1. Steps to Construct ANY Confidence Interval:
“PANIC” P:
Parameter of Interest (what are you looking for?) A:
Assumptions (what are the conditions?) N:
Name the type of interval (what type of data do we have?) I:
Interval (Finally! You can calculate!) C:
Conclusion in context (I am ___% confident the true parameter lies between ________ and _________)

2. The true percent of Americans who fill out their tax forms correctly
Example #1
A news release by the IRS reported 90% of all Americans fill out their tax forms correctly. A random sample of 1500 returns revealed that 1200 of them were correctly filled out. Calculate a 92% confidence interval for the proportion of Americans who correctly fill out their tax forms. Is the IRS correct in their report? P:
The true percent of Americans who fill out their tax forms correctly

3. Says randomly selected
SRS:
Says randomly selected
Normality:
0.80
Yes, safe to assume an approximately normally distribution
Independence:
It is safe to assume that there are more than 15,000 people who file their taxes

4. N:
One Sample Proportion Interval I:
Z* = ?

5. 92% 0.04 0.04 0.04 0.92 Confidence Level (C) Upper tail prob. Z* Value

6. Confidence Level (C) Upper tail prob. Z* Value
92%
 1.75
0.04
0.04
0.04
0.92
Z=?
Z=?

7. N:
One Sample Proportion Interval I:

8. C:
I am 92% confident the true percent of Americans who fill out their tax forms correctly is between 78.19% and 81.8%
Is the IRS correct in their report?
No,
90% is not in the interval!

9. Sample size for a Desired Margin of Error
If we want the margin of error in a level C confidence interval for p to be m, then we need n subjects in the sample, where:
p* =
An estimate for
Note: If p is unknown use the most conservative value of p = Since n is the sample size, it must be a whole number!!! Round up!
n 

10. Example #2
You wish to estimate with 95% confidence; the proportion of computers that need repairs or have problems by the time the product is three years old. Your estimate must be accurate within 3.5% of the true proportion.
a. Find the sample size needed if a prior study found that 19% of computers needed repairs or had problems by the time the product as three years old.

11. Example #2
You wish to estimate with 95% confidence; the proportion of computers that need repairs or have problems by the time the product is three years old. Your estimate must be accurate within 3.5% of the true proportion.
b. If no preliminary estimate is available, find the most conservative sample size required.

12. Example #2
You wish to estimate with 95% confidence; the proportion of computers that need repairs or have problems by the time the product is three years old. Your estimate must be accurate within 3.5% of the true proportion.
c. Compare the results from a and b.
Using 0.5 makes the sample size very large, ensuring that enough people will be surveyed.

13. Confidence Interval for a Population Mean ( known)
(Z-Interval)
estimate  margin of error
estimate  critical value  standard error 

14. Properties of Confidence Intervals for Population Mean
The interval is always centered around the statistic
The higher the confidence level, the wider the interval becomes
If you increase n, then the margin of error decreases

15. Stat – Tests – ZInterval
Calculator Tip:
Z-Interval
Stat – Tests – ZInterval
Data: If given actual values
Stats: If given summary of values

16. Interpreting a Confidence Interval:
What you will say:
I am C% confident that the true parameter is captured in the interval _____ to ______
What it means:
If we took many, many, SRS from a population and calculated a confidence interval for each sample, C% of the confidence intervals will contain the true mean

17. CAUTION!
Never Say:
The interval will capture the true mean C% of the time.
It either does or does not!

18. Conditions for a Z-Interval:
SRS
(problem should say)
(CLT or population approx normal)
2. Normality
(Population 10x sample size)
3. Independence

19. Steps to Construct ANY Confidence Interval:
PANIC P:
Parameter of Interest (what are you looking for?) A:
Assumptions (what are the conditions?) N:
Name the type of interval (what type of data do we have?) I:
Interval (Finally! You can calculate!) C:
Conclusion in context (I am ___% confident the true parameter lies between ________ and _________)

20. Example #1
Serum Cholesterol-Dr. Paul Oswick wants to estimate the true mean serum HDL cholesterol for all of his year old female patients. He randomly selects 30 patients and computes the sample mean to be Assume from past records, the population standard deviation for the serum HDL cholesterol for year old female patients is =13.4.
Construct a 95% confidence interval for the mean serum HDL cholesterol for all of Dr. Oswick’s year old female patients. P:
The true mean serum HDL cholesterol for all of Dr. Oswick’s year old female patients.

