Statistical Intervals for a Single Sample

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

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

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

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

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

Example (Continued)

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

Repeated construction of a confidence interval for .

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

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

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

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

A Large-Sample Confidence Interval for 

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

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.

Example

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

t distribution

Percentage points of the t distribution .

The Confidence Interval on  if σ is unknown

One-sided Confidence Interval on  if σ is unknown

Example Box and Whisker plot Normal probability plot

Example

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.

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

Course evaluation

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

Probability density functions of χ2 distribution

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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

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

One-Sided Confidence Bounds

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 = 0.0153. Large variance results in overfill or under fill. A 95% upper CI is given by

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.

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))

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

One-Sided Confidence Bounds

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:

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) =

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

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?

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