Statistical Intervalsfor a Single Sample

Chapter 8

LEARNING OBJECTIVES

• Construct confidence intervals on the mean of a normal distribution

• Construct confidence intervals on the variance and standard deviation of a normal distribution

• Construct confidence intervals on a population proportion

Confidence Interval

• Learned how a parameter can be estimated from sample data

• Confidence interval construction and hypothesis testing are the two fundamental techniques of statistical inference

• Use a sample from the full population to compute the point estimate and the interval

Confidence Interval On The Mean of a Normal Distribution, Variance Known

– From sampling distribution, L and U P (L ≤ μ ≤ U)= 1-α

– Indicates probability of 1-α that CI will contain the true value of μ

– After selecting the sample and computing l and u, the CI for μ

l ≤ μ ≤ u– l and u are called the lower- and upper-

confidence limits

Confidence Interval On The Mean of a Normal Distribution, Variance Known

• Suppose X1, X2, , Xn is a random sample from a normal distribution

• Z has a standard normal distribution

• Writing zα/2 for the z-value• Hence

• Multiplying each term

• A 100(1-α )% CI on μ when variance is known

nzxnzx // 2/2/

n

XZ

1- α

zα/2-zα/2

1}

/2/2/ z

n

XzP

Example

• A confidence interval estimate is desired for the gain in a circuit on a semiconductor device

• Assume that gain is normally distributed with standard deviation of 20

a) Find a 95% CI for μ when n=10 and b) Find a 95% CI for μ when n=25 and c) Find a 99% CI for μ when n=10 andd) Find a 99% CI for μ when n=25 and

1000x1000x1000x1000x

Examplea) 95% CI for

α=0.05, Z 0.05/2 =Z 0.025 = 1.96. Substituting the values

Confidence interval

b) 95% CI for

c) 99% CI for

d) 99% CI for

96.1,100020,10 zxn

4.12126.987

//

nzxnzx

96.1,100020,25 , zxn

8.10078.992

//

nzxnzx

58.2,100020,10 , zxn

3.10167.983

//

nzxnzx

58.2,100020,25 , zxn

3.10107.989

//

nzxnzx

Choice of Sample Size

• (1-α)100% C.I. provides an estimate• Most of the time, sample mean not equal to μ • Error E =

• Choose n such that zα/2/√n = E

• Solving for n

• Results: n = [(Zα/2σ)/E]2

• 2E is the length of the resulting C.I.

X

X

Example• Consider the gain estimation problem in

previous example

• How large must n be if the length of the 95% CI is to be 40?

• Solution– α =0.05, then Zα/2 = 1.96

– Find n for the length of the 95% CI to be 40

One-Sided Confidence Bounds

• Two-sided CI gives both a lower and upper bound for μ

• Also possible to obtain one-sided confidence bounds for μ

• A 100(1-α )% lower-confidence bound for μ

• A 100(1-α )% upper-confidence bound for μ

1/ nZX

nZXu /

A Large-Sample Confidence Interval for μ

• Assumed unknown μ and known • Large-sample CI• Normality cannot be assumed and n ≥ 40• S replaces the unknown σ

• Let X1, X2,…, Xn be a random sample with unknown μ and 2

• Using CLT:

• Normally distributed

• A 100(1-α )% CI on μ:

2

nS

X

/

n

SZx

n

SZx 2/2/

C.I. on the Mean of a Normal Distribution, Variance Unknown

• Sample is small and 2 is unknown

• Wish to construct a two-sided CI on μ

• When 2 is known, we used standard normal distribution, Z

• When 2 is unknown and sample size ≥40– Replace with sample standard deviation S

• In case of normality assumption, small n, and unknown σ, Z becomes T=(X-μ)/(S/√n)

• No difference when n is large

The t Distribution• Let X1, X2,..., Xn be a random sample from a normal

distribution with unknown μ and 2

• The random variable

• Has a t-distribution with n-1 d.o.f• No. of d.o.f is the number of observation that can be

chosen freely• Also called student’s t distribution• Similar in some respect to normal distribution• Flatter than standard normal distribution =0 and 2=k/(k-2)

nS

XT

/

The t Distribution

• Several t distributions• Similar to the standard

normal distribution• Has heavier tails than the

normal• Has more probability in the

tails than the normal• As the number d.o.f

approaches infinity, the t distribution becomes standard normal distribution

The t Distribution

• Table IV provides percentage points of the t distribution

• Let tα,k be the value of the random variable T with k (d.o.f)

