Confidence Intervals: Developing Intervals with Z and t-Distributions, Study notes of Business Statistics

Download Confidence Intervals: Developing Intervals with Z and t-Distributions and more Study notes Business Statistics in PDF only on Docsity! 1 Chapter 7 Contents 1 Outline 1 2 Introduction 1 2.1 Overview of Chapter 7 . . . . . . . . . . . . . . . . . . . . . . 1 3 Chapter 7 2 3.1 Confidence Interval Applications . . . . . . . . . . . . . . . . . 2 3.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 3.3 Confidence Interval Properties . . . . . . . . . . . . . . . . . . 3 3.4 One Sample Z-confidence interval . . . . . . . . . . . . . . . . 4 3.5 One Sample t-confidence interval . . . . . . . . . . . . . . . . 6 3.6 Two Independent Samples t-confidence interval . . . . . . . . 10 3.7 Two Dependent Samples t-confidence interval . . . . . . . . . 13 3.8 One Sample Proportion z-confidence interval . . . . . . . . . . 17 2 Introduction 2.1 Overview of Chapter 7 Goal and Objectives • To develop confidence intervals estimates • To understand that each interval has two end-points (lower and upper bound) • Interpret the confidence interval • To compute the confidence interval: a point estimate ± the margin of error • To determine the sample size 3 Chapter 7 3.1 Confidence Interval Applications Applications of Estimation in Business • Store inventory value • Manufacture process • Distribution process • Drug delivery • Auditor 3.2 Definitions Definitions • Point estimate (pt.est) is a single sample statistic that estimates the population parameter, such as, the mean or proportion. • Interval estimate of the true population mean takes into account the sampling distribution of the point estimate where we have an upper bound and a lower bound. Definitions • CI - confidence interval • CV - critical value • ME - margin of error • SE - standard error • SD - standard deviation • pt.est. - point estimate • Zα 2 - normal distribution critical value(use invnorm) • tn−1 - students t distribution critical value with n-1 degrees of freedom (use math solver or the invT) 2 Example • ME = CV × SE • The critical value (CV) for 95% CI means that the area under the curve of one tail is 5% ÷ 2 or 0.025; therefore, zpercentile = invNorm(.025) = −1.96 or invNorm(.975) = +1.96 • SE = S√ n = 15√ 25 = 3 • ME = ±1.96× 3 = ±5.88 Example • Since the confidence interval is the pt.est±ME • CI = 365± 5.88 = (359.12, 370.88). • Therefore, we are 95% confident that the population mean is between 359 and 371. • Since 368, the value that is printed on the box indicates the manufac- turing process is working properly, is within the interval, there is no reason to conclude that anything is wrong with the process. Example - z-CI Let’s use the TI-83 / TI-84: • STAT → TESTS → Zinterval → STATS → σ : 15→ x̄ : 365→ n : 25→ Clevel : .95→ CALCULATE • READOUT • Zinterval→ (359.12, 370.88)→ x̄ = 365→ n = 25 • Since 368, the target of the package, is within the interval; production should continue. 5 Note • The value of Z selected for constructing such a confidence interval is called the critical value for the distribution. • There are different critical values for each level of confidence, 1 − α, where α = significance level (SL). • Frequently Used Z-CV Signif. Level Level of Conf. 2-tailed CV 10% 90% 1.645 5% 95% 1.96 1% 99% 2.5758 Note • This leads us to an interesting challenge. There is a trade off between the width of the confidence interval and the level of confidence. • 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. 3.5 One Sample t-confidence interval One Sample t-confidence interval Let’s look at 1-sample t-confidence interval History of the Student’s t William Gosset, an employee of Guinness Breweries in Ireland, had a pre- occupation 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 ”Stu- dent.” Therefore, his contribution is still known as Students t-Distribution. 