AP Stat Mock Exam Registration Deadline: Day 54 Agenda THQ#5 Due Today DG 23 --- 20 min Begin Ch 10.2 (skipping Ch 10.1) AP Stat Mock Exam Registration Deadline: Fri, Mar 30, 2018

Advanced Placement Statistics Section 10.2: Estimating A Population Mean EQ: How do you use confidence intervals to estimate a population mean?

Part I: One Sample PROCEDURE Recall: Recall Estimates :

In class p. 648 #27 (a, b)

Confidence Intervals --- interval that we think captures the true population parameter See Formula Sheet:

Confidence Intervals --- interval that we think captures the true population parameter

Confidence Level --- probability that the value of a parameter falls within a specified interval of values for repeated samples. The 95 percent confidence level means that if one were to repeatedly take random samples of the same size from the population and construct a 95 percent confidence interval from each sample, then in the long run 95 percent of those intervals would succeed in capturing the actual value of the population parameter.

The 95 percent confidence level means that if one were to repeatedly take random samples of the same size from the population and construct a 95 percent confidence interval from each sample, then in the long run 95 percent of those intervals would succeed in capturing the actual value of the population parameter. ***Confidence Level is NOT interpreted as: “There is a 95% chance a confidence interval captures the true population parameter.

That is NOT confidence level !! Confidence Level  probability that the interval captures the parameter A confidence interval either will capture (100%) or will not capture (0%) true parameter. That is NOT confidence level !!

One-Sample t Confidence Intervals: Critical Value **Read from t-distribution table Standard Error Point Estimate Margin of Error

Two-Sample t Confidence Intervals: Critical Value **Read from calculator output Standard Error Point Estimate Margin of Error

For 1-Sample Means: Use values for t For 1-Sample Means: Use values for t* use (n – 1) called degrees of freedom (df) NOTE: When the actual df does not appear in Table C, use the next lowest value for df. This guarantees a wider confidence interval than we need, a conservative approach.

Confidence Intervals for Means are Based on t-distributions: As degrees of freedom increase, the t-distribution approaches a Normal Distribution.

In class p. 649 # 28 -1.796 1.796 -2.045 2.045 -1.333 1.333

Conditions for Using t Procedures: Randomness --- usually stated in problem Independence --- pop > 10(sample)

Conditions for Using t Procedures: Large Counts --- usually told population is Normal distribution in problem If Not Told, You Must Base Normality on Sample Size n < 15 Graph Data --- data appear Normal 15 < n < 30 Graph Data ---NO outliers or skewness n > 30 Even if population distribution is skewed can refer to CLT

Giving back Ch 7 & 8 Test, THQ#5, and DG 23. Day 55 Agenda: Giving back Ch 7 & 8 Test, THQ#5, and DG 23.

Graphs to check Normality: histogram, boxplot, Normal Probability Plot Ex. The following are weights(in lbs) of randomly selected vehicles. We are not told that the population of vehicle weights is Normally distributed. 2950 4000 3300 3350 3500 3550 2900 3250 Place your data in L1

Straight line approximately Normal Create NPP using last graph in PLOT1 ZOOM9 Sketch Since the points on the Normal Probability Plot lie close to a ____________ _______, the plot indicates that this distribution is ________________. Straight line approximately Normal

What statistics do you need to calculate to describe this distribution? N(___, ___) The distribution would be N(_____, ___).

You are expected to complete all steps of this template when finding a confidence interval. You should KNOW the following interpretation of confidence levels.

We’ll always follow the template. In Class Practice p. 657 #33b We’ll always follow the template. State: We will create a 95% confidence interval for a 1-sample mean to estimate the true population mean change in reasoning scores of all preschool children. Plan: Parameter of Interest: µ = the true population mean change in reasoning scores of all preschool children

Independence ---- all preschool children > 10(34) Plan: Conditions Randomness --- Problem does not state preschool children selected randomly. No reason to assume this was not a random sample. Independence ---- all preschool children > 10(34) Condition met for independence Large Counts --- n = 34 34 > 30 CLT says sample size large enough for Normal Approximation

Do:

Using the Graphing Calculator to Determine Confidence Intervals:

If you work it using graphing calc: If you work it LONG-HAND:

Do: We are 95% confident the true population mean change in reasoning scores for all preschool children lies in the interval 2.5521 points to 4.6839 points.

YOU WILL NEVER USE THESE FUNCTIONS ON YOUR CALCULATOR:

OUTSIDE CLASS ASSIGNMENT: p. 649 – 650 #30 - 32

Letter from Dave Starnes (author of our book) explaining how to write df from calculator.

in calc

Technology Calculates Degrees of Freedom and STORES t Technology Calculates Degrees of Freedom and STORES t* for 2 sample t-distribution Found on p.792 in you textbook. YOU DON’T EVER HAVE TO DO THIS!

