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Agenda1. AP Stats Exam Overview2. AP FRQ Scoring & FRQ: 2016 #13. Distributions Review4. FRQ: 2015 #65. Distribution Mash-Up6. Inference: PHANTOMS vs PANIC7. Clay Dice Activity8. FRQ: 2017 #29. FRQ: 2017 #6
AP Stats Exam
Exam Format
• Two 90 minute sections
• Reference Sheet and Calculator!
• First section: 40 multiple choice questions
• Second section: 6 free response questions (FRQs)
– First 5 FRQs = 12 minutes each
– FRQ # 6 : Investigative task for 30 minutes
Scoring
• Exam is scored out of 100 points.
• Each MC is worth 1.25 points
• Each FRQ (#1-5) is worth 7.5 points
• FRQ #6 is worth 15 points
What’s on the Exam?Exploring Data (20-30%): Describing patterns and departures from patterns (CH. 1-3)
Sampling and Experimentation (10-15%): Planning and conducting a study (CH. 4)
Anticipating Patterns (20-30%): Exploring random phenomena using probability and simulation (CH. 5-7)
Statistical Inference (30-40%): Estimating population parameters and testing hypotheses (CH 8-12)
FRQ Scoring
• Each question is scored from 1 to 4.
• The raw score is then multiplied by 1.875 to determined the scaled score.
• The numerical score is derived from a series of:
– Essentially Correct
– Partially Correct
– Incorrect
FRQ Scoring
Sample Scale for a 4 part question.
FRQ Scoring
AP Stats 2017 Scores
Thursday, May 17, 2018
NOON
FRQ: 2016 #1You have 12 minutes. Go!!!
SolutionPart a:The distribution of Robin’s tip amounts is skewed to the right. There is a gap between the largest tip amount (between $20 to 22.50) and the second largest tip amount (between $12.50 and $15 interval). The largest tip amount appears to be an outlier. The median tip amount is between $2.50 and $5. Robin’s tip amounts vary from a minimum of $0 to $2.50 to a maximum of between $20 and $22.50. About 78% of tip amounts are between $0 and $5.
Scoring
Solution
Part b:
If the $8 tip had really been $18, the total would increase by $10. Then, we would divide by 60, so the increase to the mean would be about 17 cents.
The median would not change, since both $8 and $18 are greater than the current median.
Scoring
Scoring Guidelines
Distributions
• Normal
• Binomial
• Geometric
• Sampling
Normal Distributions• All Normal curves are symmetric, single-peaked, and bell-
shaped
• A Specific Normal curve is described by giving its mean µand standard deviation σ.
Two Normal curves, showing the mean µ and
standard deviation σ.
Normal Distributions
• We abbreviate the Normal distribution with mean µ and standard deviation σ as N(µ,σ).
• Any particular Normal distribution is completely specified by two numbers: its mean µ and standard deviation σ.
• The mean of a Normal distribution is the center of the symmetric Normal curve.
• The standard deviation is the distance from the center to the change-of-curvature points on either side.
Although there are many different sizes and shapes of Normal curves, they all have properties in common.
The 68-95-99.7 Rule
The 68-95-99.7 Rule (“The Empirical Rule”)
In the Normal distribution with mean µ and standard deviation σ:
•Approximately 68% of the observations fall within σ of µ.
•Approximately 95% of the observations fall within 2σ of µ.
•Approximately 99.7% of the observations fall within 3σ of µ.
Binomial v. Geometric
• The primary difference between a binomial random variable and a geometric random variable is what you are counting.
• A binomial random variable counts the number of "successes" in n trials.
• A geometric random variable counts the number of trials up to and including the first "success."
Binomial vs. Geometric
The Binomial Setting The Geometric Setting
1. Each observation falls intoone of two categories.
2. The probability of successis the same for each observation.
3. The observations are allindependent.
4. There is a fixed number nof observations.
4. The variable of interest is the number of trials required to obtain the 1st
success.
1. Each observation falls intoone of two categories.
2. The probability of successis the same for each observation.
3. The observations are allindependent.
FRQ Answers Must Include:1. Name of distribution
Geometric, Binomial2. Parameters
Binomial: X (define variable), n & pGeometric: X (define variable), p
3. Probability StatementEx: P (X = 7) or P (X ≥ 3)
4. Calculation and p-valueCalculator notation is okay, but needs to be labeled.
5. Solution interpreted in context.
Twenty-five percent of the customers entering a grocery store between 5 p.m. and 7 p.m. use an express checkout. Consider five randomly selected customers, and let X denote the number among the five who use the express checkout. What is the probability that two customers used the express check out?
Let’s Practice…
Twenty-five percent of the customers entering a grocery store between 5 p.m. and 7 p.m. use an express checkout. Consider five randomly selected customers, and let X denote the number among the five who use the express checkout. hat is the probability that two customers used the express check out?
