Unit 3: Foundations for inference Lecture 4: Review /...

Unit 3: Foundations for inferenceLecture 4: Review / Synthesis

Statistics 101

Thomas Leininger

June 6, 2013

Announcements

1 Announcements

2 A few detailsHypothesis testingConfidence intervals

3 ReviewPopulations vs. samplingSampling strategiesContingency tablesConditional probabilityBinomial distributionHypothesis testing and confidence intervals

Statistics 101

U3 - L4: Review / Synthesis Thomas Leininger

Announcements

Announcements

MT tomorrow: you bring your cheat sheet & calculator, I willprovide 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

1 Announcements



Statistics 101


A few details Hypothesis testing

1 Announcements



Statistics 101



What you need to know about HTs:

Format for answering a hypothesis test question:

1 State the null and alternative hypotheses2 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 hypothesis5 Interpret your decision in context of the problem





1 State the null and alternative hypotheses

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 hypothesis5 Interpret your decision in context of the problem





1 State the null and alternative hypotheses2 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 hypothesis5 Interpret your decision in context of the problem





1 State the null and alternative hypotheses2 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 hypothesis5 Interpret your decision in context of the problem





1 State the null and alternative hypotheses2 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





1 State the null and alternative hypotheses2 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 hypothesis5 Interpret your decision in context of the problem


A few details Confidence intervals

1 Announcements



Statistics 101



What you need to know about CIs:

Format for answering a confidence interval question:

1 Find and state the critical value (z?)2 Calculate the standard error3 Calculate the confidence interval4 Interpret your confidence interval in context of the problem





1 Find and state the critical value (z?)

2 Calculate the standard error3 Calculate the confidence interval4 Interpret your confidence interval in context of the problem





1 Find and state the critical value (z?)2 Calculate the standard error

3 Calculate the confidence interval4 Interpret your confidence interval in context of the problem





1 Find and state the critical value (z?)2 Calculate the standard error3 Calculate the confidence interval

4 Interpret your confidence interval in context of the problem





1 Find and state the critical value (z?)2 Calculate the standard error3 Calculate the confidence interval4 Interpret your confidence interval in context of the problem


Review

1 Announcements



Statistics 101


Review Populations vs. sampling

1 Announcements



Statistics 101



Populations vs. sampling

Two statistics students at UCLA conducted an energy efficiencysurvey of graduate student apartments. There were seven universityapartment buildings, and the students randomly selected three to beincluded in the study. In each building, they randomly chose 25apartments.





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 sevenbuildings.

(c) All UCLA graduate student apartments in the three sampledbuildings.

(d) All UCLA graduate student apartments that were sampled in thethree buildings.





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.


Review Sampling strategies

1 Announcements



Statistics 101


Review Sampling strategies

What is the difference between stratified and cluster sampling? Whymight we choose either of these methods over a simple random sam-ple?

Stratified: homogenous strata

Cluster: heterogenous strata, but clusters are similarSRS is ideal, but

To control for how various characteristics of the population arerepresented in the sample, we might choose stratified samplingCluster sampling may be preferable for economical reasons


Review Contingency tables

1 Announcements



Statistics 101



Contingency tables

Table from earlier describing agreement with parents’ policial viewsand view on legalizing marijuana:

Parent PoliticsLegalize MJ No Yes TotalNo 11 40 51Yes 36 78 114Total 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 other-wise?



Contingency tables




When does P(A and B) = P(A) ∗ P(B)?

What formula do I use other-wise?



Contingency tables




When does P(A and B) = P(A) ∗ P(B)? What formula do I use other-wise?


Review Conditional probability

1 Announcements



Statistics 101



Testing for AIDS – with counts

Suppose that the proportion of people infected with AIDS in a largepopulation is 0.01. If AIDS is present, a certain medical test is positivewith probability 0.997 (called the sensitivity of the test). If AIDS is notpresent, the test is negative with probability 0.985 (called the specificityof the test). If a person tests positive, what is the probability that theyhave AIDS?

