chapter 7 estimates and sample sizes 1. estimation: an introduction we have come a long way. we...

CHAPTER 7

ESTIMATES AND

SAMPLE SIZES

1

ESTIMATION: AN INTRODUCTION

We have come a long way. We started by learning “what is statistics and the two areas of applied statistics.” In Chapter 1, we learned that:1. Descriptive statistics consists of methods for organizing,

displaying, and describing data by using tables, graphs, and summary measures.

2. Inferential statistics consists of methods that used samples to make decisions or predictions about the population.

In Chapters 2 and 3, we focused on descriptive statistics and learned how to draw tables, how to graph data, and how to calculate numerical summary measures such as mean, median, mode, variance, and standard deviation.

Now in Chapters 7, we will focus on inferential statistics. We begin by discussing estimation.

Introduction

2

ESTIMATION: AN INTRODUCTION

Definition Estimation is a process for assigning value(s) to a population parameter based on information collected from a sample.There are many real-life examples in which “estimation” is used. A few of them are, for example, to estimate the:

1. Mean of fuel consumption for a particular model car.2. Proportion of students that completed MAT 12 course with a

passing grade for the past 10 years.3. Proportion of female high school students that dropped out of

school because of pregnancy.4. Percentage of all California lawyers disbarred for committing a

criminal offense.

3

ESTIMATION: AN INTRODUCTIONOf course we can conduct a census to find the true mean or proportion of the population in 1 through 4. However, for what we now know about census, it would be:

1. Expensive.2. Difficult to reach or contact every member of

the population.3. Time consuming.

So, because of the problem with census, a representative sample is generally drawn from the population and the appropriate sample statistic is calculated. Then,

1. A value is assigned to the population parameter based on the calculated value of the sample statistic.

2. The value assigned to the population parameter based on the value of sample statistic is called an estimate of the population parameter.

4

ESTIMATION: AN INTRODUCTIONFor example, the Mathematics Department draws a sample of 50 students from all students who have taken MAT 12 for the past 10 years. The department records the number of students that passed and failed the course, and calculated the sample proportion, , of students who passed the course to be 0.65. So,• If the department assigns the value of sample proportion, , to

the population proportion, p, then 0.65 is called an estimate of p and is called the estimator.

p̂

p̂

p̂

SummaryEstimation procedure involves:• Draw a sample from the population.• Collect required information from each element of the sample.• Calculate the value of sample statistic.• Assign the value to corresponding population parameter.

Note: The sample must be a simple random sample.5

7-2 ESTIMATING A POPULATION PROPORTION

The estimated value of population parameter can either be based on a point estimate or an interval estimate.

Point Estimate - DefinitionA point estimate is the value of sample statistic used to estimate population parameter.

So, suppose we used the sample proportion, , as a point estimate of p, then we can say that the proportion of all students that have taken MAT 12 course with a passing grade for the past 10 years is about 0.65. That is,

p̂

Point estimate of population parameter Value of corresponding sample statistic

We discussed in Chapter 6 that the value of sample statistic varies from one sample to another that are of the same size and drawn from the population. Therefore,

1. The value assigned to the population proportion, p , based on a point estimate depends on the sample drawn.

2. The value assigned to population parameter is almost always different from the true value of population parameter. 6

An Interval EstimationDefinition:An interval estimate is an interval build around the point estimate and then a probabilistic statement is made that the built interval contains the corresponding population parameter.

Therefore following on to our example, rather than saying that the proportion of all students that have taken MAT 12 in the last 10 years is 0.65, we would:

1. Add and subtract a number to 0.65 to obtain an interval and then2. Say that the interval contains the population proportion, p.

Now, let us add and subtract 0.2 to 0.65. Then we obtain an interval

(0.65 0.2 to 0.65 0.2) (0.45 to 0.85) 1. We state that the population proportion, p, is likely to be contained in

the interval 0.45 to 0.85.2. We also state that the proportion of all students that have taken MAT

12 with a passing grade in the past 10 years is between 0.45 and .85.