6 t-based CI of the Mean This is the results of a Summer II Quiz. Summer II Quiz Data 8 11 13 19 21 23 25 25 25 28 31 35 39 47 • Construct a 95% CI for Summer II Quiz Data of 14 students • Given: x̄ = 25, SD = 10.777, n = 14; want a 95% CI? t-based CI of the Mean Example Construct a 95% CI for Summer II Quiz Data of 14 students Given: x̄ = 25, SD = 10.777, n = 14; want a 95% CI? • SE = • SD√ n = 10.777√ 14 = 2.8803 • CV = • invT (.025, 14− 1) = −2.1604 t-based CI of the Mean Construct a 95% CI for Summer II Quiz Data of 14 students Given: x̄ = 25, SD = 10.777, n = 14, SE = 2.8803, CV = ±2.1604; want 95% CI. • ME = CV × SE where CV = t 1−CL 2 ; • ME = ±2.1604× 2.8803 = ±6.2225 • pt.est = 25 • 95% CI = pt.est±ME • 95% CI → (18.777, 31.223) 7 iClicker Question A random sample of twenty-five factory workers were asked how many vacation days they take per year. The average of the sample was 22.85 days, and the standard deviation of the sample was 5.80 day. Construct a 95% confidence interval for the average number of vacation days per year taken by factory workers? a) (20.9, 24.8) b) (20.5, 25.2) c) (21.0, 25.0) d) 22.4, 23.3) e) (22.5, 23.2) iClicker Question A random sample of twenty-five factory workers were asked how many vacation days they take per year. The average of the sample was 22.85 days, and the standard deviation of the sample was 5.80 day. What sample size is needed to obtain a margin of error of 2 days per year with a 99% confidence interval? a) 55 b) 33 c) 56 d) 23 e) 32 3.6 Two Independent Samples t-confidence interval Two Independent Samples t-confidence interval Let’s look at 2 independent samples t-confidence interval 10 Comparing Mean of two Independent Samples • We are not limited to comparing an average to a constant. Suppose we want to compare the means of two independent samples. This often happens in the workplace. • Remember CI is pt.est±ME pt.est = d̄ = x̄1 − x̄2, ME = CV × SE where CV = tn1+n2−2, SE = √ SE21 + SE 2 2 Example diff of 2 indep. groups A statistics student designed a study 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 manu- factures: 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 (minutes) when the sound stopped. AA batteries Generic Brand Name x̄ 206 187.4 S 10.3 14.6 N 6 6 Example diff of 2 indep. groups Given: want a 95 % CI a) What is the pt.est =? a) pt.est = x̄1 − x̄2 = 206− 187.4 = 18.6 b) What is the standard error? b) pooled SD = √ (n1−1)s21+(n2−1)s22 n1+n2−2 = 12.6 SE = √ 12.62 6 + 12.6 2 6 = 7.27 c) What is the 95% CV? c) CV = tn1+n2−2 = invT (.025, 10) = 2.2281 11 Example diff of 2 indep. groups Given: want a 95% CI d) What is the ME? d) ME = CV × SE = 2.2281× 7.27 = 16.1983 e) What is the 95% CI? e) 95% CI → pt.est±ME → (2.35, 34.85) f) Does this confidence interval suggest that generic AA batteries will last longer than brand-name AA batteries? f) Yes, because zero is not within the interval Example diff of 2 indep. groups Given: want a 95% CI g) Interpret the 95% CI. g) We are 95% confident that the true average difference is between 2 and 35. Example: 2-sample t-CI Let’s use the TI-83 / TI-84: • STAT → TESTS → 2− SampTint→ STATS → x̄1 : 206.0 → S1 : 10.3 → n1 : 6 → x̄2 : 187.4 → S2 : 14.6 → n2 : 6→ Clevel : .95→ Pooled : Y es→ CALCULATE • READOUT • 2− sampTint→ (2.35, 34.85)→ Sxp : 12.6343 • Since zero is not within this interval, we can conclude that there is a difference between the two mean. 12 Example: 2-related means t-CI Let’s use the TI-83 / TI-84; the data is on page 92. • STAT → EDIT Enter data into L1 and L2 and place cursor on L3, do 2nd1− 2nd2→ (i.e., L1 − L2) • STAT → TESTS → tInterval → DATA → List: → L3 Freq: 1 → Clevel: .95 → CALCULATE • READOUT • tInterval→ (0.3189, 15.573)→ x̄ = 7.946→ sx = 6.1426→ n = 5 • Since zero is not within this interval, we can conclude that there is a difference between the two mean. page 104 #13: Comparing Means of 2 related groups Some stock market analysts have speculated that parts of West Michigan Telecom might