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

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

5. Part I: One Sample PROCEDURE
Recall:
Recall Estimates :

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

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

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

9. 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.

10. 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.

11. 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 !!

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

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

15. 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.

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

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

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

20. 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 < Graph Data --- data appear Normal
15 < n < 30 Graph Data ---NO outliers or skewness
n > Even if population distribution is
skewed can refer to CLT

21. 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.

22. 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

23. Straight line approximately Normal
Create NPP using last graph in PLOT 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

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

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

28. 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

29. 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 = > CLT says sample
size large enough
for Normal Approximation

30. Do:

31. Using the Graphing Calculator to Determine Confidence Intervals:

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

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

35. YOU WILL NEVER USE THESE FUNCTIONS
ON YOUR CALCULATOR:

36. OUTSIDE CLASS ASSIGNMENT:
p. 649 – #

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

38. in calc

39. 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!

40. 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

41. 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

42. *** 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

43. RECALL: Creating NPP
Away Games
Home Games

44. 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.

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

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

47. 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 to 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.

48. 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.

49. 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

50. 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.

51. 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 to points.

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

53. 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

54. 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

55. #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

56. 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

57. 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

58. 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 to disruptive behaviors.

59. #35b
No, this was an observational study. There could be other reasons for the dementia patient’s change in behavior.

60. #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.

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