3. The 0.45 is called the lower limit and 0.85 is called the upper limit.4. The number we subtracted and added to the point estimate is called

margin of error.7

An Interval Estimation

Confidence Interval Point estimate Margin of error

5. The value of margin of error depends on:a. Standard deviation, , of the sample

proportion, .b. Level of confidence that we like to attach to

the interval.6. So,

a. The larger is , the greater is margin of error.

b. To ensure that the population proportion is contained in the interval, we have to use a higher confidence level.

c. We add a probabilistic statement so the interval is based on the confidence level.

d. An interval constructed based on the confidence level is called a confidence interval.

7. Confidence interval is defined as

p̂ p ˆ .65p

.45 .85

p̂p̂

p̂

8

An Interval Estimation8. The confidence level associated with a confidence interval is

defined as

Confidence level (1 )100% orit is called confidence coefficient when expressed as probability and expressed as: Confidence level (1 )

significance level.

This formula means that we have (1 )100% confidence that the interval contains the true population proportion.

9

7.3-7.4 ESTIMATION OF A POPULATION MEAN: KNOWN

The three possible cases on how to construct a confidence interval for population mean with known are as follows:I. We use standard normal distribution to construct the confidence

interval for with if:

1. Standard deviation is known.2. Sample size is small, n<303. Population is normally distributed or at least close to normal

distribution provided there is no outliers.II. We use standard normal distribution to construct the confidence

interval for with if:

1. Standard deviation is known.2. Sample size is large, 3. By central limit theorem, the sampling distribution of the sample

mean is approximately normal. However, we may not be able to use standard normal distribution if the population distribution is very different from normal distribution.

assuming that 0.05x n n N

assuming that 0.05x n n N

30n

10

ESTIMATION OF A POPULATION MEAN: KNOWN

III. We use a nonparametric method to construct the confidence interval if:

a. Standard deviation is known.b. Sample size is small, n<30c. Population is not normally distributed or is unknown.

The rest of this section will deal with Cases I and II. We will not cover the 3rd case.

FormulaThe (1 )100% confidence interval for under Cases I and II isdefined as,

(1 )100% confidence interval ,

where,

and the margin of error, E

x

x x

x z

zn

11


Three Possible Cases

12


Let us revisit the definition of confidence level. Remember that

confidence level is the area under the standard normal

curve of between two points on both sides and of equal distance from .

Confidence level (1 ) where is the significance level.

(1 )%x

13


How to determine z given confidence level

1. To find the 2 locations for z, first the a. Area between the 2 z’s isb. Since z1 and z2 are the same distance from

the mean, , then the sum of areas to the left of z1 and right of z2 is

c. Since the area to the left of z1 and the area to the right of z2 are equal, then:

2. Using table A-2, we can find the values of z1 and z2 that correspond to the required area.

3. Note that the values of z1 and z2 are the same, but they have opposite signs.

(1 )

1 (1 ) 1

2

Area to the left of z2

Area to the left of z2

14


Interpretation of confidence levelLet us consider 20 samples of the same size taken from the same population. Then,

1. Let us calculate the sample mean, for each sample.

2. Let us then calculate the confidence interval for around each sample mean, , based on a confidence level of 90%.

3. The normal curve of the sampling distribution for is shown to the right.

4. In the context of this example, we say that 90% of the intervals such as for x1 and x2 will include , and 10% such as the interval around x3 will not.

x

x

x

15


Width of a confidence IntervalAs stated previously, the confidence interval is defined as,

1. z which depends on the confidence level

2. and n because

(1 )100% confidence interval ,

where is margin of error. Then the width of the confidence

interval depends on , which in turn depends on:

x

x

x

x z

z

z

Since is out of control of the investigators, then the width of confidence level can only be controlled by using z and n. Thus, the width is controlled by the following relationships:

x

n

1. The value of z increases as the confidence level increases.2. The value of z decreases as the confidence level decreases.3. With n remaining constant, the higher the confidence level, the larger the

width of a confidence interval.4. An increase in the sample size causes a decrease in the width of confidence

levelIn conclusion, we can reduce the width of a confidence interval by lowering confidence level or increase sample size.

