1. Statistical Intervals for a Single Sample
Chapter 8
Statistical Intervals for a Single Sample

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-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

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 zα/2 for the z-value
Hence
Multiplying each term
A 100(1-α )% CI on μ when variance is known
1- α
zα/2
-zα/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 n=10 and
Find a 95% CI for μ when n=25 and
Find a 99% CI for μ when n=10 and
Find a 99% CI for μ when n=25 and

7. Example
a) 95% CI for
α=0.05, Z 0.05/2 =Z = Substituting the values
Confidence interval
b) 95% CI for
c) 99% CI for
d) 99% CI for

8. 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.

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
α =0.05, then Zα/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-α )% lower-confidence bound for μ
A 100(1-α )% 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 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:

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 σ, Z becomes T=(X-μ)/(S/√n)
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 student’s t distribution
Similar in some respect to normal distribution
Flatter than standard normal distribution
=0 and 2=k/(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 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

16. 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

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 s=0.25
Find a 99% lower confidence bound on Izod impact strength

18. Solution Find the value of tα/2,n-1
α=0.01and n=20, then the value of tα/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→∞, 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 X2α,k be the value of the random variable X2 with k (d.o.f)
Then, X2α,k

22. 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

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 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 X20.01, 14 =29.14

25. A Large Sample C.I. For A Population 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-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)

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 = z = 1.96 and = as the initial estimate of p