Binomial DistributionN= 5P = 0.25
X = # of people use expressP (X =2)
Twenty-five percent of the customers entering a grocery store between 5 p.m. and 7 p.m. use an express checkout. Consider five randomly selected customers, and let X denote the number among the five who use the express checkout. What is the probability that two customers used the express check out?
binompdf(5, 0.25, 2) = .26375= number of customers0.25= probability of success2= number of successes desired
There is a 26.37% chance that exactly 2 customers will use the express checkout lane between 5pm and 7pm.
A bag of Starbrust candies can be considered an SRS of the whole population of Starbrust candies. Since there are 4 flavors, the probability that each Starbrust is cherry flavored is 0.25. Each bag of Starbrust contains 200 candies. Suppose you buy one bag of Starbrust.
1. What is the probability of getting exactly 60 cherry flavored Starbrusts?
2. What is the probability of getting at least 60 cherry flavored Starbrusts?
Let’s Practice…
A bag of Starbrust candies can be considered an SRS of the whole population of Starbrust candies. Since there are 4 flavors, the probability that each Starbrust is cherry flavored is 0.25. Each bag of Starbrust contains 200 candies. Suppose you buy one bag of Starbrust.
1. What is the probability of getting exactly 60 cherry flavored Starbrusts?0.017 = Binompdf (200, 0.25, 60) OR nCk(pk)(1-p)n-k
2. What is the probability of getting at least 60 cherry flavored Starbrusts?0.0516 = binomcdf (60, 200, 0.25)
Let’s Practice…
stapplet.comProbability
Normal Approximation
Sampling Distribution
Population Distributions vs. Sampling Distributions
There are actually three distinct distributions involved when we sample repeatedly and measure a variable of interest.
1) The population distribution gives the values of the variable for all the individuals in the population.
2) The distribution of sample data shows the values of the variable for all the individuals in the sample.
3) The sampling distribution shows the statistic values from all the possible samples of the same size from the population.
Sample Proportion Formulas
n
ppp
)1(ˆ
The sample size MUST be less than 10% of the total population.
Sample Means Formulas
Notes: The sample size must be less than 10% of the population to satisfy the independence condition. The mean and standard deviation of the sample mean are true no matter the same of the population distribution.
x
Sample Distributions & Normality:
Sample Distributions & Normality:
Sample Distributions & Normality:
HOW LARGE IS LARGE ENOUGH?
If the Population shape is….
Minimum Sample Size to assume Normal
Normal 0
Slightly Skewed 15
Heavily Skewed 30
Unknown 30
FRQ: 2015 #6You have 22 minutes. Go!!!
Scoring Rubric
Scoring Rubric
Scoring Rubric
Scoring Rubric
Distributions Mash Up
SKIP: Distributions Mash Up
2B.
3B (bonus!!)
6 A
Distributions Mash Up
1. a. 0.019
2. a. normalcdf ( 140, inf, 120, 10.5) = 0.0287
c. binompdf (3, .5, 0) = 0.064
3. a. normpdf (850, inf, 840, 7.9) = 0.064
4. a. $0.70
b. 715 plays
c. Normcdf ( 500, inf, 700, 92.79) = 0.9844
Distributions Mash Up
5. a. geometcdf (.1, 4, 100) = 0.729
b. binompdf (20, .1, 2) = 0.285
c. binomcdf (104, .1, 21) = 0.001368
6. b. normcdf (4, inf, 3.9, 1.1
40) = 0.282659
OR normcdf (160, inf, 156, 6.987)
7. a. 5 questions
b. 18 + .2(7) - .8(7)(.25)
c. binomcdf (7, 0.2, 3, 7) = 0.148
PHANTOMS vs
PANIC
The Process: Confidence Intervals
Parameter: = true mean….
p = true proportion….
Assess Conditions:Random
Normal
Independent
Name Interval:
Interval: Use your calculator. “We are ____% confident
that the interval from ____ to ____ will capture the true mean/proportion of (context)…”
Conclude in Context: Answer the question being asked.
Carrying Out a Significance Test
P Parameters
H Hypothesis
A Assess Conditions
N Name the Test
T Test Statistic (Calculate)
O Obtain P-value
M Make a decision
S State conclusion
Parameter: must be in present tense. Past tense seems that it refers to sample not
population. FRQ 2015 #4: A researcher conducted a medical study to investigate whether taking a low-dose aspirin reduces the chance of developing colon cancer. As part of the study, 1,000 adult volunteers were randomly assigned to one of two groups. Half of the volunteers were assigned to the experimental group that took a low-dose aspirin each day, and the other half were assigned to the control group that took a placebo each day…..At the significance level α= 0.05, do the data provide convincing statistical evidence that taking a low-dose aspirin each day would reduce the chance of developing colon cancer among all people similar to the volunteers?
Sample Solution
Pa= proportion develop colon cancer & take aspirin
Pp = proportion develop colon cancer and take placebo
Clay Dice Activity
FRQ 2017 #2You have 12 minutes, GO!!!
2017 #2 Solutions
2017 #2 Solutions
2017 #2 Solutions
2017 #2 Solutions
FRQ: 2017 #6You have 25 minutes. Go!!!
