1. Statistical Intervals for a Single Sample

2. Interval Estimation
Why do we need it?
Problems of point estimators.

3. Confidence Interval on the Mean of a Normal Distribution, Variance Known

4. Confidence Interval on the Mean of a Normal Distribution, Variance Known

5. Confidence Interval on the Mean of a Normal Distribution, Variance Known
Since Z follows a standard normal distribution, we can write:
Lower-confidence limit
upper-confidence limit
This a 100(1-α)% CI on μ

6. Find a point estimate and a 95% CI for μ
Example
The impact energy causing fracture of some metallic alloy is a normal random variable with σ = 1J.
A sample of 10 readings:
64.1, 64.7, 64.5, 64.6, 64.5, 64.3, 64.6, 64.8, 64.2, 64.3.
Find a point estimate and a 95% CI for μ

7. Example (Continued)

8. Interpreting a Confidence Interval
The confidence interval is a random interval
The appropriate interpretation of a CI interval is:
If an infinite number of samples are collected and a CI is constructed from each sample, then 100(1-)% of the CIs will contain the true value of .
We do not know if the CI contains , but the method used generates CIs that contains  100(1-)% of the time
Demo

9. Repeated construction of a confidence interval for .

10. Precision of Estimation and Sample Size
1 – α large
the CI is wide
the estimation of μ is less precise.
Choose n such that:

11. Example As 1-α increases zα/2 increases hence n increases
As σ increases, n increases
As E decreases, n increases

12. One-sided Confidence Bounds
A 100(1-α)% upper-confidence bound for µ is
A 100(1-α)% lower-confidence bound for µ is

13. A Large-Sample Confidence Interval for 
If X1, X2, … Xn are independent r.v.
where Xi ~f(µ, σ2), then
If s is used to estimate σ, then
is approximately normal

14. A Large-Sample Confidence Interval for 

15. A sample of fish was selected from 53 lakes.
Example
A sample of fish was selected from 53 lakes.
Mercury concentration in the muscle tissue was measured (ppm). The mercury concentration values are

16. Check normality of the data
The plots indicate that the distribution of mercury concentration in not normal.
Since n > 40, the assumption of normality in not necessary.

17. Example

18. Confidence Interval on the Mean of a Normal Distribution, Variance Unknown
The t distribution
Let X1, X2, … Xn be independent and Xi ~N(µ, σ2)
then
has a t distribution with n – 1 degrees of freedom

19. t distribution

20. Percentage points of the t distribution.

21. The Confidence Interval on  if σ is unknown

22. One-sided Confidence Interval on  if σ is unknown

23. Example
Box and Whisker plot
Normal probability plot

24. Example

25. Example
To estimate the GPA of KFUPM students a sample of 25 GPA’s was collected.
The sample mean and standard deviation are 2.8 and 0.5 respectively.
Since σ is unknown we use the t distribution.

26. Assumptions of this Section
x1, x2, …. xn are independent normal random variables or
n >> 0

27. Course evaluation

28. Confidence Interval on the σ2 and σ of a Normal Distribution
Chi-square distribution, χ2
Let Zi, i =1,…,k be independent standard normal r.v.
Then
follows χ2 distribution with k degrees of freedom

29. Probability density functions of χ2 distribution

30. Help Session on Thursday at 1 PM in Room 22-119

31. Distribution of S2
Suppose X1, X2, …Xn is a random sample from a normal distribution.
Let S2 be the sample variance.
What is the distribution of S2 ?
The random variable
has a χ2 distribution with n-1 degrees of freedom

32. Confidence Interval on the σ2 and σ of a Normal Distribution
Demo

33. One-Sided Confidence Bounds

34. Large variance results in overfill or under fill.
Example
The amount of liquid in detergent container is a normal random variable.
A sample of 20 bottles results in a sample variance of fill volume of s2 =
Large variance results in overfill or under fill.
A 95% upper CI is given by

35. A Large-Sample Confidence Interval For a Population Proportion
Examples:
Estimate the proportion of students who score more than 85% in Ram 2.
Estimate the percentage of defective parts in a production line.
Estimate the proportion of parts that are within specifications.

36. A sample of n units is taken from an infinite population.
Main idea
A sample of n units is taken from an infinite population.
X is the number of units that satisfy the condition
A point estimate of p is X/n
X is a binomial random variable.
E(X) = np, V(X) = np(1 - p)
If np > 5 and n(1-p) > 5, then X~N(np, np(1-p))

37. Main idea
The quantity is called the standard error of the point estimator .

38. One-Sided Confidence Bounds

39. Example
In a random sample of 85 bearings, 10 have a surface finish that is rougher than the specifications allow.
A point estimate of the proportion of bearings in the population that exceed the specification is
A 95% two-sided C.I. for p is:

40. The sample size for a specified value E is given by
Choice of Sample Size
The sample size for a specified value E is given by
The maximum value p(1 – p) =

41. Example Consider the previous example.
How large a sample is required if we want to be 95% confident that the error in estimating p is less than 0.05?
Using

42. Example Consider the previous example.
How large a sample is required if we want to be 95% confident that the error in estimating p is less than 0.05 regardless of the estimate of p?

43. Final Exam Material
Ch 2: Sections 2-3(addition rules) to 2-5(multiplication and total prob)
Ch 3: All sections except for the geometric, -ve binomial, and hypergeometric
Ch 4: All sections except for the normal approximation, Erlang, gamma, Weibull, and lognormal
Ch 7: All sections except 7.3.4(mean square error) and 7.4(methods of point estimation)
Ch 8: All sections except 8-7 tolerance and prediction intervals