Confidence Intervals for Two Means

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1 Confidence Intervals for Two Means Section 7.5 Cathy Poliak, Ph.D. Office in Fleming 11c Department of Mathematics University of Houston Lecture Cathy Poliak, Ph.D. Office in Fleming 11c Sections (Department 7.5 of Mathematics University of Lecture Houston 21 ) / 26

2 Outline 1 Matched Pairs For Means 2 Two Population Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department 7.5 of Mathematics University of Lecture Houston 21 ) / 26

3 Popper Set Up Fill in all of the proper bubbles. Make sure your ID number is correct. Make sure the filled in circles are very dark. This is popper number 17. Cathy Poliak, Ph.D. Office in Fleming 11c Sections (Department 7.5 of Mathematics University of Lecture Houston 21 ) / 26

4 Popper #17 Questions Match the proper confidence interval for the examples given: a) ˆp ± z ˆp(1 ˆp) n b) x ± z σ n c) x ± t df 1. An experimenter flips a coin 100 times and gets 56 heads. Find the 96.5% confidence interval for the probability of flipping a head with this coin. 2. An SRS of 24 students at UH gave an average height of 6.1 feet and a standard deviation of.3 feet. Construct a 90% confidence interval for the mean height of students at UH. 3. The gas mileage for a certain model of car is known to have a standard deviation of 4 mi/gallon. A simple random sample of 49 cars of this model is chosen and found to have a mean gas mileage of 27.5 mi/gallon. Construct a 96.5% confidence interval for the mean gas mileage for this car model. Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department 7.5 of Mathematics University of Lecture Houston 21 ) / 26 s n

5 Popper #17 Questions 4. As the length of the confidence interval for the population mean increases, the degree of confidence in the interval s actually containing the population mean a) does not change b) decreases c) increases 5. What will reduce the width of a confidence interval? a) Increase variance. b) Increase confidence level. c) Decrease variance. d) Decrease number in sample. Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department 7.5 of Mathematics University of Lecture Houston 21 ) / 26

6 Example of Matched Pairs In low-speed crash test of five BMW cars, the repair costs were computed for a factory-authorized repair center and an independent repair facility. The results are as follows. Authorized repair center $797 $571 $904 $1147 $418 Independent Repair center $523 $488 $875 $911 $297 We want to estimate the mean of the difference between the two repair centers. Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department 7.5 of Mathematics University of Lecture Houston 21 ) / 26

7 Inference for Matched Pairs The previous popper question is a matched pair. We are looking at the same car. The subject units are exactly the same for both responses. We calculate the differences first and find the mean and standard deviation of the differences. Then this problem is the same as a one-sample confidence interval. We first find the differences from each observation. The point estimate is xd = mean of the differences. The standard deviation is s d = the standard deviation of the differences. ( Then the margin of error is m = t sd n ). The confidence interval is xd ± t ( sd n ). Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department 7.5 of Mathematics University of Lecture Houston 21 ) / 26

8 Crash Test Repair Costs We want to determine a 95% confidence interval for the difference in the repair cost of the authorized repair center and the independent repair center. Authorized repair center $797 $571 $904 $1147 $418 Independent Repair center $523 $488 $875 $911 $297 Differences $274 $83 $29 $236 $121 Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department 7.5 of Mathematics University of Lecture Houston 21 ) / 26

9 R code > auth=c(797,571,904,1147,418) > indep=c(523,488,875,911,297) > t.test(auth,indep,conf.level = 0.95, paired = TRUE) Paired t-test data: auth and indep t = , df = 4, p-value = alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: sample estimates: mean of the differences Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department 7.5 of Mathematics University of Lecture Houston 21 ) / 26

10 TI84(83) STAT Edit Input one list into L1 and another list into L2. At L3 subtract L1 - L2. Then compute the confidence interval by STAT TEST 8:TInterval / 26

11 TI84(83) Screen Shots / 26

12 Example 2 The effect of exercise on the amount of lactic acid in the blood was examined in an article for an exercise and sport magazine. Eight males were selected at random from those attending a week-long training camp. Blood lactate levels were measured before and after playing three games of racquetball, as shown in the table below. To make it easier I have already calculated the differences (After - Before) Player Before After Differences / 26

13 Two Population problems The goal of inference is to compare the responses in two groups. Each group is considered to be a sample from a distinct population. The responses in each group are independent of those in the other group / 26

14 Assumptions for Difference of Two Means 1. Both samples must be independent SRSs from the populations of interest. 2. Both sets of data must come form normally distributed populations / 26