• Then, tα,k is an upper-tail 100α percentage point of the t distribution with k

The t Confidence Interval on μ

• A 100(1-α ) % C.I. on the mean of a normal distribution with unknown 2

• tα/2,n-1 is the upper 100α/2 percentage point of the t distribution with n-1 d.o.f

nStxnStx nn // 1,2/1,2/

Example

• An Izod impact test was performed on 20 specimens of PVC pipe

• The sample mean is 1.25 and the sample standard deviation is s=0.25

• Find a 99% lower confidence bound on Izod impact strength

Solution

• Find the value of tα/2,n-1

• α=0.01and n=20, then the value of tα/2,n-1 =2.878

054.2445.0

20

25.0878.225.1

20

25.0878.225.1

19,005.019,005.0

n

stx

n

stx

Chi-square Distribution

• Sometimes C.I. on the population variance is needed• Basis of constructing this C.I.

• Let X1, X2,..,Xn be a random sample from a normal distribution with μ and 2

• Let S2 be the sample variance• Then the random variable:

• Has a chi-square (X2) distribution with n-1 d.o.f.

2

22 )1(

Sn

X

Shape of Chi-square Distribution• The mean and variance

of the X2 are k and 2k• Several chi-square

distributions• The probability

distribution is skewed to the right

• As the k→∞, the limiting form of the X2 is the normal distribution

Percentage Points of Chi-square Distribution

• Table III provides percentage points of X2 distribution

• Let X2α,k be the value of the random variable X2 with k

(d.o.f)

• Then, X2α,k

kX

k duufXXP,

2,

2

2

)()(

C.I. on the Variance of A Normal Population

• A 100(1-α)% C.I. on 2

• X2 α/2,n-1 and X2

1-α/2,n-1 are the upper and lower 100α/2 percentage points of the chi-square distribution with n-1 degrees of freedom

21,2/1

22

1,2/2

2 )1()1(

nn X

sn

X

sn

One-sided C.I.

• A 100(1 )% lower confidence bound or upper confidence bound on 2

21,1

222

1,2

2 )1( and

)1(

nn X

sn

X

sn

Example

• A rivet is to be inserted into a hole. A random sample of n=15 parts is selected, and the hole diameter is measured

• The sample standard deviation of the hole diameter measurements is s=0.008 millimeters

• Construct a 99% lower confidence bound for 2

• Solution– For = 0.01 and X2

0.01, 14 =29.14

2

22

00003075.0

14.29

)008.0(14

A Large Sample C.I. For APopulation Proportion

• Interested to construct confidence intervals on a population proportion

• =X/n is a point estimator of the proportion• Learned if p is not close to 1 or 0 and if n is relatively

large• Sampling distribution of is approximately normal• If n is large, the distribution of

• Approximately standard normalnpp

pp

pnp

npXZ

)1(

ˆ

)1(

p̂

p̂

Confidence Interval on p• Approximate 100 (1-α) % C.I. on the proportion p of the population

where zα/2 is the upper α/2 percentage point of the standard normal distribution

• Choice of sample size– Define the error in estimating p by – E=– 100(1-α)% confident that this error less than

– Thus

n = (Zα/2/E)2p(1-p)

n

ppzpp

n

ppzp

)ˆ1(ˆˆ

)ˆ1(ˆˆ 2/2/

pp ˆp̂

n

ppz

)1(2/

n

ppzE

)1(2/

Example• Of 1000 randomly selected cases of lung cancer,

823 resulted in death within 10 years• Construct a 95% two-sided confidence interval on

the death rate from lung cancer• Solution

– 95% Confidence Interval on the death rate from lung cancer

832.01000

832ˆ p 1000n 96.12/ z

8552.08088.01000

)168.0(832.096.1832.0

1000

)168.0(832.096.1832.0

)ˆ1(ˆˆ

)ˆ1(ˆˆ 2/2/

p

p

n

ppzpp

n

ppzp

Example• How large a sample would be required in previous

example to be at least 95% confident that the error in estimating the 10-year death rate from lung cancer is less than 0.03?

• Solution– E = 0.03, = 0.05, z/2 = z0.025 = 1.96 and = 0.823 as the

initial estimate of p

62.596

)832.01(832.003.0

96.1

)ˆ1(ˆ

2

2

2/

ppE

zn