Announcements Midterm II Review Sta 101 - Fall 2018 ▶ Today’s office hours changed to 2 - 3pm ▶ Office hours Wednesday 2 - 3pm Duke University, Department of Statistical Science ▶ No office hours on Thursday Dr. Abrahamsen Slides posted at https://stat.duke.edu/courses/Fall18/sta101.002 1 Midterm 2 Exam Format ▶ When: Thursday, Nov 8 - In class ▶ Covers HT from Unit 3, Units 4 and Unit 5 ▶ What to bring: ▶ 3 written questions - 60 pts – Scientific calculator (graphing calculator ok, No Phones!) ▶ 5 T/F questions - 2 pts each – Cheat sheet (can be typed) ▶ Provided: Z, t and χ 2 tables ▶ 10 multiple choice questions - 3 pts each 2 3
Unit 4.1 - Inference for Numerical Variables ▶ Two mean testing problems What should you know? – Independent means – Paired (dependent) means ▶ Conditions – Independence – Approximate Normality 4 5 All other details of the inferential framework is the same... Clicker question HT : test statistic = point estimate − null A study examining the relationship between weights of school SE children and absences found a 95% confidence interval for the difference between the average number of days missed by CI : point estimate ± critical value × SE overweight and non-overweight children ( µ overweight − µ non − overweight ) to be 1.3 days to 2.8 days. According to this interval, we are 95% confident that overweight children on average miss Independent means: One mean: Paired means: df = min ( n 1 − 1 , n 2 − 1) df = n − 1 df = n diff − 1 1. 1.3 days fewer to 2.8 days more HT: HT: HT: 2. 1.3 to 2.8 days more H 0 : µ 1 − µ 2 = 0 H 0 : µ = µ 0 H 0 : µ diff = 0 3. 1.3 to 2.8 days fewer x 1 − ¯ x 2 T df = ¯ x − µ x diff − 0 T df = ¯ T df = ¯ sdiff √ s s 2 s 2 4. 1.3 days more to 2.8 days fewer √ n √ ndiff 1 2 n 1 + n 2 than non-overweight children. CI: CI: CI: s s diff x ± t ⋆ x diff ± t ⋆ √ ¯ ¯ s 2 n 1 + s 2 x 1 − ¯ x 2 ± t ⋆ √ n √ n diff ¯ df df 1 2 df n 2 6 7
Unit 4.2 - Bootstrapping Bootstrap interval, standard error ▶ Bootstrapping works as follows: For a random sample of 20 Horror movies, the dot plot below (1) take a bootstrap sample - a random sample taken with replacement shows the distribution of 100 bootstrap medians of the Rotten from the original sample, of the same size as the original sample Tomatoes audience scores. The median of the original sample is (2) calculate the bootstrap statistic - a statistic such as mean, median, 43.5 and the bootstrap standard error is 4.88. Estimate the 90% proportion, etc. computed on the bootstrap samples bootstrap confidence interval for the median RT score of horror (3) repeat steps (1) and (2) many times to create a bootstrap distribution - a distribution of bootstrap statistics movies using the standard error method. ▶ The XX% bootstrap confidence interval can be estimated by – the cutoff values for the middle XX% of the bootstrap distribution, ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● OR ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● – point estimate ± t ⋆ SE boot 30 35 40 45 50 55 bootstrap medians 8 9 Unit 4.3: Power Example - Medical history surveys Decision fail to reject H 0 reject H 0 H 0 true Type 1 Error, α 1 − α A medical research group is recruiting people to complete short surveys about their Truth medical history. For example, one survey asks for information on a person’s family H A true Type 2 Error, β Power, 1 − β history in regards to cancer. Another survey asks about what topics were discussed during the person’s last visit to a hospital. So far, on average people complete an average of 4 surveys, with the standard deviation of 2.2 surveys. The ▶ Type 1 error is rejecting H 0 when you shouldn’t have, and the research group wants to try a new interface that they think will encourage new probability of doing so is α (significance level) enrollees to complete more surveys, where they will randomize a total of 300 ▶ Type 2 error is failing to reject H 0 when you should have, and enrollees to either get the new interface or the current interface (equally distributed between the two groups). What is the power of the test that can detect an increase the probability of doing so is β (a little more complicated to of 0.5 surveys per enrollee for the new interface compared to the old interface? calculate) Assume that the new interface does not affect the standard deviation of completed ▶ Power of a test is the probability of correctly rejecting H 0 , and surveys, and α = 0 . 