be worth more than the whole. For example, the company’s communication systems in Ann Arbor and Detroit can be sold to other com- munications companies. Suppose that a stock market analyst chose nine (9) acquisition experts and asked each to predict the return (in percent) on in- vestment (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: page 104 #13: Comparing Means of 2 related groups ROI in percent 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 Difference 3 4 4 -3 1 5 3 7 5 Given: x̄ = 3.22, and s = 2.8626, What is a 95% CI? • SE = sdiff√ n = 2.8626√ 9 = 0.9542 • CV = tα 2 ,n−1 = invT (.025, 8) = 2.3060 • ME = 2.306× .9542 = 2.2004 • 95%CI → pt.est±ME → ( 1.0218, 5.4226) 15 page 104 #13: Comparing Means of 2 related groups • Does this confidence interval suggest a difference between breaking up the company or not? Yes, b/c zero is NOT within CI. • Interpret the 95% CI? We are 95% confident that the true difference among the experts is between 1.0 and 5.4. Example: 2-related means t-CI Let’s use the TI-83 / TI-84: • STAT → EDIT Enter data into L1 and L2 and place cursor on L3, do 2nd2− 2nd1→ (i.e., L2 − L1) • STAT → TESTS → tInterval → DATA → List : L3 → Freq : 1 → Clevel : .95→ CALCULATE • READOUT • tInterval→ (1.02, 5.42)→ x̄ = 3.22→ sx : 2.8626→ n = 9 • Since zero is not within this interval, we can conclude that there is a difference between the two mean. iClicker Question In order to measure the effect of a storewide sales campaign on non-sale items, the Director of Sales of a regional supermarket chain took a random sample 12 pairs of stores that were matched according to average weekly sales volume. One store of each pair (the experimental group) was exposed to the sales campaign, and the other member of the pair (the control group) was not. The average difference between the two groups and the standard 16 deviation are 4.56 and 4.29, respectively. What is the 95% margin of error? a. 4.4482 b. 2.7257 c. 5.4514 d. 2.2241 e. 2.2010 3.8 One Sample Proportion z-confidence interval One Sample Proportion z-confidence interval Let’s look at 1-sample proportion z -confidence interval Estimating the Population Proportion using Intervals • Suppose we want to estimate the proportion of a population using intervals. • Remember CI is pt.est±ME • Therefore, pt.est = success sampleSize = x n • ME = CV × SE • This test works well if n× p > 5 and n× (1− p) > 5 Example: CI for proportion 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.” 17 page 105 #17 Given: p̂ = 0.3, z0.025 = ±1.96, SE = 0.0102 d) ME = CV × SE = d) 1.96× 0.0102 = 0.0201 e) 95%CI → e) pt.est±ME → ( .2799, .3201) f) Interpret the 95% CI? f) We are 95% confident that the true proportion is between .28 and .32. iClicker Question The owner of a restaurant serving continental food wants to study charac- teristics of customers of her restaurant. In particular, she decides to focus on two variables; the amount of money spent by customers and whether or not customers order dessert. The results from a random sample of 40 customers are as follows: • Amount spent: x̄ = $48.53, S = $8.15 • 15 customers purchased dessert What is a 90% confidence interval estimate of the population proportion of customers who purchase dessert? a. (0.249, 0.501) b. (0.265, 0.485) c. 0.178, 0.572) d. (0.197, 0.553) e. 0.225, 0.525) 20 iClicker Question The owner of a restaurant serving continental food wants to study charac- teristics of customers of her restaurant. In particular, she decides to focus on two variables; the amount of money spent by customers and whether or not customers order dessert. The results from a random sample of 40 customers are as follows: • Amount spent: x̄ = $48.53, S = $8.15 • 15 customers purchased dessert If she wants to have 95% confidence of estimating the true proportion of customers who purchase dessert to within ±0.075, what is sample size is needed? a. 160 b. 68 c. 113 d. 161 e. 112 Question Questions? 21