Let’s assume there are 1 million individuals in this population.How many are expected to have AIDS, and how many are notexpected to have AIDS?

Have AIDS: 1, 000, 000 × 0.01 = 10, 000Don’t have AIDS: 1, 000, 000 × 0.99 = 990, 000

From http:// www.pitt.edu/∼nancyp/ stat-1000-s07/ week6.pdf .


http://www.pitt.edu/~nancyp/stat-1000-s07/week6.pdf




Let’s assume there are 1 million individuals in this population.

How many are expected to have AIDS, and how many are notexpected to have AIDS?









Have AIDS: 1, 000, 000 × 0.01 = 10, 000

Don’t have AIDS: 1, 000, 000 × 0.99 = 990, 000





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




Question

How many of the people with AIDS would we expect to test positive?

(a) 30F

(b) 9,850

(c) 9,970 → 10, 000 × 0.997 = 9970

(d) 987,030

(e) 997,000




1,000,000

noAIDS:

990,000

0.99

AIDS:10,000

0.01




1,000,000

noAIDS:

990,000

0.99

AIDS:10,000

10,000 × 0.003 = 30

−0.003

10,000 × 0.997 = 9,970+

0.997

0.01




1,000,000

noAIDS:

990,000990,000 × 0.985 = 975,150

−0.985

990,000 × 0.015 = 14,850+

0.0150.99

AIDS:10,000

10,000 × 0.003 = 30

−0.003

10,000 × 0.997 = 9,970+

0.997

0.01




1,000,000

noAIDS:

990,000990,000 × 0.985 = 975,150

−0.985

990,000 × 0.015 = 14,850+

0.0150.99

AIDS:10,000

10,000 × 0.003 = 30

−0.003

10,000 × 0.997 = 9,970+

0.997

0.01

P(AIDS|+) =9, 970

9, 970 + 14, 850≈ 0.40



Testing for AIDS – with probabilities

no AIDS

P(no AIDS & −) = 0.99 × 0.985

−0.985

P(no AIDS & +) = 0.99 × 0.015+

0.0150.99

AIDS

P(AIDS & −) = 0.01 × 0.003

−0.003

P(AIDS & +) = 0.01 × 0.997+

0.997

0.01

P(AIDS|+) =0.01 × 0.997

0.01 × 0.997 + 0.99 × 0.015≈ 0.40


Review Binomial distribution

1 Announcements



Statistics 101



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



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 → k successes in ntrials

(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



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:

0.43 × 0.62 ≈ 0.023

But this isn’t the only possible scenario:

1. MMMHH2. MMHMH

3. MHMMH4. HMMMH

5. HMMHM6. HMHMM

7. HHMMM8. MHMHM

9. MHHMM10. MMHHM

Each one of these scenarios has 3 Ms and 2 Hs, therefore theprobability of each scenario is 0.023.

Then, the total probability is 10 × 0.023 = 0.23.



... concisely


(53

)× 0.43 × 0.62 =

5!3! × 2!

× 0.43 × 0.62

= 10 × 0.023

= 0.23


Review Hypothesis testing and confidence intervals

1 Announcements



Statistics 101



A random sample of 36 female college-aged dancers was obtained and theirheights (in inches) were measured. Provided below are some summarystatistics and a histogram of the distribution of these dancers’ heights. Theaverage height of all college-aged females is 64.5 inches. Do these dataprovide convincing evidence that the average height of female college-ageddancers is different from this value?

n 36

mean 63.6 inches

sd 2.13 inches60 61 62 63 64 65 66 67

01

23

45

H0 : µ = 64.5HA : µ , 64.5

x̄ = 63.6, s = 2.13, n = 36, α = 0.05



x̄ ∼ N(mean = 64.5, SE =

2.13√

36= 0.355

)

Z =63.6 − 64.5

0.355= −2.54

p−value = 2×P(Z < −2.54) = 2×0.0055 = 0.01163.6 64.5 65.4

Since p-value < 0.05, reject H0. The data provide convincingevidence that the average height of female college-aged dancers isdifferent than 64.5 inches.


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