Lecture 10: Interval Estimationweb.iku.edu.tr/~eyavuz/dersler/probability/10.pdfLecture 10: Interval...

Introduction The Estimation of Means

Lecture 10: Interval Estimation

Assist. Prof. Dr. Emel YAVUZ DUMAN

Introduction to Probability and StatisticsIstanbul Kultur UniversityFaculty of Engineering


Outline

1 Introduction

2 The Estimation of Means


Introduction

An estimate of a population parameter given by a single number iscalled a point estimate of the parameter.


Introduction

An estimate of a population parameter given by a single number iscalled a point estimate of the parameter. An estimate of apopulation parameter given by two numbers between which theparameter may be considered to lie is called an interval estimate ofthe parameter.


Introduction

An estimate of a population parameter given by a single number iscalled a point estimate of the parameter. An estimate of apopulation parameter given by two numbers between which theparameter may be considered to lie is called an interval estimate ofthe parameter.

Example 1

If we say that a distance is 5.28 feet, we are giving a pointestimate. If, on the other hand, we say that the distance is5.28 ± 0.03 feet, i.e., the distance lies between 5.25 and 5.31 feet,we are giving an interval estimate. A statement of the error orprecision of an estimate is often called its reliability.


Let μS and σS be the mean and standard deviation (standarderror) of the sampling distribution of a statistic S .


Let μS and σS be the mean and standard deviation (standarderror) of the sampling distribution of a statistic S .Then, if thesampling distribution of S is approximately normal (which as wehave seen is true for many statistics if the sample size n ≥ 30),


Let μS and σS be the mean and standard deviation (standarderror) of the sampling distribution of a statistic S .Then, if thesampling distribution of S is approximately normal (which as wehave seen is true for many statistics if the sample size n ≥ 30),wecan expect to find S lying in the intervals μS − σS to μS + σS ,μS − 2σS to μS + 2σS or μS − 3σS to μS + 3σS about 68.27%,95.45%, and 99.73% of the time, respectively.


Equivalently we can expect to find, or we can be confident offinding,


Equivalently we can expect to find, or we can be confident offinding, in the intervals to S − σS to S + σS , S − 2σS to S + 2σSor S − 3σS to S + 3σS about 68.27%, 95.45%, and 99.73% of thetime, respectively.


Equivalently we can expect to find, or we can be confident offinding, in the intervals to S − σS to S + σS , S − 2σS to S + 2σSor S − 3σS to S + 3σS about 68.27%, 95.45%, and 99.73% of thetime, respectively. Because of this, we call these respectiveintervals the 68.27%, 95.45%, and 99.73% confidence intervals forestimating μS (i.e., for estimating the population parameter, in thecase of an unbiased S).


Equivalently we can expect to find, or we can be confident offinding, in the intervals to S − σS to S + σS , S − 2σS to S + 2σSor S − 3σS to S + 3σS about 68.27%, 95.45%, and 99.73% of thetime, respectively. Because of this, we call these respectiveintervals the 68.27%, 95.45%, and 99.73% confidence intervals forestimating μS (i.e., for estimating the population parameter, in thecase of an unbiased S).The end numbers of these intervals(S ± σS , S ± 2σS , S ± 3σS ) are then called the 68.27%, 95.45%,and 99.73% confidence limits.


Equivalently we can expect to find, or we can be confident offinding, in the intervals to S − σS to S + σS , S − 2σS to S + 2σSor S − 3σS to S + 3σS about 68.27%, 95.45%, and 99.73% of thetime, respectively. Because of this, we call these respectiveintervals the 68.27%, 95.45%, and 99.73% confidence intervals forestimating μS (i.e., for estimating the population parameter, in thecase of an unbiased S).The end numbers of these intervals(S ± σS , S ± 2σS , S ± 3σS ) are then called the 68.27%, 95.45%,and 99.73% confidence limits. Similarly, S ± 1.96σS andS ± 2.575σS are 95% and 99% (or 0.95 and 0.99) confidence limitsfor μS .


