copyright ©2011 nelson education limited large-sample estimation chapter 8

Copyright ©2011 Nelson Education Limited

Large-Sample Estimation

CHAPTER 8CHAPTER 8


Key ConceptsKey Concepts

I. Types of EstimatorsI. Types of Estimators

1. Point estimator: a single number is calculated to estimate the population parameter.

2. Interval estimatorInterval estimator: two numbers are calculated to form an interval that contains the parameter.


Key ConceptsKey ConceptsII. Properties of Good EstimatorsII. Properties of Good Estimators

1. Unbiased: the average value of the estimator equals the parameter to be estimated.2. Minimum variance: of all the unbiased estimators, the best estimator has a sampling distribution with the smallest standard error.3. The margin of error measures the maximum distance between the estimator and the true value of the parameter.

THIS CHAPTER – SAMPLE SIZE IS LARGE ENOUGH TO USE CLT!!


Interval EstimationInterval Estimation• Since we don’t know the value of the parameter,

consider which has a variable center.

• Only if the estimator falls in the tail areas will the interval fail to enclose the parameter. This happens only 5% of the time.

Estimator 1.96SEEstimator 1.96SE

WorkedWorkedWorked

Failed

APPLETAPPLETMY


Confidence Intervals:Confidence Intervals:

€

Confidence interval for a population mean (n >= 30) :

x ± zα / 2

s

n

€

Confidence interval for a population mean (n >= 30) :

x ± zα / 2

s

n

€

Confidence interval for a population proportion p

(np,nq >= 5) : ˆ p ± zα / 2

ˆ p ̂ q

n

€

Confidence interval for a population proportion p

(np,nq >= 5) : ˆ p ± zα / 2

ˆ p ̂ q

n

Estimator +/- 1.96 SE


ExampleExample• Of a random sample of n = 150 college students, 104 of

the students said that they had played on a soccer team during their K-12 years. Estimate the proportion of college students who played soccer in their youth with a 98% confidence interval.


Estimating the Difference Estimating the Difference between Two Meansbetween Two Means

•Sometimes we are interested in comparing the means of two populations.

•The average growth of plants fed using two different nutrients.•The average scores for students taught with two different teaching methods.

•To make this comparison,

€

A random sample of size n1 drawn from

population 1 with mean μ1 and variance σ 12 .

€



€



€





•We compare the two averages by making inferences about -, the difference in the two population averages.

•If the two population averages are the same, then 1- = 0.•The best estimate of 1- is the difference in the two sample means,

€

X1 − X 2

€

X1 − X 2


Sampling Distribution of Sampling Distribution of

€

1. The mean of X1 − X 2 is μ1 − μ2,the difference in

the population means.

2. We assume that the two samples are independent!!!

3. The standard deviation of X1 − X 2 is SE =σ 1

2

n1

+σ 2

2

n2

.

4. If the sample sizes are large, the sampling distribution

of X1 − X 2 is approximately normal, and SE can be estimated

as SE =s1

2

n1

+s2

2

n2

.

€

1. The mean of X1 − X 2 is μ1 − μ2,the difference in

the population means.

2. We assume that the two samples are independent!!!

3. The standard deviation of X1 − X 2 is SE =σ 1

2

n1

+σ 2

2

n2

.


of X1 − X 2 is approximately normal, and SE can be estimated

as SE =s1

2

n1

+s2

2

n2

.

€

X1 − X 2

€

X1 − X 2


Estimating Estimating 11--•For large samples (n1, n2 >=30), point estimates and their margin of error as well as confidence intervals are based on the standard normal (z) distribution.

2

22

1

21

2121

1.96 :Error ofMargin

:-for estimatePoint

n

s

n

s

xx

2

22

1

21

2121

1.96 :Error ofMargin

:-for estimatePoint

n

s

n

s

xx

€

Confidence interval for μ1 - μ2 :

(x 1 − x 2) ± zα / 2

s12

n1

+s2

2

n2

€

Confidence interval for μ1 - μ2 :

(x 1 − x 2) ± zα / 2

s12

n1

+s2

2

n2


ExampleExample

• Compare the average daily intake of dairy products of men and women using a 95% confidence interval.

Avg Daily Intakes Men Women

Sample size 50 50

Sample mean 756 762

Sample Std Dev 35 30


Example, continuedExample, continued

• Could you conclude, based on this confidence interval, that there is a difference in the average daily intake of dairy products for men and women?

• The confidence interval contains the value 11--= 0= 0.. Therefore, it is possible that 11 = = You would not want to conclude that there is a difference in average daily intake of dairy products for men and women.

Confidence interval for mean difference: (-18,78, 6.78)


Estimating the Difference Estimating the Difference between Two Proportionsbetween Two Proportions

•Sometimes we are interested in comparing the proportion of “successes” in two binomial populations.