Ex. Confidence Interval for 2 Sample Means Do the “Cameron Crazies” at Duke home games help the Blue Devils play better defense? Below are the points allowed by Duke (men) at home and on the road for the conference games from a recent season. Pts allowed at home 44 56 54 75 101 91 81 Pts allowed on road 58 70 74 80 67 65 79

This problem is looking for the difference of two means This problem is looking for the difference of two means. Notice 2 distinct populations. State: We will create a 95% confidence interval for a 2-sample mean to estimate the true population difference in mean points allowed by the Duke men’s basketball team for home and away games. Plan: Parameters of Interest: µH = the true population mean points allowed by the Duke men’s basketball team at home games µA = the true population mean points allowed by the Duke men’s basketball team at away games

*** Note on Conditions: must show for both populations Home Games Away Games Randomness --- The problem does not state the games were selected randomly. No reason assume otherwise The problem does not state the games were selected randomly. No reason assume otherwise. Independence --- all Duke home games > 10(8) Condition met for independence Independence all Duke road games > 10(8) Condition met Large Counts --- samples size not large enough; See NPP Large Counts --- samples size not large enough; See NPP

RECALL: Creating NPP Away Games Home Games

Randomness --- The problem does not state the games were selected randomly. No reason assume otherwise The problem does not state the games were selected randomly. No reason assume otherwise. Independence --- all Duke home games > 10(8) Condition met for independence Independence all Duke road games > 10(8) Condition met Large Counts --- samples size not large enough; See NPP Large Counts --- samples size not large enough; See NPP NPP shows linear trend. Normal approximation appropriate. NPP shows linear trend. Normal approximation appropriate.

Using the Graphing Calculator to Determine Confidence Intervals: NEVER Pool in Conf Int

Do: 9.28 in calc t* (-19.13, 18.378)

What value is CAPTURED in this interval? Do: We are 95% confident the true difference in mean points allowed for Duke home games and Duke away games lies in the interval -19.13 to 18.378 points. NOTE: What value is CAPTURED in this interval? What does this possibly say about the Cameron Crazies influence on the opposing team? Since 0 is captured in this confidence interval, there may be NO difference in the points Duke allows at home games and the points Duke allows at road games. We have evidence to conclude that the Cameron Crazies DO NOT give the team home court advantage.

We will create a 95% confidence interval for a 1-sample mean to Ex. A biology student at a major university is writing a report about bird watchers. She has developed a test that will score the abilities of a bird watcher to identify common birds. She collects data from a random sample of 20 people that classify themselves as bird watchers (data shown below). Find a 90% confidence interval for the mean score of the population of bird watchers. State: We will create a 95% confidence interval for a 1-sample mean to estimate true population mean score birdwatchers make on a test to identify birds.

Independence all bird watchers > 10(20) Ex. A biology student at a major university is writing a report about bird watchers. She has developed a test that will score the abilities of a bird watcher to identify common birds. She collects data from a random sample of people that classify themselves as bird watchers (data shown below). Find a 90% confidence interval for the mean score of the population of bird watchers. Plan: Parameter: µ = the true population mean score on the bird identification test Randomness --- It was stated the researcher collected data from a random sample of people who classify themselves as birdwatchers. Independence all bird watchers > 10(20) Condition met for independence

Large Counts--- The sample size is only 20 birdwatchers, so the CLT cannot be applied The boxplot shows little skewness and no outliers The NPP appears to be linear so we will assume an approximately normal distribution of scores.

Do: Conclusion: We are 90% confident the true population mean score on the bird identification ability test for the population of persons who classify themselves as bird watchers lies in the interval 6.1846 to 7.6854 points.

Now we need to look at one last type of means problem. In class assignment: p. 657 #35

Day 58 Agenda: DG 24 --- 20 minutes Finish p. 657 #35 together mean diff vs diff of means You will need to answer #36 and do the birdwatcher example on your own. Answers are in ppt. To stay on schedule, we must cover Ch 10.3 today

Mean Difference ---- two samples from the same population This problem as asking for the mean difference. Not to be confused with difference in the means. #35 Mean Difference ---- two samples from the same population Difference in the Means --- two samples from two distinct populations

#35 State: We will create 95% confidence interval for 1 sample mean to estimate the true population mean difference in number of disruptive behaviors on moon days and for all other days for dementia patients in nursing homes. Plan: Parameter of Interest: μdiff = the true population mean difference in number of disruptive behaviors on moon days and for all other days for dementia patients in nursing homes Randomness --- The problem does not state the dementia patients were randomly selected. No reason to assume otherwise Independence all dementia patients> 10(15) Condition Met in nursing homes

Plan: Large counts --- Sample size of 20 dementia patients not large enough to apply CLT. Boxplot shows no apparent outliers or severe skewness NPP shows linearity; Normal Distribution appropriate

NOTE on mean diff problems You are looking for a mean difference, so we need to calculate _____________________________ in L3 the difference of L1 and L2

Do: We are 95% confident the true population mean difference in number of disruptive behaviors for dementia patients on moon days and for all other days lies in the interval 1.624 to 3.2414 disruptive behaviors.

#35b No, this was an observational study. There could be other reasons for the dementia patient’s change in behavior.

#36a We have the ENTIRE population of US presidents. Inference Procedure is not needed. Do not use t procedures. #36b Thirty-two students in AP Statistics are not representative of the entire population of all students. Do not use t procedures. #36c Sample size is n = 20 and stem plot is severely skewed left. Based on the distribution of the stemplot , do not use t procedures.

Remember to do p. 649 #30 – 32 Do p. 659 #38 – 40 Assignment for Con Int for Means: Remember to do p. 649 #30 – 32 Do p. 659 #38 – 40