The t Distribution
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 upperconfidence 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 Z X n • Writing zα/2 for the z-value • Hence P z / 2 X z / 2 } 1 / n 1- α -zα/2 zα/2 • Multiplying each term x z / 2 / n x z / 2 / n • A 100(1-α )% CI on μ when variance is known 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 a) b) c) d) Find a 95% CI for μ when n=10 and x 1000 Find a 95% CI for μ when n=25 and x 1000 Find a 99% CI for μ when n=10 and x 1000 Find a 99% CI for μ when n=25 and x 1000 Example a) 95% CI for α=0.05, Z 0.05/2 =Z 0.025 = 1.96. Substituting the values n 10, 20 x 1000, z 1.96 Confidence interval x z / n x z / n 987.6 1212.4 b) 95% CI for , n 25, 20 x 1000, z 1.96 x z / c) 99% CI for n x z / n 992.8 1007.8 , n 10, 20 x 1000, z 2.58 x z / n x z / n 983.7 1016.3 d) 99% CI for , n 25, 20 x 1000, z 2.58 x z / n x z / n 989.7 1010.3 Choice of Sample Size • • • • • • • (1-α)100% C.I. provides an estimate Most of the time, sample X mean not equal to μ Error E = X 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 X Z / n 1 • A 100(1-α )% upper-confidence bound for μ u X Z / n A Large-Sample Confidence Interval for μ • Assumed unknown μ and known 2 • 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: X S/ n • Normally distributed • A 100(1-α )% CI on μ: x Z / 2 S S x Z / 2 n n 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 T X S/ n • 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 uppertail 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 x t / 2,n 1S / n x t / 2,n1S / n • 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 s s x t0.005,19 x t0.005,19 n n 0.25 0.25 1.25 2.878 1.25 2.878 20 20 0.445 2.054 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: X 2 (n 1) S 2 2 • 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 P( X 2 X 2,k ) f (u)du X 2 ,k C.I. on the Variance of A Normal Population • A 100(1-α)% C.I. on 2 2 ( n 1) s 2 ( n 1 ) s 2 2 2 X / 2,n 1 X 1 / 2,n 1 • 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 ( n 1) s 2 2 2 X ,n 1 and 2 ( n 1 ) s 2 2 X 1 ,n 1 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 14(0.008) 2 2 29.14 0.00003075 2 A Large Sample C.I. For A Population Proportion • Interested to construct confidence intervals on a population proportion • p̂ =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 p̂ is approximately normal • If n is large, the distribution of Z X np np (1 p ) pˆ p p (1 p ) n • Approximately standard normal Confidence Interval on p • Approximate 100 (1-α) % C.I. on the proportion p of the population pˆ z / 2 pˆ (1 pˆ ) pˆ (1 pˆ ) p pˆ z / 2 n n 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 p̂ – E= p pˆ – 100(1-α)% confident that this error less than z / 2 – Thus E z / 2 p(1 p) n p(1 p) n 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 pˆ 832 0.832 1000 pˆ z / 2 0.832 1.96 n 1000 pˆ (1 pˆ ) p pˆ z / 2 n z / 2 1.96 pˆ (1 pˆ ) n 0.832(0.168) 0.832(0.168) p 0.832 1.96 1000 1000 0.8088 p 0.8552 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 z n pˆ (1 pˆ ) E 1.96 0.832(1 0.832) 0.03 596.62 2 /2 2
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