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

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 1- α zα/2 -zα/2

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

Example a) 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

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.

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

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:

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)

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

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

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.

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

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

One-sided C.I. A 100(1 )% lower confidence bound or upper confidence bound on 2

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

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

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)

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

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