Equivalently we can expect to find, or we can be confident offinding, in the intervals to S − σS to S + σS , S − 2σS to S + 2σSor S − 3σS to S + 3σS about 68.27%, 95.45%, and 99.73% of thetime, respectively. Because of this, we call these respectiveintervals the 68.27%, 95.45%, and 99.73% confidence intervals forestimating μS (i.e., for estimating the population parameter, in thecase of an unbiased S).The end numbers of these intervals(S ± σS , S ± 2σS , S ± 3σS ) are then called the 68.27%, 95.45%,and 99.73% confidence limits. Similarly, S ± 1.96σS andS ± 2.575σS are 95% and 99% (or 0.95 and 0.99) confidence limitsfor μS . The percentage confidence is often called the confidencelevel.


Equivalently we can expect to find, or we can be confident offinding, in the intervals to S − σS to S + σS , S − 2σS to S + 2σSor S − 3σS to S + 3σS about 68.27%, 95.45%, and 99.73% of thetime, respectively. Because of this, we call these respectiveintervals the 68.27%, 95.45%, and 99.73% confidence intervals forestimating μS (i.e., for estimating the population parameter, in thecase of an unbiased S).The end numbers of these intervals(S ± σS , S ± 2σS , S ± 3σS ) are then called the 68.27%, 95.45%,and 99.73% confidence limits. Similarly, S ± 1.96σS andS ± 2.575σS are 95% and 99% (or 0.95 and 0.99) confidence limitsfor μS . The percentage confidence is often called the confidencelevel. The numbers 1.96, 2.575, etc., in the confidence limits arecalled critical values.


Equivalently we can expect to find, or we can be confident offinding, in the intervals to S − σS to S + σS , S − 2σS to S + 2σSor S − 3σS to S + 3σS about 68.27%, 95.45%, and 99.73% of thetime, respectively. Because of this, we call these respectiveintervals the 68.27%, 95.45%, and 99.73% confidence intervals forestimating μS (i.e., for estimating the population parameter, in thecase of an unbiased S).The end numbers of these intervals(S ± σS , S ± 2σS , S ± 3σS ) are then called the 68.27%, 95.45%,and 99.73% confidence limits. Similarly, S ± 1.96σS andS ± 2.575σS are 95% and 99% (or 0.95 and 0.99) confidence limitsfor μS . The percentage confidence is often called the confidencelevel. The numbers 1.96, 2.575, etc., in the confidence limits arecalled critical values. From confidence levels we can find criticalvalues, and conversely.


Definition 2 (Confidence Interval)

If θ1 and θ2 are values of the random variables Θ1 and Θ2 suchthat

P(Θ1 < θ < Θ2) = 1− α

for some specified probability 1− α, we refer to the interval

θ1 < θ < θ2

as a (1− α)100% confidence interval for θ. The probability 1− αis called the degree of confidence, and the endpoints of the intervalare called the lower and upper confidence limits.


Definition 2 (Confidence Interval)

If θ1 and θ2 are values of the random variables Θ1 and Θ2 suchthat

P(Θ1 < θ < Θ2) = 1− α

for some specified probability 1− α, we refer to the interval

θ1 < θ < θ2

as a (1− α)100% confidence interval for θ. The probability 1− αis called the degree of confidence, and the endpoints of the intervalare called the lower and upper confidence limits.

For instance, when α = 0.05, the degree of confidence is 0.95 andwe get 95% confidence interval.


Outline

1 Introduction

2 The Estimation of Means


The Estimation of Means

We know that the sampling distribution of X for a random samplesof size n from a normal population with mean μ and the varianceσ2 is a normal distribution with

μX = μ and σ2X=

σ2

n.




μX = μ and σ2X=

σ2

n.

Thus we, can write

P(|Z | < zα/2) = 1− α

where

Z =X − μ

σ/√n

and zα/2 is such that the integral of the standard normal densityfrom zα/2 to ∞ equals α/2.




μX = μ and σ2X=

σ2

n.