•The germination rates of untreated seeds and seeds treated with a fungicide.•The proportion of male and female voters who favor a particular candidate for prime minister.

•To make this comparison,

.parameter with 1 population binomial

fromdrawn size of sample randomA

1

1

p

n.parameter with 1 population binomial


1

1

p

n

.parameter with 2 population binomial


2

2

p

n.parameter with 2 population binomial


2

2

p

n

The two samples are independent!



•We compare the two proportions by making inferences about p-p, the difference in the two population proportions.

•If the two population proportions are the same, then p1-p = 0.•The best estimate of p1-p is the difference in the two sample proportions,

€

ˆ p 1 − ˆ p 2 =X1

n1

−X2

n2

€

ˆ p 1 − ˆ p 2 =X1

n1

−X2

n2


The Sampling The Sampling Distribution of Distribution of 1 2ˆ ˆp p

€

1. The mean of ˆ p 1 − ˆ p 2 is p1 − p2,the difference in

the population proportions.

2. The two samples are independent.

3. The standard deviation of ˆ p 1 − ˆ p 2 is SE =p1q1

n1

+p2q2

n2

.


of ˆ p 1 − ˆ p 2 is approximately normal, and SE can be estimated

as SE =ˆ p 1ˆ q 1n1

+ˆ p 2 ˆ q 2n2

.

€

1. The mean of ˆ p 1 − ˆ p 2 is p1 − p2,the difference in

the population proportions.

2. The two samples are independent.

3. The standard deviation of ˆ p 1 − ˆ p 2 is SE =p1q1

n1

+p2q2

n2

.


of ˆ p 1 − ˆ p 2 is approximately normal, and SE can be estimated

as SE =ˆ p 1ˆ q 1n1

+ˆ p 2 ˆ q 2n2

.


Estimating Estimating pp11--pp•For large samples, point estimates and their margin of error as well as confidence intervals are based on the standard normal (z) distribution.

2

22

1

11

2121

ˆˆˆˆ1.96 :Error ofMargin

ˆˆ :for estimatePoint

n

qp

n

qp

pp-pp

2

22

1

11

2121

ˆˆˆˆ1.96 :Error ofMargin

ˆˆ :for estimatePoint

n

qp

n

qp

pp-pp

2

22

1

112/21

21

ˆˆˆˆ)ˆˆ(

:for interval Confidence

n

qp

n

qpzpp

pp

2

22

1

112/21

21

ˆˆˆˆ)ˆˆ(

:for interval Confidence

n

qp

n

qpzpp

pp


ExampleExample

• Compare the proportion of male and female university students who said that they had played on a soccer team during their K-12 years using a 99% confidence interval.

Youth Soccer Male Female

Sample size 80 70

Played soccer 65 39


Example, continuedExample, continued

• Could you conclude, based on this confidence interval, that there is a difference in the proportion of male and female university students who said that they had played on a soccer team during their K-12 years?

• The confidence interval does not contains the value pp11--pp = 0= 0.. Therefore, it is not likely that pp11= = pp You would conclude that there is a difference in the proportions for males and females.

Confidence interval: (0.06, 0.44)


Choosing the Sample SizeChoosing the Sample Size

• The total amount of relevant information in a sample is controlled by two factors:- The sampling plansampling plan or experimental designexperimental design: the procedure for collecting the information- The sample size sample size nn: the amount of information you collect.

• In a statistical estimation problem, the accuracy of the estimation is measured by the margin of errormargin of error or the width of the confidence interval.width of the confidence interval.


1. Determine the size of the margin of error, B, that you are willing to tolerate.

2. Choose the sample size by solving for n or n n 1

n2 in the inequality: 1.96 SE B, where SE is a

function of the sample size n.

3. For quantitative populations, estimate the population standard deviation using a previously calculated value of ss or the range approximation Range / 4.Range / 4.

4. For binomial populations, use the conservative approach and approximate p using the value pp .5 .5.

Choosing the Sample SizeChoosing the Sample Size


ExampleExampleA producer of PVC pipe wants to survey wholesalers who buy his product in order to estimate the proportion who plan to increase their purchases next year. What sample size is required if he wants his estimate to be within .04 of the actual proportion with probability equal to .95?


Key ConceptsKey ConceptsIII. Large-Sample Point EstimatorsIII. Large-Sample Point Estimators

To estimate one of four population parameters when the sample sizes are large, use the following point estimators with the appropriate margins of error.


Key ConceptsKey ConceptsIV. Large-Sample Interval EstimatorsIV. Large-Sample Interval Estimators

To estimate one of four population parameters when the sample sizes are large, use the following interval estimators.

copyright ©2011 nelson education limited large-sample estimation chapter 8

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