Unit 3: Foundations for inference Lecture 4: Review / Synthesis Statistics 101 Thomas Leininger June 6, 2013
Announcements Announcements MT tomorrow: you bring your cheat sheet & calculator, I will provide tables Project proposal due Tuesday Requested topics for today? Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 2 / 17
A few details Hypothesis testing What you need to know about HTs: Format for answering a hypothesis test question: State the null and alternative hypotheses 1 Check the conditions 2 Calculate the Z -statistic (and standard error) 3 Calculate the p -value (double if two-sided hypothesis) 4 Reject or fail to reject the null hypothesis 5 Interpret your decision in context of the problem 6 Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 3 / 17
A few details Confidence intervals What you need to know about CIs: Format for answering a confidence interval question: Check conditions 1 Find and state the critical value ( z ⋆ ) 2 Calculate the standard error 3 Calculate the confidence interval 4 Interpret your confidence interval in context of the problem 5 Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 4 / 17
Review Populations vs. sampling Populations vs. sampling Two statistics students at UCLA conducted an energy efficiency survey of graduate student apartments. There were seven university apartment buildings, and the students randomly selected three to be included in the study. In each building, they randomly chose 25 apartments. What population does their results actually apply to? (a) All graduate student apartments in the world. (b) All UCLA graduate student apartments in this group of seven buildings. (c) All UCLA graduate student apartments in the three sampled buildings. (d) All UCLA graduate student apartments that were sampled in the three buildings. Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 5 / 17
Review Populations vs. sampling Populations vs. sampling Two statistics students at UCLA conducted an energy efficiency survey of graduate student apartments. There were seven university apartment buildings, and the students randomly selected three to be included in the study. In each building, they randomly chose 25 apartments. What type of study is this? (a) an observational study using simple random sampling. (b) a double-blind experiment. (c) an observational study using cluster sampling. (d) an observational study using stratified sampling. Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 6 / 17
Review Sampling strategies What is the difference between stratified and cluster sampling? Why might we choose either of these methods over a simple random sam- ple? Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 7 / 17
Review Contingency tables Contingency tables Table from earlier describing agreement with parents’ policial views and view on legalizing marijuana: Parent Politics Legalize MJ No Yes Total No 11 40 51 Yes 36 78 114 Total 47 118 165 Let’s put this information into a Venn diagram. When does P ( A and B ) = P ( A ) ∗ P ( B ) ? What formula do I use otherwise? Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 8 / 17
Review Conditional probability Testing for AIDS – with counts Suppose that the proportion of people infected with AIDS in a large population is 0.01. If AIDS is present, a certain medical test is positive with probability 0.997 (called the sensitivity of the test). If AIDS is not present, the test is negative with probability 0.985 (called the specificity of the test). If a person tests positive, what is the probability that they have AIDS? Let’s assume there are 1 million individuals in this population. How many are expected to have AIDS, and how many are not expected to have AIDS? Have AIDS: 1 , 000 , 000 × 0 . 01 = 10 , 000 Don’t have AIDS: 1 , 000 , 000 × 0 . 99 = 990 , 000 From http://www.pitt.edu/ ∼ nancyp/stat-1000-s07/week6.pdf . Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 9 / 17
Review Conditional probability Testing for AIDS – with counts (cont.) Question How many of the people with AIDS would we expect to test positive? (a) 30F (b) 9,850 (c) 9,970 (d) 987,030 (e) 997,000 Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 10 / 17
Review Conditional probability Testing for AIDS – with counts (cont.) 10,000 × 0.997 = 9,970 + 7 9 9 0 . AIDS: 10,000 − 0 . 0 0 3 10,000 × 0.003 = 30 0 . 01 1,000,000 990,000 × 0.015 = 14,850 + 0 . 99 no 5 1 0 0 . AIDS: − 990,000 0 . 9 8 5 990,000 × 0.985 = 975,150 9 , 970 P ( AIDS | +) = 9 , 970 + 14 , 850 ≈ 0 . 40 Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 11 / 17
Review Conditional probability Testing for AIDS – with probabilities P ( AIDS & +) = 0 . 01 × 0 . 997 + 7 9 9 0 . AIDS − 0 . 0 0 3 P ( AIDS & − ) = 0 . 01 × 0 . 003 0 . 01 P ( no AIDS & +) = 0 . 99 × 0 . 015 + 0 . 99 5 1 0 0 . no AIDS − 0 . 9 8 5 P ( no AIDS & − ) = 0 . 99 × 0 . 985 0 . 01 × 0 . 997 P ( AIDS | +) = 0 . 01 × 0 . 997 + 0 . 99 × 0 . 015 ≈ 0 . 40 Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 12 / 17
Review Binomial distribution Question Which of the following probabilities should be calculated using the Bi- nomial distribution? Probability that (a) a basketball player misses 3 times in 5 shots (b) train arrives on the time on the third day for the first time (c) height of a randomly chosen 5 year old is greater than 4 feet (d) a randomly chosen individual likes chocolate ice cream best Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 13 / 17
Review Binomial distribution Why Binomial? Suppose the probability of a miss for this basketball player is 0.40. What is the probability that she misses 3 times in 5 shots? One possible scenario is that she misses the first three shots, and makes the last two. The probability of this scenario is: 0 . 4 3 × 0 . 6 2 ≈ 0 . 023 But this isn’t the only possible scenario: 1. MMM HH 3. M H MM H 5. H MM H M 7. HH MMM 9. M HH MM 2. MM H M H 4. H MMM H 6. H M H MM 8. M H M H M 10. MM HH M Each one of these scenarios has 3 M s and 2 Hs, therefore the probability of each scenario is 0 . 023. Then, the total probability is 10 × 0 . 023 = 0 . 23. Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 14 / 17
Review Binomial distribution concisely... Suppose the probability of a miss for this basketball player is 0.40. Assuming each of her shots are independent, what is the probability that she misses exaclty 3 times in 5 shots? � 5 � 5 ! × 0 . 4 3 × 0 . 6 2 3 ! × 2 ! × 0 . 4 3 × 0 . 6 2 = 3 = 10 × 0 . 023 = 0 . 23 Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 15 / 17
Review Hypothesis testing and confidence intervals A random sample of 36 female college-aged dancers was obtained and their heights (in inches) were measured. Provided below are some summary statistics and a histogram of the distribution of these dancers’ heights. The average height of all college-aged females is 64.5 inches. Do these data provide convincing evidence that the average height of female college-aged dancers is different from this value? 5 n 36 4 3 mean 63.6 inches 2 1 sd 2.13 inches 0 60 61 62 63 64 65 66 67 H 0 : µ = 64 . 5 H A : µ � 64 . 5 ¯ x = 63 . 6 , s = 2 . 13 , n = 36 , α = 0 . 05 Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 16 / 17
Review Hypothesis testing and confidence intervals ¯ x ∼ N ( mean = , SE = ) Z = p − value = 63.6 64.5 65.4 Since p-value < 0.05, reject H 0 . The data provide convincing evidence that the average height of female college-aged dancers is different than 64.5 inches. Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 17 / 17
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