21. A:
SRS:
Says randomly selected
Normality:
Approximately normal by the
CLT (n  30)
Independence:
I am assuming that Dr. Oswick has 300 patients or more. N:
One sample Z-Interval

22. I:

23. C:
I am 95% confident the true mean serum HDL cholesterol for all of Dr. Oswick’s year old female patients is between and

24. 53 is contained in the interval.
Example #1
Serum Cholesterol-Dr. Paul Oswick wants to estimate the true mean serum HDL cholesterol for all of his year old female patients. He randomly selects 30 patients and computes the sample mean to be Assume from past records, the population standard deviation for the serum HDL cholesterol for year old female patients is =13.4.
b. If the US National Center for Health Statistics reports the mean serum HDL cholesterol for females between years old to be  = 53, do Dr. Oswick’s patients appear to have a different serum level compared to the general population? Explain.
No,
53 is contained in the interval.

25. Lower confidence level
Example #1
Serum Cholesterol-Dr. Paul Oswick wants to estimate the true mean serum HDL cholesterol for all of his year old female patients. He randomly selects 30 patients and computes the sample mean to be Assume from past records, the population standard deviation for the serum HDL cholesterol for year old female patients is =13.4.
c. What two things could you do to decrease your margin of error?
Increase n
Lower confidence level

26. The true mean weight of Snickers 1-oz Fun-size candy bars
Example #2
Suppose your class is investigating the weights of Snickers 1-ounce Fun-Size candy bars to see if customers are getting full value for their money. Assume that the weights are Normally distributed with standard deviation = ounces. Several candy bars are randomly selected and weighed with sensitive balances borrowed from the physics lab. The weights are
ounces. Determine a 90% confidence interval for the true mean, µ. Can you say that the bars weigh 1oz on average? P:
The true mean weight of Snickers 1-oz Fun-size candy bars

27. A:
SRS:
Says randomly selected
Normality:
Approximately normal because the population is approximately normal
Independence:
I am assuming that Snickers
has 80 bars or more in the 1-oz size N:
One sample Z-Interval

28. I:

29. C:
I am 90% confident the true mean weight of Snickers 1-oz Fun-size candy bars is between and ounces. I am not confident that the candy bars weigh as advertised at the 90% level.

30. Choosing a Sample Size for a specific margin of error
Note: Always round up! You can’t have part of a person! Ex: rounds up to 164.

31. What is the sample mean income?
Example #3
A statistician calculates a 95% confidence interval for the mean income of the depositors at Bank of America, located in a poverty stricken area. The confidence interval is $18,201 to $21,799.
What is the sample mean income?

32. b. What is the margin of error?
Example #3
A statistician calculates a 95% confidence interval for the mean income of the depositors at Bank of America, located in a poverty stricken area. The confidence interval is $18,201 to $21,799.
b. What is the margin of error? m
m = 21,799 – 20,000
m = 1,799

33. Example #4
A researcher wishes to estimate the mean number of miles on four-year-old Saturn SCI’s. How many cars should be in a sample in order to estimate the mean number of miles within a margin of error of  1000 miles with 99% confidence assuming =19,700.

34. 8.3 – Estimating a Population Mean
In the previous examples,
we made an unrealistic assumption that the population standard deviation was known and could be used to calculate confidence intervals.