15 Two-sample t If the population standard deviations σ 1 and σ 2 is unknown the sample standard deviations s 1 and s 2 is used. When we use the sample standard deviations we use the two-sample t statistic t = ( x 1 x 2 ) (µ 1 µ 2 ) s 2 1 n 1 + s2 2 n 2 with k degrees of freedom approximated by software or the smaller value of n 1 1 or n / 26

16 Approximate Degrees of Freedom The reality is that the previous model is not really Student s t, but only something close. So the calculators and other software such as R uses an approximate degrees of freedom called Satterthwaite degrees of freedom. Calculated df = ( ) s n 1 + s2 2 n 2 ( ) 1 s n 1 1 n n 2 1 ( s 2 2 n 2 ) 2 This is only to show what degrees of freedom R and the calculators are using. If we do this by hand use the smaller of n 1 1 or n / 26

17 Interval Estimation of µ 1 µ 2 1. Point Estimate: x 1 x 2 2. Confidence level: 1 α 3. Critical value: t with degrees of freedom of n 1 1 or n 2 1 whichever is smaller. In R: t = qt(c + α/2,df). 4. Margin of Error: E = t s 2 1 n 1 + s2 2 n 2 5. Confidence Interval: point estimate ± margin of error / 26

18 Example: Check out A well known grocery store chain performed a study to determine whether the average purchase through a self-checkout facility was less than the average purchase at the traditional checkout stand. To conduct the test, a random sample of 125 customer transactions at the self-checkout was obtained and a second random sample of 125 transactions from customers using traditional checkout process was obtained. The following statistics were computed from each sample Self-Checkout Traditional Checkout x 1 = $45.68 x 2 = $78.49 s 1 = $58.20 s 2 = $62.45 n 1 = 125 n 2 = 125 Develop a 90% confidence interval of the difference between the different checkouts / 26

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20 Using Software to determine confidence interval In TI-83(84): STAT TESTS 0:2-SampTInt In R: ( x 1 x 2 )-qt((1 + C)/2)*sqrt(s 2 1 /n 1 + s 2 2 /n 2) for lower value of confidence interval, ( x 1 x 2 )+qt((1 + C)/2)*sqrt(s 2 1 /n 1 + s 2 2 /n 2) for upper value of confidence interval. > ( )-qt(1.9/2,124)*sqrt(58.2^2/ ^2/125) [1] > ( )+qt(1.9/2,124)*sqrt(58.2^2/ ^2/125) [1] / 26

21 TI84(83) / 26

22 Example 2 for Difference of Two Population Means The following chart gives the statistics for Males and Females of mathematics test scores of male and female eighth graders. Males Females Sample Size 1,764 1,739 Sample Mean Standard deviation Determine a 95% confidence interval for the difference of the mean test scores between male and female eighth graders. 2. Can we say that there is evidence to say that there is a difference in the test scores of male and female eighth graders? / 26

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24 Confidence Intervals for One Parameter The intervals are calculated depending on what you are given. The following table gives you a step by step approach: Parameter µ given σ µ not given σ p proportions 1. Estimate x = x n x = x n ˆp = x n 2. Confidence Level C = 1 α this is given in the problem 3. Critical value z t with df = n 1 z 4. Standard error σ n s n ˆp(1 ˆp) n 5. Margin of error Critical value Standard error 6. Confidence interval Point estimate ± Margin of error / 26

25 Confidence Intervals for the Difference of Two Parameters The intervals are calculated depending on what you are given and what you want to find. The following table gives you a step by step approach: Parameter µ d µ 1 µ 2 p 1 p 2 proportions 1. Estimate x d = d x n 1 x 2 ˆp 1 ˆp 2 2. Confidence Level C = 1 α this is given in the problem 3. Critical value t with t with df smaller z df = n 1 of n 1 1 or n Standard error s n d s s2 2 ˆp1 (1 ˆp 1 ) + ˆp 2(1 ˆp 2 ) n 1 n 2 n 1 n 2 5. Margin of error Critical value Standard error 6. Confidence interval Point estimate ± Margin of error / 26

26 Popper #17 Question 6. If the 90% confidence limits for the population mean are 47 and 53, which of the following could be the 98% confidence limits a) [46, 54] b) [49, 53] c) [48, 55] d) [50, 50] e) [46, 51] / 26

Density Curves & Normal Distributions

Density Curves & Normal Distributions Sections 4.1 & 4.2 Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Department of Mathematics University of Houston Lecture 9-2311 Cathy Poliak, Ph.D. cathy@math.uh.edu