Chapter 11 Inference About a Population 11.1 Inference About a Population Mean When the Population Standard Deviation is Unknown Recall: By the central limit theorem, when σ is known xis normally distributed if: • • the sample is drawn from a normal population, or the population is not normal but the sample is sufficiently large 2 When σ is unknown, we use s instead, and has the t-distribution x Zt x −µ = σs n Pop dist: Near Normal σ known n ≤ 30 Pop dist: Non-Normal μ? Z-dist ? n > 30 Z-dist n ≤ 30 Pop dist: σ unknown Near Normal n > 30 Z-dist Pop dist: Non-Normal tn-1-dist ? t = x −µ s Using the t-table n The Student- t distribution is mound-shaped, and symmetrical around zero D of freedom = n2 > n1 D of freedom = n1 Example 1: The productivity of newly hired trainees is studied It is believed that trainees can process and distribute more than 450 packages per hour within one week of hiring Can we conclude that this belief is correct, if the mean productivity observation of 50 trainees is 460.38 and the standard deviation is 38.83 Step 1:H0:µ = 450; H1:µ > 450 Step 2: α = 0.05 Step 3: n= 50, use tn-1 = t49 Step 4: Reject Region t ≥ tα,n-1 ≅ t.05,50 = 1.676 Cf 1.645 for the Z-distribution Step 5: x−µ t= s n 1.676 Critical value 460.38 − 450 = = 1.89 38.83 50 Confidence interval estimator of µ when σ x ± tα is unknown s n d.f = n − Example 2: An investor is trying to estimate the return on investment in companies that won quality awards last year A random sample of 83 such companies is selected, yields x = 15.02 s = 68.98 Construct a 95% confidence interval for the mean return s = 68.98 = 8.31 x = 15.02 s = 68.98 s = 68.98 = 8.31 x ± t α 2,n−1 s 8.31 ≅ 15.02 ± 1.990 = [13.205,16.835] n 83 t.025,82≅ t.025,80 Checking the required conditions: The Student t distribution is robust, which means that if the population is non-normal, the results of the t-test and confidence interval estimate are still valid provided that the population is “not extremely non-normal” 14 12 10 30 25 20 15 10 400 425 450 475 500 525 550 575 More -4 14 22 30 More Example Assume the content of a can of Bubbly cola is Normally distributed Design a test to see whether its manufacturer adequately fills their 1-liter (i.e., 1000 mls) bottles H0 : μ = 1, 000mls (adequately) HA: μ < 1, 000mls (1-side test) (inadequately) α =.10 (This might be just a class-room project) n = 25 (small sample size) -> Use the t24 -distribution Reject H0 at α = 10 Reject region: T < −1.32 −1.32 Critical value 10 11.2 Inference About a Population Variance This statistic is (n − 1)s σ 2 If the population is normally distributed, it has a Chi-squared distribution, with df = n-1: χ n2−1 The Chi-squared distribution 11 Testing the population variance – Left hand tail test Example 3: A container-filling machine is considered to fill liter containers consistently if the variance of the filling is less than cc (.001 liter) A random sample of 25 1-liter fills was taken, and s =.6333 Do these data support the belief that the variance is less than 1cc at 5% significance level? 12 2 Step 1:H0:σ = 1; H1:σ < Step 2: α = 0.05 χ n2−1 = χ 24 Step 3: n= 25, use χ12−α ,n −1 = χ 295, 24 = 13.85 Step 4: Reject Region Χ 24 Step 5: (n − 1) s χ = 13.85 σ Critical value (25 − 10)(.6333) = = 15.20 95 Step 6: Do not reject the null hypothesis 13 Testing the population variance – Right hand tail test; Two tail test; A right hand tail test: A two tail test H0: σ = value H1: σ > value H0: σ = value H1: σ ≠ value Rejection region Rejection region: χ ≥ χ 2α,n−1 χ ≤ χ12− α 2,n−1 or χ ≥ χ 2α 2, n−1 14 Estimating the population variance From the following probability statement 2 P(χ 1-α/2 < χ < χ α/2) = 1-α 2 we have (by substituting χ = [(n - 1)s ]/σ ) (n − 1)s χ 2α / 2 5], is approximately normally distributed, with µ = p and σ = p(1 - p)/n 18 Test statistic for p pˆ − p Z= p(1 − p) / n where np > and n(1 − p) > Interval estimator for p (1-α confidence level) pˆ ± z α / pˆ(1 − pˆ) / n provided npˆ > and n(1 − pˆ) > 19 Example 11.5 (Predicting the winner in election day): Voters are asked by a certain network to participate in an exit poll in order to predict the winner on election day Based on the data presented in Xm12.5.xls (where 1=Democrat, and 2=Republican), can the network conclude that the republican candidate will win the state college vote? 20 Step 1: H0: p = 5; H1: p > Step 2: α = 0.05 Step 3: n= 765, use the Z-dist Step 4: Reject Region Step 5: z= Z  p− p p (1 − p ) / n 1.645 532 − = = 1.77 5(1 − 5) / 765 Critical value Step 6: Reject the null hypothesis 21 Estimating the Proportion Example (marketing application): In a survey of 2000 TV viewers at 11.40 p.m on a certain night, 226 indicated they watched “The Tonight Show” Estimate the number of TVs tuned to the Tonight Show in a typical night, if there are 100 million potential television sets Use 95% confidence level pˆ ± zα / pˆ (1 − pˆ ) / n = 113 ± 1.96 113(.887) / 2000 = 113 ± 014 1-.113 = 887 226/2000 = 113 22 Flowchart of Techniques Describe a Population Data Type? Interval Nominal Type of descriptive measurement? Central Location t test & estimator of u z test & estimator of p Variability 2 X Χ test & estimator of d 12.23 ... television sets Use 95% confidence level pˆ ± zα / pˆ (1 − pˆ ) / n = 113 ± 1.96 113 (.887) / 2000 = 113 ± 014 1- .113 = 887 226/2000 = 113 22 Flowchart of Techniques Describe a Population Data Type?... Use the t24 -distribution Reject H0 at α = 10 Reject region: T < −1.32 −1.32 Critical value 10 11. 2 Inference About a Population Variance This statistic is (n − 1)s σ 2 If the population is normally... distributed, it has a Chi-squared distribution, with df = n-1: χ n2−1 The Chi-squared distribution 11 Testing the population variance – Left hand tail test Example 3: A container-filling machine