35. Standard Error:
When the standard deviation of a statistic is estimated from the data
When we know  we can use the Z-table to make a confidence interval. But, when we don’t know it, then we have to use something else!
(Calculator Bingo activity p. 502)

36. Properties of the t-distribution:
σ is unknown
Degrees of Freedom = n – 1
More variable than the normal distribution (it has fatter tails than the normal curve)
Approaches the normal distribution when the degrees of freedom are large (sample size is large).
Area is found to the right of the t-value

37. Properties of the t-distribution:
If n < 15, if population is approx normal, then so is the sample distribution. If the data are clearly non-Normal or if outliers are present, don’t use!
If n > 15, sample distribution is normal, except if population has outliers or strong skewness
If n  30, sample distribution is normal, even if population has outliers or strong skewness

41. Use invT on calculator:
Go to 2nd – VARS - #4 invT
Type in: invT((1+C)/2, n-1)
Example #1: Suppose you want to construct a 90% confidence interval for the mean of a Normal population based on SRS of size 10. What critical value t* should you use?
Degrees of freedom = n – 1 = 10-1 = 9
Calculate: invT((1+.90)/2, 9) = 1.833
t* = 1.833

42. Degrees of Freedom (n-1)
Example #2
Practice finding t* n
Degrees of Freedom (n-1)
Confidence Interval t*
n = 10
99% CI
n = 20
90% CI
n = 40
95% CI
n = 30 9

43. Example #2 Practice finding t* 9 3.250 19 n Degrees of Freedom
Confidence Interval t*
n = 10
99% CI
n = 20
90% CI
n = 40
95% CI
n = 30 9
3.250 19

44. Example #2 Practice finding t* 9 3.250 19 1.729 39 n
Degrees of Freedom
Confidence Interval t*
n = 10
99% CI
n = 20
90% CI
n = 40
95% CI
n = 30 9
3.250 19
1.729 39

45. Example #2 Practice finding t* 9 3.250 19 1.729 39 2.042 29 n
Degrees of Freedom
Confidence Interval t*
n = 10
99% CI
n = 20
90% CI
n = 40
95% CI
n = 30 9
3.250 19
1.729 39
2.042 29

46. Example #2 Practice finding t* 9 3.250 19 1.729 39 2.042 29 2.756 n
Degrees of Freedom
Confidence Interval t*
n = 10
99% CI
n = 20
90% CI
n = 40
95% CI
n = 30 9
3.250 19
1.729 39
2.042 29
2.756

47. Calculator Tip:
Finding P(t)
2nd – Dist – tcdf( lower bound, upper bound, degrees of freedom)

48. Go to: Stat – Tests – TInterval
One-Sample t-interval:
Calculator Tip:
One sample t-Interval
Go to: Stat – Tests – TInterval
Data: If given actual values
Stats: If given summary of values

49. Conditions for a t-interval:
SRS
(problem should say)
(population approx normal and n<15, or moderate size (15≤ n < 30) with moderate skewness or outliers, or large sample size n ≥ 30)
2. Normality
3. Independence
(Population 10x sample size)

50. Robustness:
The probability calculations remain fairly accurate when a condition for use of the procedure is violated
The t-distribution is robust for large n values, mostly because as n increases, the t-distribution approaches the Z-distribution. And by the CLT, it is approx normal.

51. The true mean waste generated per person per day.
Example #3
As part of your work in an environmental awareness group, you want to estimate the mean waste generated by American adults. In a random sample of 20 American adults, you find that the mean waste generated per person per day is 4.3 pounds with a standard deviation of 1.2 pounds. Calculate a 99% confidence interval for  and explain it’s meaning to someone who doesn’t know statistics. P:
The true mean waste generated per person per day.

52. A:
SRS:
Says randomly selected
Normality:
15<n<30. We must assume the population doesn’t have strong skewness. Proceeding with caution!
Independence:
It is safe to assume that there are more than 200 Americans that create waste. N:
One Sample t-interval

53. I:
df =
20 – 1 = 19

55. I:
df = 20 – 1 = 19

56. C:
I am 99% confident the true mean waste generated per person per day is between and pounds.