05 . the probability of doing so is 1 − β ▶ In hypothesis testing, we want to keep α and β low, but there are inherent trade-offs. 10 11
Calculating power Problem 1 The preceeding question can be rephrased as – How likely is it that we can reject a null hypothesis of H 0 : µ new − µ current = 0 if the new interface results in an increase of 0.5 surveys per enrollee, on Which values of ( ¯ x new interface − ¯ x old interface ) represent sufficient average? evidence to reject H 0 ? Let’s break this down intro two simpler problems: H 0 : µ new − µ current = 0 H A : µ new − µ current > 0 1. Problem 1: Which values of (¯ x new − ¯ x current ) represent sufficient evidence to reject this H 0 ? n new = n current = 150 2. Problem 2: What is the probability that we would reject this H 0 if ¯ x new − ¯ x current had come from a distribution with µ new − µ current = 0 . 5 , i.e. what is the probability that we can obtain such an observed difference from this distribution? 12 13 Problem 1 - cont. Problem 1 - cont. Clicker question What is the lowest t -score that will allow us to reject the null Clicker question hypothesis in favor of the alternative? Which values of ( ¯ x new − ¯ x current ) represent sufficient evidence to reject H 0 ? H 0 : µ new − µ current = 0 H 0 : µ new − µ current = 0 H A : µ new − µ current > 0 H A : µ new − µ current > 0 n new = n current = 150 , α = 0 . 05 , s new = 2 . 2 = s current = 2 . 2 n new = n current = 150 , α = 0 . 05 x new − ¯ x current < − 0 . 42 (a) ¯ (a) 1.65 x new − ¯ x current > − 0 . 42 (b) ¯ (b) 1.66 x new − ¯ x current < 0 . 42 (c) ¯ (c) 1.96 x new − ¯ x current > 0 . 42 0.05 (d) ¯ (d) 1.98 x new − ¯ x current > 1 . 66 (e) ¯ (e) 2.63 t* = ? 14 15
Problem 2 Problem 2 - cont. Clicker question What is the probability that we would reject this H 0 if ¯ x new − ¯ x current had come from a distribution with µ new − µ current = 0 . 5 , i.e. what is the probability that we can obtain Clicker question such an observed difference from this distribution? What is β , the Type 2 error rate? H 0 : µ new − µ current = 0 H A : µ new − µ current > 0 (a) 5% n new = n current = 150 , α = 0 . 05 , s new = 2 . 2 = s current = 2 . 2 (b) 38% (c) 62% (a) 5% (d) 80% (b) 38% (e) 95% (c) 62% (d) 80% (e) 95% 16 17 Unit 4.4: Analysis of VAriance (ANOVA) ANOVA tests for some difference in means of many different groups Null hypothesis: H 0 : µ placebo = µ purple = µ brown = . . . = µ peach = µ orange . ▶ ANOVA tests for some difference in means of many different Clicker question groups Which of the following is a correct statement of the alternative ▶ Conditions hypothesis? 1. Independence : (a) within group: sampled observations must be independent (a) For any two groups, including the placebo group, no two group (b) between group: groups must be independent of each other means are the same. 2. Approximate normality : distribution should be nearly normal within each group (b) For any two groups, not including the placebo group, no two 3. Equal variance : groups should have roughly equal variability group means are the same. (c) Amongst the jelly bean groups, there are at least two groups that have different group means from each other. (d) Amongst all groups, there are at least two groups that have different group means from each other. 18 19
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