16

Determining the Sample Size for the Estimation of MeanBecause of the problems associated with conducting a census or even a sample survey, we need to find a way to determine a sample size that will produce required results without wasting unnecessary effort or financial resources on surveying larger sample size.

2 2

2

z z zE E n n E n z n

En n

zn

ESo, to find the appropriate sample size, n, we need: Confidence level Width of a confidence interval

So, having a predetermined margin of error, we can find the sample size that will produce the required results.Note that if is not known, one could take a small sample and calculate sample standard deviation, s, and then use the s in lieu of in the formula.

17

ESTIMATION OF A POPULATION MEAN: KNOWNExample #1 – Problem 8.10

Example #1 – Solution

1

a) Given: 1 .90 .10, .052

From Table IV, the value of z that corresponds to the area .05 to the left of z is 1.65 or

1.64. Also, the value of z that corresponds to the ar

2ea .05 to the right of z is 1.64 or 1.65.

Thus, the value of z that corresponds to a confidence level of 90% is 1.64 or 1.65.

Find z for each of the following confidence levelsa) 90% b) 95%

1

b) Given: 1 .95 .05, .0252

From Table IV, the value of z that corresponds to the area .025 to the left of z is 1.96

Also, the value of z that corresponds to the area .025

2

to the right or 0.975 to the left of z is 1.96

Thus, the value of z that corresponds to a confidence level of 95% is 1.96.

z1 z2

.05.05

.90

z1 z2

.025.025

.9518

ESTIMATION OF A POPULATION MEAN: KNOWNExample #2


For a data set obtained from a sample n = 81 and =48.25. It is known that = 4.8.

a) What is the point estimate of ? b) Make a 95% confidence interval forc) What is the margin of error of estimate for part b?

x

25.48 of estimatePoint

of estimatepoint theis What a)

d.distributenormally is population ,8.4 ,25.48 ,81 :Given

x

xn

19

ESTIMATION OF A POPULATION MEAN: KNOWNExample #2 – Solution

05.1)5333(.96.1 E

error of margin theisWhat c)

49.30 to20.47 05.125.48

)5333(.96.125.48for interval confidence The

533333.081

8.4 Thus,

1.96. is curve theof taileach in 0.025 ofarea the to

scorrespond that z of value theIV, Table from Thus, interval. confidence make

toondistributi normal use can we thenddistributenormally is population Since

025.02

05.

2 is curve

normal theof taileach in areas the Hence,95%. is level confidence The

.for interval confidence 95%a Makeb)


x

x

x

z

zxn

xn

20


Example #3

The standard deviation for population is = 14.8. A sample of 25 observations selected from this population gave a mean equal to 143.72. The population is known to have a normal distribution.a) Make a 99% confidence interval for b) Construct a 95% confidence interval for c) Determine a 90% confidence interval for d) Does the width of the confidence intervals constructed in parts a through c decrease as the confidence level decreases? Explain your answer.

21


151.36 to08.136 64.772.143

)96.2(58.272.143for interval confidence 99% theThus,

2.58.or 2.57 is curve theof each tailin 0.005 of area the to


on todistributi normal usecan then weddistributenormally is population Since

005.02

01.

2 is curve

normal theof each tailin areas theHence, 99%. is level confidence The

.for interval confidence 99% a Make a)

96.225

8.14

then,,population ddistributenormally a fromdrawn is sample Since


x

xzx

n

xn


22


149.52 to92.137 80.572.143


1.96. is curve theof each tailin 0.025 of area the to



025.02

05.

2 is curve


.for interval confidence 95% a Make b)

96.225

8.14



x

xzx

n

xn


23


interval. theof

width theand valuez theis so decreases, level confidence theas because Yes, d)

148.60 to84.138 88.472.143


1.65.or 1.64 is curve theof each tailin 0.05 of area the to



05.02

10.

