σ 2 : ESTIMATES, CONFIDENCE INTERVALS, AND TESTS Business Statistics

Transcription

1 σ : ESTIMATES, CONFIDENCE INTERVALS, AND TESTS Business Statistics

2 CONTENTS Estimating other parameters besides μ Estimating variance Confidence intervals for σ Hypothesis tests for σ Estimating standard deviation Old exam question Further study

3 ESTIMATING OTHER PARAMETERS BESIDES μ Besides estimates of μ, the estimation of other parameters may be needed variance σ (of any distribution) proportion π (of a Bernoulli process) median, minimum, maximum, etc. Today: variance of a normally distributed population point estimates confidence intervals hypothesis test

4 ESTIMATING VARIANCE Suppose: we have a population X with mean μ (known) and variance σ (known) (distribution unknown) we draw a random sample of size n (X 1, X,, X n ) It can be shown that n E 1 n i=1 = n 1 n So what? X i തX = n 1 n E 1 n i=1 σ n X i μ

5 ESTIMATING VARIANCE So what? 1 σ n n i=1 X i തX is not a good estimator of σ it will be wrong by a factor n 1 n in statistical terminology: it is a biased estimator That means: asymptotically (if n ) it will be correct (because n 1 n 1) we can easily remove the bias by dividing by n 1 this yields S = 1 σ n n 1 i=1 X i തX so S is an unbiased estimator of σ n That s of course why we defined s with n 1 in the denominator

6 ҧ ESTIMATING VARIANCE So s provides a point estimate for σ Can we also develop a confidence interval for σ? so construct CI σ,1 α = σ lower Recall the case for μ: the CLT says that ത X μ σ/ n ~N 0,1 which implies that CI μ,1 α =, σ upper xҧ z α/ σn, x ҧ + z α/ X μ and Student says that ത ~t S/ n n 1 which implies that CI μ,1 α = ቂx t n 1,α/ sn, ҧ t n 1,α/, s n ቃ x + Should we use z (CLT) or t ( Student ) for CI σ,1 α? σ n

7 ESTIMATING VARIANCE Both z and t are not appropriate here! X μ Just like ത ~N 0,1... σ/ n X μ... and ത ~t S/ n n it can be shown that n 1 S σ ~χ n 1 Pronounce as kai-square where χ is the χ -distribution with n 1 degrees of freedom requirement: the population X is normally distributed

8 ESTIMATING VARIANCE Some notes on the use of the χ -distribution like the t-distribution, but unlike the z-distribution, it has degrees of freedom in contrast to the z- and t-distribution, it is asymmetric in contrast to the z- and t-distribution, it is positive only for positive arguments

9 ESTIMATING VARIANCE Finding critical values with the χ -distribution more difficult due to asymmetry Example with α = 0.05 and df = 39 χ 39,0.975 = 3.65; χ 39,0.05 = 58.1 With z and t, we had z = z 0.05 etc. With χ it is different!

10 CONFIDENCE INTERVALS FOR σ Recall the relationship between the distribution of തX and the confidence interval for μ X μ We have ത ~N 0,1 σ/ n and therefore CI μ,1 α = We have ത X μ S/ n ~t n 1 and therefore CI μ,1 α = And now n 1 S σ ~χ n 1 and therefore CI σ,1 α = xҧ z lower σn, x ҧ + z upper xҧ t lower sn, ҧ n 1 s χ, upper n 1 s χ lower x + t upper σ s n n upper for the lower limit and lower for the upper limit? See next slide...

11 CONFIDENCE INTERVALS FOR σ Because P P n 1 S σ n 1 S σ σ P σ < ~χ n 1, we can calculate > χ upper = α n 1 S < 1 χ upper = α n 1 S = α χ upper And in the same way P σ > n 1 S = α χ lower So χ upper defines the lower limit and χ lower upper limit! defines the

12 EXERCISE 1 From a normal population, we sample n = 10 values and find s = 4.5. a. Give a point estimate of σ b. Give an 95% confidence interval estimate of σ

13 HYPOTHESIS TESTS FOR σ Step 1 H 0 : σ = σ 0 ; H 1 : σ σ 0 ; α = 0.05 Step sample statistic S ; reject for low and for high values Step 3 null distribution Step 4 n 1 S σ 0 value of test statistic χ calc χ lower Step 5 reject H 0 if χ calc ~χ n 1 ; normal population required = = χ n 1;1 α/ ; χ upper χ lower n 1 s σ 0 = χ n 1;α/ or χ calc ; critical values are χ upper

14 HYPOTHESIS TESTS FOR σ Using the p-value approach Steps 1-3 identical Step 4 p-value from computer output Step 5 reject H 0 if p value < α

15 HYPOTHESIS TESTS FOR σ Example A manufacturer of golf balls claims that they control the weights of the golf balls accurately so that the variance of the weights is not more than 1 mg A random sample of 31 golf balls yields a sample standard deviation of 1.31 mg Is there sufficient evidence to reject the claim at 5%?

16 HYPOTHESIS TESTS FOR σ Step 1: H 0 : σ 1; H 1 : σ > 1; α = 0.05 Step : sample statistic: S ; reject for too large values Step 3: n 1 S null distribution: ~χ σ n 1 ; assume normally distributed population (because nothing is said about it we have to assume it) Step 4: n 1 s = = σ 1 χ upper = χ crit,30,0.05 = (χ lower not needed) Step 5: reject H 0 ; there is evidence for concluding that σ > 1 mg

17 ESTIMATING STANDARD DEVIATION First estimate s and CI σ = σ lower, σ upper Next use s = s and σ = σ Again, only valid if the population is normally distributed the CLT is not of any help because we are not investigating the average but the variance

18 EXERCISE From a population, we sample n = 0 values and find s = 4.5. We want to test H 0 : σ = 8. a. Find the value of the test statistic. b. Any requirements on the data?

19 OLD EXAM QUESTION 6 May 014, Q1k

20 FURTHER STUDY Doane & Seward 5/E 8.10, 9.8 Tutorial exercises week 3 variance hypothesis test variance confidence interval