Confidence Intervals: Statistical Methods and Applications - Prof. Jung C. Wang, Study notes of Business Statistics

Download Confidence Intervals: Statistical Methods and Applications - Prof. Jung C. Wang and more Study notes Business Statistics in PDF only on Docsity! Chapter 7. Confidence Intervals J.C. Wang Department of Statistics Western Michigan University Goal and Objectives Goal: to learn confidence intervals Objectives: I To understand that each interval has two end-points (lower and upper bound) and Interpret the confidence interval I To compute the confidence interval: a point estimate ± the margin of error I To determine the sample size Outline Introduction Applications, Definations and Notation Confidence Intervals Computation z-Confidence Intervals t-Confidence Intervals Sample Size Determination An Example Comparing Two Populations Comparing Means of Two Independent Populations Comparing Means of Two Dependent Populations Confidence Interval for Proportion Confidence Interval for Population Proportion Sample Size Determination Applications of Estimation in Business examples I Store inventory value I Manufacture process I Distribution process I Drug delivery I Auditor Critical Value I z for normal distribution I t for students t-distribution I The students t-distribution has n − 1 degrees of freedom, df = n − 1 z-Critical Value I Notation: zα/2 = upper (100× α/2)th standard normal percentile I That is: P(Z > zα/2) = α/2 ≡ P(Z ≤ zα/2) = 1− α/2 So, zα/2 = invNorm(1− α/2) I Example 95% confidence interval will give 2.5% in each tail of the bell-shaped curve; therefore, the z-CVal, zcv = z.025 = invNorm(1−.025) = invNorm(.975) = 1.96. z-Critical Value continued 1 − α zα 2 α 2 α 2 area to the left of zα 2 is 1 − α 2 t-Critical Value using TI calculators 1. math −→ solver −→ tcdf(L,U,D)− A/T , where I L = tcv (to be solved) I U = 9999 I D = df = n − 1 I A = α (error rate) I T = number of tails = 2 for c.i. 2. or use invT(1− α/2,df ) Cereal Box Packaging Example Consider a cereal packaging plant in Battle Creek that is concerned with putting 368 gram of cereal into a box. I What are the costs associated with putting too much cereal in a box? I What are the costs associated with putting too little cereal in a box? Cereal Box Packaging Example continued I Suppose sample size n = 25 I Suppose sample average x = 365 grams I Suppose SD is a process SD; therefore, σ = 15 grams I Suppose we want a 95% confidence interval I Therefore, the critical value is zcv = 1.96 Problem When SD is Unknown We have been dealing with N(µ, σ) where σ (population or process SD) is known. What happens when standard deviation (σ) is not from a population or process SD? Is this requirement rigid? Can we compute standard deviation from the sample? Let us review some history first. History of the Student t Distribution William Gosset, an employee of Guinness Breweries in Ireland, had a preoccupation with making statistical inferences about the mean when SD was unknown. Since the employees of the company were not allowed to publish their scientific work under their own name. He chose the pseudonym “Student.” Therefore, his contribution is still known as Student’s t-Distribution. Comparing Standard Normal Curve with t curves −3 −2 −1 0 1 2 3 4 0.0 0.1 0.2 0.3 0.4 Comparison of Standard Normal with t Curves x de ns ity N(0,1) t1 t5 t10 t-Confidence Interval for the Mean summer II quiz example Construct a 95% CI for the mean score for Summer II Quiz Data of 14 students Given: 95% CL, x = 25, s = 10.777, n = 14, SE = 10.777√ 14 = 2.8803 CVal = tα/2,n−1 = t.025,13 = invT(1− .025,13) = 2.1604 ME = 2.1604× 2.8803 = 6.2225 pt .est = 25 95%CI = pt .est ±ME = (18.778,31.222) t-Confidence Interval for the Mean using TI calculators I Do this: STAT→ TESTS ↓ TInterval→ STATS ↓ x:25 ↓ Sx:10.777 ↓ n:14 ↓ C-Level:.95 ↓ CALCULATE I READOUT: Tinterval (18.778, 31.222) x = 25 n = 14 I We are 95% confident that the true mean quiz score is between 18.8 and 31.2. Sample Size Determination based on confidence intervals I What sample size should we use for the average quiz score determination if we want 95% confidence, ME =5, and σ = 10.777 I n = z2σ2 ME2 == 1.962 × 10.7772 52 = 17.8 ≈ 18 Example battery example A statistics student designed an experiment to see if there was any real difference in battery life between brand-name AA batteries and generic AA batteries. He used six