2 is curve


.for interval confidence 90% a Make c)

96.225

8.14


x

xzx

n


24

ESTIMATION OF A POPULATION MEAN: KNOWNExample #4 For a population, the value of the standard deviation is 4.96. A sample of 32 observations taken from this population produced the following data.

74 85 72 73 86 81 77 60 83 78 79 88 76 73 84 78 81 72 82 81 79 83 88 86 78 83 87 82 80 84 76 74

a) What is the point estimate of b) Make a 99% confidence interval for c) What is the margin or error of estimate for part b?


4688.479 of estimatePoint

of estimatepoint theis What a)

8768.32

96.4 Thus, rem.limit theo central by the 30 n

large, is size sample because normalely approximat is x ofon distributi

sampling theknown,not is xfor on distributi sampling heAlthough t

4688.7932

2543 data,given thefrom and 32 :Given

x

x

n

xn

25


2621.2)2.58(.8768E

error ofmargin theis What c)

81.73 to21.77 2621.24688.79

)8768(.58.24688.79for interval confidence 99% theThus,

2.58.or 2.57 is curve theof each tailin 0.005 of area the toscorrespond that

z of value theIV, Table from Thus, interval. confidence make on todistributi

normal usecan then wed,distributenormally ely approximat is sample Since

005.02

01.

2 is curve


.for interval confidence 99% a Make b)

8768.32

96.4

then,d,distributenormally ely approximat is sample Since

x

x

x

z

zx

n


26

ESTIMATION OF A POPULATION MEAN: KNOWNExample #5For a population data set, = 14.50.a) What should the sample size be for a 98% confidence interval for to have a margin of error of estimate equal to 5.50?b) What should the sample size be for a 95% confidence interval for to have a margin of error of estimate equal to 4.25?


3873.37)5.5(

)50.14()33.2(

Thus, 2.33. is curve normal under the taileach

in 0.01 ofarea the toscorrespond that z of value theIV, Table From

01.02

02.

2 is curve normal under the taileach in areas The

size. sample find 5.50, Eand 98% level confidence that theGiven a)

50.14 :Given

2

22

2

22

E

zn

27

ESTIMATION OF A POPULATION MEAN: KNOWNExample #5 – Solution

4571.44)25.4(

)50.14()96.1(

Thus, 1.96. is curve normal under the taileach

in 0.025 ofarea the toscorrespond that z of value theIV, Table From

025.02

05.

2 is curve normal under the taileach in areas The

size. sample find 4.25, Eand 95% level confidence that theGiven b)

50.14 :Given

2

22

2

22

E

zn

28


Example #6

Inside the Box Corporation makes corrugated cardboard boxes. One type of these boxes states that the breaking capacity of this box is 75 pounds. Fifty-five randomly selected such boxes were loaded until they break. The average breaking capacity of these boxes was found to be 78.52 pounds. Suppose that the standard deviation of the breaking capacities of all such boxes is 2.63 pounds. Calculate a 99% confidence interval for the average breaking capacity of all boxes of this type.

29


pounds 79.43 to61.77 9147.52.78

)3546(.58.252.78for interval confidence 99% theThus,

2.58.or 2.57 is curve normal the

under each tailin 0.005 of area the toscorrespond that z of value theIV, Table

from Thus, .for interval confidence make on todistributi normal usecan

then wed,distributenormally ely approximat is x ofon distributi sampling Since

005.02

01.

