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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
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
A few details Hypothesis testing
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
A few details Hypothesis testing
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 3 / 17
A few details Hypothesis testing
What you need to know about HTs:
Format for answering a hypothesis test question:
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 3 / 17
A few details Hypothesis testing
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 3 / 17
A few details Hypothesis testing
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 3 / 17
A few details Hypothesis testing
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 hypothesis
5 Interpret your decision in context of the problem
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 3 / 17
A few details Hypothesis testing
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 3 / 17
A few details Confidence intervals
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
A few details Confidence intervals
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 4 / 17
A few details Confidence intervals
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 4 / 17
A few details Confidence intervals
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 error
3 Calculate the confidence interval4 Interpret your confidence interval in context of the problem
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 4 / 17
A few details Confidence intervals
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 interval
4 Interpret your confidence interval in context of the problem
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 4 / 17
A few details Confidence intervals
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 4 / 17
Review
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
Review Populations vs. sampling
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
Review Populations vs. sampling
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.
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 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.
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 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 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 5 / 17
Review Sampling strategies
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
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
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? 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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 6 / 17
Review Contingency tables
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
Review Contingency tables
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?
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 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?
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 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?
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 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?
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 7 / 17
Review Conditional probability
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
Review Conditional probability
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 .
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 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 .
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 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 .
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 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, 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 8 / 17
Review Conditional probability
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 .
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 8 / 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 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 → 10, 000 × 0.997 = 9970
(d) 987,030
(e) 997,000
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 9 / 17
Review Conditional probability
Testing for AIDS – with counts (cont.)
1,000,000
noAIDS:
990,000
0.99
AIDS:10,000
0.01
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 10 / 17
Review Conditional probability
Testing for AIDS – with counts (cont.)
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 10 / 17
Review Conditional probability
Testing for AIDS – with counts (cont.)
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 10 / 17
Review Conditional probability
Testing for AIDS – with counts (cont.)
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 10 / 17
Review Conditional probability
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 11 / 17
Review Conditional probability
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 11 / 17
Review Binomial distribution
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
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 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 → 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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 12 / 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:
0.43 × 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.
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:
0.43 × 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.
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:
0.43 × 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.
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:
0.43 × 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.
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:
0.43 × 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.
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 13 / 17
Review Binomial distribution
... concisely
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?
(53
)× 0.43 × 0.62 =
5!3! × 2!
× 0.43 × 0.62
= 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.What is the probability that she misses 3 times in 5 shots?
(53
)× 0.43 × 0.62 =
5!3! × 2!
× 0.43 × 0.62
= 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.What is the probability that she misses 3 times in 5 shots?
(53
)× 0.43 × 0.62 =
5!3! × 2!
× 0.43 × 0.62
= 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.What is the probability that she misses 3 times in 5 shots?
(53
)× 0.43 × 0.62 =
5!3! × 2!
× 0.43 × 0.62
= 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.What is the probability that she misses 3 times in 5 shots?
(53
)× 0.43 × 0.62 =
5!3! × 2!
× 0.43 × 0.62
= 10 × 0.023
= 0.23
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 15 / 17
Review Binomial distribution
... concisely
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?
(53
)× 0.43 × 0.62 =
5!3! × 2!
× 0.43 × 0.62
= 10 × 0.023
= 0.23
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 15 / 17
Review Binomial distribution
... concisely
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?
(53
)× 0.43 × 0.62 =
5!3! × 2!
× 0.43 × 0.62
= 10 × 0.023
= 0.23
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 15 / 17
Review Binomial distribution
... concisely
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?
(53
)× 0.43 × 0.62 =
5!3! × 2!
× 0.43 × 0.62
= 10 × 0.023
= 0.23
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 15 / 17
Review Hypothesis testing and confidence intervals
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
Review Hypothesis testing and confidence intervals
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 16 / 17
Review Hypothesis testing and confidence intervals
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 16 / 17
Review Hypothesis testing and confidence intervals
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
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 16 / 17
Review Hypothesis testing and confidence intervals
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.
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 17 / 17
Review Hypothesis testing and confidence intervals
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.
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 17 / 17
Review Hypothesis testing and confidence intervals
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.
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 17 / 17
Review Hypothesis testing and confidence intervals
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.
Statistics 101 (Thomas Leininger) U3 - L4: Review / Synthesis June 6, 2013 17 / 17