Mrs. Daniel Alonzo & Tracy Mourning Sr. High sdaniel@dadeschools.net AP Stats Review Mrs. Daniel Alonzo & Tracy Mourning Sr. High sdaniel@dadeschools.net

Agenda AP Stats Exam Overview AP FRQ Scoring & FRQ: 2016 #1 Distributions Review FRQ: 2015 #6 Distribution Mash-Up Inference: PHANTOMS vs PANIC Clay Dice Activity FRQ: 2017 #2 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 Sample Scale for a 3 part question.

AP Stats 2017 Scores

Thursday, May 17, 2018 NOON

FRQ: 2016 #1 You have 12 minutes. Go!!!

Solution Part 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

Two Normal curves, showing the mean µ and standard deviation σ. Normal Distributions All Normal curves are symmetric, single-peaked, and bell-shaped A Specific Normal curve is described by giving its mean µ and 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.

The 68-95-99.7 Rule Although there are many different sizes and shapes of Normal curves, they all have properties in common. 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 Each observation falls into one of two categories. 1. Each observation falls into one of two categories. The probability of success is the same for each observation. 2. The probability of success is the same for each observation. The observations are all independent. The observations are all independent. There is a fixed number n of observations. 4. The variable of interest is the number of trials required to obtain the 1st success.

FRQ Answers Must Include: Name of distribution Geometric, Binomial Parameters Binomial: X (define variable), n & p Geometric: X (define variable), p Probability Statement Ex: P (X = 7) or P (X ≥ 3) Calculation and p-value Calculator notation is okay, but needs to be labeled. Solution interpreted in context.

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. What is the probability that two customers used the express check out?

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 Distribution N= 5 P = 0.25 X = # of people use express P (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) = .2637 5= number of customers 0.25= probability of success 2= number of successes desired There is a 26.37% chance that exactly 2 customers will use the express checkout lane between 5pm and 7pm.

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? 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)

stapplet.com Probability 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. The population distribution gives the values of the variable for all the individuals in the population. The distribution of sample data shows the values of the variable for all the individuals in the sample. The sampling distribution shows the statistic values from all the possible samples of the same size from the population.

Sample Proportion Formulas 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.

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 Slightly Skewed 15 Heavily Skewed 30 Unknown

FRQ: 2015 #6 You 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 a. 0.019 a. normalcdf ( 140, inf, 120, 10.5) = 0.0287 c. binompdf (3, .5, 0) = 0.064 a. normpdf (850, inf, 840, 7.9) = 0.064 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: μ = 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 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 #2 You have 12 minutes, GO!!!

2017 #2 Solutions

2017 #2 Solutions

2017 #2 Solutions

2017 #2 Solutions

FRQ: 2017 #6 You have 25 minutes. Go!!!