2 is curve

normal under the each tailin areas theHence, 99%. is level confidence The

3546.55

63.2

then,d,distributenormally

is x ofon distributi sampling that theassumecan we30, n large, is sample Since

55 ,63.2 78.52, x :Given

x

xzx

n

n


30

ESTIMATION OF A POPULATION MEAN: NOT KNOWN

The three possible cases on how to construct a confidence interval for population mean when is unknown are as follows:I. We use t distribution to construct the confidence interval for if:

1. Standard deviation, , is unknown.2. Sample size is small, n<303. Population is normally distributed.

II. We use t distribution to construct the confidence interval for if:1. Standard deviation, , is unknown.2. Sample size is large,

III. We use a nonparametric method to construct the confidence interval for if:

1. Standard deviation, , is unknown.2. Sample size is small, n <30. 3. Population is not normally distributed.

30n

31

ESTIMATION OF A POPULATION MEAN: NOT KNOWN

Three Possible Cases

32

The t Distribution

• The t distribution is also called student’s t distribution. • It is similar to the normal distribution because it has:

1. A bell-shape curve, 2. A total area of 1.0 under the curve, and3. A population mean, , of zero

• It is different from the normal distribution curve because:1. It has a lower height and wider spread,2. The units are denoted by t, and3. It’s population standard deviation, , is defined as

4. df is the degree of freedom, and is defined as the number of observations that can be chosen freely. It is denoted as

• t distribution depends only one parameter, df . • As the sample size becomes larger, the t distribution approaches

the standard normal distribution.

)2/( dfdf

size sample is n where,1ndf

33

Figure 8.5 The t distribution for df = 9 and the standard normal distribution.

34

The t Distribution

Steps to read t distribution in Table V:

1. Table A-3 lists t value for a given degree of freedom and an area in the right tail under a t distribution curve.

2. This area is the same as the area in the left tail under the t distribution curve because of symmetry.

1. Locate the value of degree of freedom under the column labeled “df”, and draw a horizontal line through the row.

2. Locate the area under one of the columns for areas in the right tail under the t distribution curve, and draw a vertical line through the column.

3. The entry where the horizontal line and vertical line meet is the required t value.

4. For example, let us find a t value for a t distribution with a sample size of 9 and an area of 0.01 in the right rail of the t distribution curve.

35

The t DistributionExample #7


Find the value of t for t distribution for each of the following, a) Area in the right tail = .05 & df = 12 b) Area in the left tail = .05 & df = 12

1.997- is ondistributifor t t valuerequired theV, Table From

025.2

and ,651661 then

,66 0.025, left tail theinArea :Given

(b)

1.782 is ondistributifor t t valuerequired theV, Table From

05.2

then,12 0.05, right tail theinArea :Given

(a)

ndf

n

df

36

The t DistributionExample #8


For each of the following, find the area in the appropriate tail of the t distribution. a) t = 2.467 & df = 28 b) t = -1.672 & df = 58 c) t = -2.670 & n = 55

0.005 is curve under the

taileach inarea required theV, Table From

541551 then,55 & 670.2 t :Given c)



58 & 1.672- t :Given b)



28 & 2.467 t :Given (a)

ndfn

df

df

37

Confidence Interval for μ Using the t Distribution In Section 4.3, we define as

However, since is normally unknown, we can estimate a sample standard deviation, s, and use it in lieu of and in place of

is calculated as,

Therefore, the (1 – α)100% confidence interval for is

Note: If df>75, we can either use:1. The entries in last row of Table V, where , or2. A normal distribution to approximate the t distribution.

xn

x

xxs

xsn

ss x

xx tserror of Marginand tsx interval confidence %100)1(

df

38

Confidence Interval for μ Using the t Distribution Example #9


Find the value of t from the t distribution table for each of the following. a) Confidence level = 99% & df = 13 b) Confidence level = 95% & n = 36

2.030 t valuerequired theV, Table From

0.0252

and 05.

351 then36,n & 95% level Confidence :Given b)

3.012 t valuerequired theV, Table From

0.0052

and 01.

then,13 & 99% level Confidence :Given a)

ndf

df

39

Confidence Interval for μ Using the t Distribution Example #10 – Problem 8.47A sample of 11 observations taken from a normally distributed population produced the following data.

-7.1 10.3 8.7 -3.6 -6.0 -7.5 5.2 3.7 9.8 -4.4 6.4a) What is the point estimate of b) Make a 95% confidence interval for c) What is the margin of error of estimate for part b?