Thus we, can write

P(|Z | < zα/2) = 1− α

where

Z =X − μ

σ/√n

and zα/2 is such that the integral of the standard normal densityfrom zα/2 to ∞ equals α/2. It follows that

P

(|X − μ| < zα/2 ·

σ√n

)= 1− α

or in other words, we have the following theorem:


Theorem 3

If X , the mean of a random sample of size n from a normalpopulation with the known variance σ2, is to be used as anestimator of the mean of the population, the probability is 1− αthat the error will be less than zα/2 · σ√

n.


A few values of zα/2 obtained from the standard normal table foreasy reference. To illustrate the notation, suppose we want todetermine the 90% degree of confidence. We set 1− α = 0.90 soα/2 = 0.05 and, from standard normal distribution table, we havez0.05 = 1.645. Therefore, when estimating μ by X , the 90% degreeof confidence is 1.645σ/

√n.

Values of zα/21− α 0.80 0.85 0.90 0.95 0.99

zα/2 1.28 1.44 1.645 1.96 2.575


Example 4

A team of efficiency experts intends to use the mean of a randomsample of size n = 150 to estimate the average mechanicalaptitude of assembly-line workers in a large industry (as measuredby a certain standardized test). If, based on experience, theefficiency expert can assume that σ = 6.2 for such data, what canthey assert with probability 0.99 about the maximum error of theirestimate?


Example 4

A team of efficiency experts intends to use the mean of a randomsample of size n = 150 to estimate the average mechanicalaptitude of assembly-line workers in a large industry (as measuredby a certain standardized test). If, based on experience, theefficiency expert can assume that σ = 6.2 for such data, what canthey assert with probability 0.99 about the maximum error of theirestimate?

Solution. Substituting n = 150, σ = 6.2, andα/2 = 0.01/2 = 0.005 so z0.005 = 2.575 into the expression for themaximum error, we get

2.575 · 6.2√150

= 1.30.

Thus, the efficiency experts can assert with probability 0.99 thattheir error will be less than 1.30.


Theorem 5

If x is the value of the mean of a random sample of size n from anormal population with the known variance σ2, then

x − zα/2 ·σ√n< μ < x + zα/2 ·

σ√n

is a (1− α)100% confidence interval for the mean of thepopulation.


Example 6

If a random sample of size n = 20 from a normal population withthe variance σ2 = 225 has the mean x = 64.3, construct a 95%confidence interval for the population mean μ.


Example 6

If a random sample of size n = 20 from a normal population withthe variance σ2 = 225 has the mean x = 64.3, construct a 95%confidence interval for the population mean μ.

Solution. Substituting n = 20, x = 64.3, σ = 15, andz0.025 = 1.96 into the confidence interval formula of Theorem 5,we get

64.3− 1.96 · 15√20

< μ < 64.3 + 1.96 · 15√20

which reduces to57.7 < μ < 70.9.


Strictly speaking, Theorems 3 and 5 require that we are dealingwith a random sample from a normal population with the knownvariance σ2.


Strictly speaking, Theorems 3 and 5 require that we are dealingwith a random sample from a normal population with the knownvariance σ2.However, by virtue of the central limit theorem, theseresults can also be used for random samples from nonnormalpopulations provided that n is sufficiently large;


Strictly speaking, Theorems 3 and 5 require that we are dealingwith a random sample from a normal population with the knownvariance σ2.However, by virtue of the central limit theorem, theseresults can also be used for random samples from nonnormalpopulations provided that n is sufficiently large;that is, n ≥ 30.


Strictly speaking, Theorems 3 and 5 require that we are dealingwith a random sample from a normal population with the knownvariance σ2.However, by virtue of the central limit theorem, theseresults can also be used for random samples from nonnormalpopulations provided that n is sufficiently large;that is, n ≥ 30. Inthat case, we may also substitute for σ the value of the samplestandard deviation.


