Statistical Intervals for a Single Sample

1
Statistical Intervalsfor a Single Sample

• Chapter 8

2
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

3
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

4
Confidence Interval On The Mean of a Normal
Distribution, Variance Known

• From sampling distribution, L and U
• P (L µ U) 1-a
• Indicates probability of 1-a 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

5
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 za/2 for the z-value
• Hence
• Multiplying each term
• A 100(1-a ) CI on µ when variance is known

1- a
za/2
-za/2
6
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
• Find a 95 CI for µ when n10 and
• Find a 95 CI for µ when n25 and
• Find a 99 CI for µ when n10 and
• Find a 99 CI for µ when n25 and

7
Example

• a) 95 CI for
• a0.05, 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

8
Choice of Sample Size

• (1-a)100 C.I. provides an estimate
• Most of the time, sample mean not equal to µ
• Error E
• Choose n such that za/2?/vn E
• Solving for n
• Results n (Za/2s)/E2
• 2E is the length of the resulting C.I.

9
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
• a 0.05, then Za/2 1.96
• Find n for the length of the 95 CI to be 40

10
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-a ) lower-confidence bound for µ
• A 100(1-a ) upper-confidence bound for µ

11
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 s
• Let X1, X2,, Xn be a random sample with unknown
  µ and ?2
• Using CLT
• Normally distributed
• A 100(1-a ) CI on µ

12
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 s, Z becomes T(X-µ)/(S/vn)
• No difference when n is large

13
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 students t distribution
• Similar in some respect to normal distribution
• Flatter than standard normal distribution
• ?0 and ?2k/(k-2)

14
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

15
The t Distribution

• Table IV provides percentage points of the t
  distribution
• Let ta,k be the value of the random variable T
  with k (d.o.f)
• Then, ta,k is an upper-tail 100a percentage point
  of the t distribution with k

16
The t Confidence Interval on µ

• A 100(1-a ) C.I. on the mean of a normal
  distribution with unknown ?2
• ta/2,n-1 is the upper 100a/2 percentage point of
  the t distribution with n-1 d.o.f

17
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 s0.25
• Find a 99 lower confidence bound on Izod impact
  strength

18
Solution

• Find the value of ta/2,n-1
• a0.01and n20, then the value of ta/2,n-1 2.878

19
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.

20
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?8, the limiting form of the X2 is the
  normal distribution

21
Percentage Points of Chi-square Distribution

• Table III provides percentage points of X2
  distribution
• Let X2a,k be the value of the random variable X2
  with k (d.o.f)
• Then, X2a,k

22
C.I. on the Variance of A Normal Population

• A 100(1-a) C.I. on ?2
• X2 a/2,n-1 and X2 1-a/2,n-1 are the upper and
  lower 100a/2 percentage points of the chi-square
  distribution with n-1 degrees of freedom

23
One-sided C.I.

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

24
Example

• A rivet is to be inserted into a hole. A random
  sample of n15 parts is selected, and the hole
  diameter is measured
• The sample standard deviation of the hole
  diameter measurements is s0.008 millimeters
• Construct a 99 lower confidence bound for ?2
• Solution
• For ? 0.01 and X20.01, 14 29.14

25
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 normal

26
Confidence Interval on p

• Approximate 100 (1-a) C.I. on the proportion p
  of the population
• where za/2 is the upper a/2 percentage point of
  the standard normal distribution
• Choice of sample size
• Define the error in estimating p by
• E
• 100(1-a) confident that this error less than
• Thus
• n (Za/2/E)2p(1-p)

27
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

28
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