0.765 1.638 2.353

The body of the table contains t values, not probabilities

- 1 = 2 confidence level: 90%  /2 = 0.05

0 2.920

8-32

t Distribution Values

With comparison to the z value

Confidence t t t z Level (10 d.f.) (20 d.f.) (30 d.f.) ____

0.80 1.372 1.325 1.310 1.28

0.90 1.812 1.725 1.697 1.64

0.95 2.228 2.086 2.042 1.96

0.99 3.169 2.845 2.750 2.58

Example

A random sample of n = 25 has x = 50 and s = 8. Form a 95% confidence interval for μ  d.f. = n – 1 = 24 , so

 /2 , n  1  t 0.025,24  2.0639

The confidence interval is x  t s n  50  (2.0639) 8 25 46.698 53.302

TINV

   NORMSINV(p) gives the z-value that puts probability (area) p to the left of that value of z. TINV(p,DF) gives the t-value that puts one-half the probability (area) to the right with DF degrees of freedom. Download and then review “

TDIST Vs. TINV ”



Only need to review “TINV”

TINV Example

 For 95% confidence intervals we use α = .05, so that we are looking t .025

 Suppose d.f. is 17  t .025,17 = t value that puts .025 to the right of t with 17 degrees of freedom. Since TINV splits α = .05 to .025, this value is =TINV( .05

,17).

In-class Practice Example

 The file

Excel file Coffee

on the class website contains a random sample of 144 German coffee drinkers and measures the annual coffee consumption in kilograms for each sampled coffee drinker. A marketing research firm wants to use this information to develop an advertising campaign to increase German coffee consumption.

 Develop a

95% , 90% and 75% confidence interval estimates

Determining Required Sample Size

  Wishful situation  High confidence level, low margin of error, and small sample size In reality, conflict among three….

  For a given sample size, a high confidence level will tend to generate a large margin of error For a given confidence level, a small sample size will result in an increased margin of error  Reducing of margin of error requires either reducing the confidence level or increasing the sample size, or both Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-38

Determining Required Sample Size

 How large a sample size do I really need?

Determining Required Sample Size When σ is known

 The required sample size can be found to reach a desired margin of error (e) and level of confidence (1  )  Required sample size,

σ known

Required Sample Size Example

If s = 45 (known), what sample size is needed to be 90% confident of being correct within ± 5?

n



z

σ

e

2 

1.645 (45)

5

2 

219.19

So the required sample size is

n = 220

If σ is unknown (most real situation)

 If σ is unknown, three possible approaches 1.

Use a value for σ that is expected to be at least as large as the true σ 2.

Select a pilot sample (smaller than anticipated sample size) and then estimate σ with the pilot sample standard deviation, s Use the range of the population to estimate the population’s Std Dev. As we know, µ ± 3σ contains virtually all of the data. Range = max – min. Thus, R = ( µ + 3σ) – (µ - 3σ) = 6σ. Therefore, σ = R/6 (or R/4 for a more conservative estimate, producing a larger sample size) Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-42

Example when

σ is unknown

Confidence Intervals for the Population Proportion, π

 An interval estimate for the population proportion ( π ) can be calculated by adding an allowance for uncertainty to the sample proportion ( p ).

 For example, estimation of the proportion of customers who are satisfied with the service provided by your company  Sample proportion, p = x/n Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-44

Confidence Intervals for the Population Proportion, π

(continued)

 Recall that the distribution of the sample proportion is approximately normal if the sample size is large, with standard deviation σ π  π(1  π) n See Chpt. 7!!

 We will estimate this with sample data:

s

p 

p(1



p) n

Confidence Interval Endpoints

  Upper and lower confidence limits for the population proportion are calculated with the formula p  z p(1  p) n where    z is the standard normal value for the level of confidence desired p is the sample proportion n is the sample size Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-46

Example



A random sample of 100 people shows that 25 are left-handed.



Form a 95% confidence interval for the true proportion of left-handers

Example



(continued)

A random sample of 100 people shows that 25 are left-handed. Form a 95% confidence interval for the true proportion of left-handers.

p  25/100  0.25

S p  p(1  p)/n  0.25(0.75) /100  0.0433

0.25



1.96 (0.0433) 0.1651

0.3349

Interpretation

 We are 95% confident that the true percentage of left-handers in the population is between 16.51% and 33.49%  Although this range may or may not contain the true proportion, 95% of intervals formed from samples of size 100 in this manner will contain the true proportion.

Changing the sample size



Increases in the sample size reduce the width of the confidence interval.

Example:  If the sample size in the above example is doubled to 200, and if 50 are left-handed in the sample, then the interval is still centered at 0.25, but the width shrinks to 0.19 0.31

Finding the Required Sample Size for Proportion Problems

Define the margin of error: e  z π(1  π) n Solve for n:

Will be in % units

n  z 2 π (1  π) e 2 π can be estimated with a pilot sample, if necessary (or conservatively use π = 0.50 – worst possible variation thus the largest sample size ) Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-51

What sample size...?

 How large a sample would be necessary to estimate the true proportion defective in a large population within 3%, with 95% confidence?

What sample size...?

(continued)

Solution: For 95% confidence, use Z = 1.96

e = 0.03

p = 0.12, so use this to estimate π n  z 2 π (1  π) e 2  (1.96) 2 (0.12)(1  0 .12) (0.03) 2  450.74

Using PHStat

 PHStat can be used for confidence intervals for the mean or proportion  Two options for the mean: known and unknown population standard deviation  Required sample size can also be found 

Download from the textbook website

PHStat Interval Options

PHStat Sample Size Options

Using PHStat (for μ, σ unknown)

Using PHStat (sample size for proportion)

How large a sample would be necessary to estimate the true proportion defective in a large population within 3%, with 95% confidence?

Chapter Summary

 Discussed point estimates  Introduced interval estimates  Discussed confidence interval estimation for the mean [ σ known]  Discussed confidence interval estimation for the mean [ σ unknown]  Addressed determining sample size for mean and proportion problems  Discussed confidence interval estimation for the proportion Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-59

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