Example 7

An industrial designer wants to determine the average amount oftime it takes an adult to assemble an “easy-to-assemble” toy. Usethe following data (in minutes), a random sample, construct a 95%confidence interval for the mean of the population sampled

17 13 18 19 17 21 29 22 16 28 21 1526 23 24 20 08 17 17 21 32 18 25 2216 10 20 22 19 14 30 22 12 24 28 11


Example 7

An industrial designer wants to determine the average amount oftime it takes an adult to assemble an “easy-to-assemble” toy. Usethe following data (in minutes), a random sample, construct a 95%confidence interval for the mean of the population sampled

17 13 18 19 17 21 29 22 16 28 21 1526 23 24 20 08 17 17 21 32 18 25 2216 10 20 22 19 14 30 22 12 24 28 11

Solution. Substituting n = 36, x = 19.92, z0.025 = 1.96, ands = 5.73 for σ into the confidence interval formula of Theorem 5,we get

19.92 − 1.96 · 5.73√36

< μ < 19.92 + 1.96 · 5.73√36

.

Thus, the 95% confidence limits are 18.05 and 21.79 minutes.


Example 8

Measurements of the diameters of a random sample of 200 ballbearings made by a certain machine during one week showed amean of 0.824 inch and a standard deviation of 0.042 inch. Find(a) 95%, (b) 99% confidence limits for the mean diameter of allthe ball bearings.


Example 8

Measurements of the diameters of a random sample of 200 ballbearings made by a certain machine during one week showed amean of 0.824 inch and a standard deviation of 0.042 inch. Find(a) 95%, (b) 99% confidence limits for the mean diameter of allthe ball bearings.

Solution. Since n = 200 is large, we can assume that X is verynearly normal.(a) Substituting n = 200, x = 0.824, z0.025 = 1.96, and s = 0.042for σ into the confidence interval formula of Theorem 5, we get

0.824 − 1.96 · 0.042√200

< μ < 0.824 + 1.96 · 0.042√200

.

Thus, the 95% confidence limits are 0.824 ± 0.006 inch.


(b) Substituting n = 200, x = 0.824, z0.005 = 2.575, and s = 0.042for σ into the confidence interval formula of Theorem 5, we get

0.824 − 2.575 · 0.042√200

< μ < 0.824 + 2.575 · 0.042√200

.

Thus, the 99% confidence limits are 0.824 ± 0.008 inch.


Example 9

A random sample of 50 mathematics grades out of a total of 200showed a mean of 75 and a standard deviation of 10. (a) What arethe 95% confidence limits for the mean of the 200 grades? (b)With what degree of confidence could we say that the mean of all200 grades is 75 ± 1?


Example 9

A random sample of 50 mathematics grades out of a total of 200showed a mean of 75 and a standard deviation of 10. (a) What arethe 95% confidence limits for the mean of the 200 grades? (b)With what degree of confidence could we say that the mean of all200 grades is 75 ± 1?

Solution. Since the population size is not very large comparedwith the sample size, we must adjust for sampling withoutreplacement. Using the formula

x − zα/2 ·σ√n

√N − n

N − 1< μ < x + zα/2 ·

σ√n

√N − n

N − 1

we find that the 95% confidence limits are

75− 1.96 · 10√50

√200− 50

200− 1< μ < 75 + 1.96 · 10√

50

√200 − 50

200− 1

thus75± 2.4.


(b) The confidence limits can be represented by

75− zα/2 ·10√50

√200− 50

200− 1< μ < 75 + zα/2 ·

10√50

√200 − 50

200− 1

or75± 1.23zα/2.

since this must be equal 75± 1, we have 1.23zα/2 = 1 orzα/2 = 0.81. The The area under the normal curve from z = 0 toz = 0.81 is 0.2910; hence, the required degree of confidence is2(0.2919) = 0.582 or 58.2%.


Thank You!!!

Lecture 10: Interval Estimationweb.iku.edu.tr/~eyavuz/dersler/probability/10.pdfLecture 10: Interval...

Documents

Transcript of Lecture 10: Interval Estimationweb.iku.edu.tr/~eyavuz/dersler/probability/10.pdfLecture 10: Interval...