pairs of AA alkaline batteries from two major battery manufactures: a well known brand name and a generic brand. He measured the length of battery life while playing a CD player continuously. He recorded the time (in minutes) when the sound stopped. Battery Example continued Generic Brand Name x 206 187.4 S 10.3 14.6 n 6 6 Want 95% CI I (a) What is the standard error? I (b) What is the 95% CVal? I (c) What is the ME? I (d) What is the 95% CI? I (e) Does this confidence interval suggest that generic AA batteries will last longer than brand-name AA batteries? I (f) Interpret the 95% CI. Battery Example continued, answers I (a) pooled SD = √ (n1 − 1)s21 + (n2 − 1)s22 n1 + n2 − 2 = 12.6, SE = √ 12.62 6 + 12.62 6 = 7.27 I (b) CVal = tn1+n2−2 = invT(.975,10) = 2.2281 I (c) ME = CVal × SE = 2.2281× 7.27 = 16.2527 I (d) 95% CI→ (2.35, 34.85) Battery Example continued, answers I (e) Does this confidence interval suggest that generic AA batteries will last longer than brand-name AA batteries? Yes, because zero is not within the interval I (f) Interpret the 95% CI. We are 95% confident that the true mean difference is between 2 and 35. Battery Example continued, using TI calculator I Do this: STAT→ TESTS ↓ 2-SampTInt→ STATS ↓ x1:206.0 ↓ Sx1:10.3 ↓ n1:6 ↓ x2:187.4 ↓ Sx2:14.6 ↓ n2:6 ↓ C-Level:.95 ↓ Pooled:Yes ↓ CALCULATE I READOUT: 2-sampTInt (2.3471, 34.853) df=10 : Sxp: 12.6342788 : I Zero is not within this interval, we can conclude that there is a difference between the two means. Comparing Means of two related groups I We are not limited to comparing two averages of independent populations. Suppose we want to compare the means of two related populations. I Recall CI is pt .est ±ME pt .est = x1 − x2 ME = CVal × SE where CVal = tα/2,n−1 = invT(1− α 2 ,n − 1) SE = sdiff√ n West Michigan Telecom Example problem 13 on page 104 Some stock market analysts have speculated that parts of West Michigan Telecom might be worth more that the whole. For example, the company’s communication systems in Ann Arbor and Detroit can be sold to other communications companies. Suppose that a stock market analyst chose nine (9) acquisition experts and asked each to predict the return (in percent) on investment (ROI) in the company held to the year 2003 if (i) it does business as usual, or (ii) if it breaks up its communication system and sells all its parts. Their predictions follow: West Michigan Telecom Example, continued Expert 1 2 3 4 5 6 7 8 9 Not Break 12 21 8 20 16 5 18 21 10 Break Up 15 25 12 17 17 10 21 28 15 I SE = sdiff/ √ n = 2.8626/ √ 9 = 0.9542 I CVal = tα/2,n−1 = t.025,8 = invT(1− .025,8) = 2.3060 I ME = 2.306× .9542 = 2.2004 I 95%CI −→ (1.0218, 5.4226) I Does this confidence interval suggest a difference between breaking up the company or not? Yes, because zero is NOT within CI. I Interpret the 95% CI: We are 95% confident that the true difference among the experts is between 1.0 and 5.4. West Michigan Telecom Example continued, using TI calculators I STAT→ EDIT and Enter data into L1 and L2 and place cursor on L3, do 2nd2 − 2nd1 (i.e., L2 − L1) then do STAT→ TESTS ↓ tInterval→ Data ↓ List:L3 ↓ C-Level:.95 ↓ CALCULATE I READOUT: TInterval (1.02, 5.42) x = 3.22 Sx: 2.8626 n = 9 I Zero is not within this interval, we can conclude that there is a difference between the two means. Confidence Interval for population proportion Suppose we want to estimate the population proportion using intervals. I Recall CI is pt .est ±ME I Therefore, use pt .est = success sampleSize = x n I ME = CVal × SE I This CI works well if n × p > 5 and n × (1− p) > 5 (Note: that is, it woks well if the expected number of successes and the expected number of failures are both greater than 5.) EAS Sensor Example If a sales clerk fails to remove the EAS sensor when an item is purchased, it can result in an embarrassing situation for the customer. A survey was conducted to study consumer reaction to such false alarms. Of 250 customers surveyed, 40 said that if they were to set off an EAS alarm because store personnel did not deactivate the merchandise, then “they would never shop at the store again.” EAS Sensor Example continued I pt .est = 40 250 = 0.16, SEp̂ = √ .16(1− .16) 250 = 0.02319 I CVal = zα/2 = z.025 = invNorm(1− .025) = 1.96 I ME = 1.96× 0.02319 = 0.04544 I 95%CI −→ (.11456, .20544) I Interpret the 95% CI: We are 95% confident that the true proportion is between 0.11 and 0.21.