4091.111

5.15 of estimatet Poin

of estimatepoint theisWhat a)

11 :Given

x

n

40

Confidence Interval for μ Using the t Distribution


8098.4tsE c)

6.22 to3.40-

8098.44091.1)1588.2(228.24091.1

tsxfor interval confidence 95%

2.228 is ,10 and .025 of areaan for t valuerequired The

025.2

and 0.05 level, confidence 95% aFor

1588.211

1600.7ss

1600.710

11)5.15(

49.534

1

)(x

s

determine tohave weunknown, is Since

for interval confidence 95% a Make b)

x

x

x

222

df

n

nn

x

x x2

-7.1 50.41

10.3 106.09

8.7 75.69

-3.6 12.96

-6.0 36.00

-7.5 56.25

5.2 27.04

3.7 13.69

9.8 96.04

-4.4 19.36

6.4 40.96

49.5342 x5.15 x

41

Confidence Interval for μ Using the t Distribution Example #11 A random sample of 16 airline passengers at the Bay City airport showed that the mean time spent waiting in line to check in at the ticket counter was 31 minutes with a standard deviation of 7 minutes. Construct a 99% confidence interval for the mean time spent waiting in line by all passengers at this airport. Assume that such waiting times for all passengers are normally distributed.


36.16 to25.84- 16.531 )75.1(947.231

tsxfor interval confidence 99%

2.947 is ,15 and .025 of areaan for t valuerequired The

005.2

and 0.01 level, confidence 99% aFor

75.116

7ss

d.distributenormally is Population

15 then 99%, level Confidence minutes, 7 s minutes, 31x 16, n Given

x

x

df

n

df

42

ESTIMATION OF A POPULATION PROPORTION: LARGE SAMPLES

ˆ where 1p

pqq p

n

p̂

p̂

ps ˆ

We already learned that for a large sample size, that is, np>5 and nq > 5, then

1. The sampling distribution of is approximately normally distributed2. The mean, , of the sampling distribution of is equal to the population proportion3. The standard deviation, , of the sampling distribution of the sample proportion, , is define as,

Since we may not know , we will need to use as an estimate of

The Confidence interval for the p =

Margin of error =

(1 )100%

p̂ p̂

p̂

p̂p̂

ˆ

ˆ ˆˆ where point estimate of p

pqs p p

n

ˆˆpp zs

p̂zs43

DETERMINING THE SAMPLE SIZE FOR THE ESTIMATION OF PROPORTION

Given the confidence level and the values of and , the sample size that will produce a predetermined maximum of error E of the confidence interval estimate of p is

p̂

2

2

ˆ ˆz pqn

E

q̂

44

DETERMINING THE SAMPLE SIZE FOR THE ESTIMATION OF PROPORTION

In case the values of and are not known

1. We make the most conservative estimate of the sample size n by using and

2. We take a preliminary sample (of arbitrarily determined size) and calculate

and from this sample. Then use these values to find n.

p̂ q̂

ˆ .5p ˆ .5q

q̂p̂

45

Example

46

Example #12Check if the sample size is large enough to use the normal distribution to

make a confidence interval for P for each of the following cases.

a. n=50, =.25,

b. N=160, =.03

Answers:a. n = (50)(.25)=12.5, and n = (50)(.75)=37.5 so, the sample size is

large enough t use the normal distribution.

b. n = (160)(.03)= 4.8 , the sample size is not large enough to use the normal distribution .

p̂

p̂

p̂ q̂

p̂

Example Example #12 A sample of 200 observation selected from a population produced a

sample proportion equal to .91. Make a 90% confidence interval for p.

Answer: n=200, =.91, =1-.91=.09, =.02023611 The 90% confidence interval for p is =.91+1.65(.02023611)= =.877 to .943

47

pzsp ˆˆ

p̂ q̂ nqpsp /ˆˆˆ

chapter 7 estimates and sample sizes 1. estimation: an introduction we have come a long way. we...

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