1. Chapter 8: Estimating With Confidence

2. 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.

3. 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!

4. 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

5. 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

9. Use invT on calculator:
Go to 2nd – VARS - #4 invT
** If you use a TI-83 it will be a program you have to run!
Type in: invT((1+C)/2, n-1)
Example #1: Suppose you want to construct a 90% confidence interval for the mean of a population on SRS size 10. What critical t* value should you use?

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

11. 9 Example #2 Practice finding t* n Degrees of Freedom
Confidence Interval t*
n = 10
99% CI
n = 20
90% CI
n = 40
95% CI
n = 30 9

13. 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

15. 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

17. 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

19. 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

20. Conditions for a t-interval:
SRS
(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)

21. 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.

22. 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.
Parameter:
The true mean waste generated per person per day.

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

24. I:
df =
20 – 1 = 19

26. I:
df = 20 – 1 = 19

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

28. Matched Pairs t-procedures:
Subjects are matched according to characteristics that affect the response, and then one member is randomly assigned to treatment 1 and the other to treatment 2. Recall that twin studies provide a natural pairing. Before and after studies are examples of matched pairs designs, but they require careful interpretation because random assignment is not used.
Apply the one-sample t procedures to the differences

29. Confidence Intervals for Matched Pairs

30. Example #4
Archaeologists use the chemical composition of clay found in pottery artifacts to determine whether different sites were populated by the same ancient people. They collected five random samples from each of two sites in Great Britain and measured the percentage of aluminum oxide in each. Based on these data, do you think the same people used these two kiln sites? Use a 95% confidence interval for the difference in aluminum oxide content of pottery made at the sites and assume the population distribution is approximately normal. Can you say there is no difference between the sites?
New Forrest
20.8 18
15.8
18.3
Ashley Trails
19.1
14.8
16.7
17.7
Difference
1.7
3.2
1.3
-2.5 .6

31. P:
μn = New Forrest percentage of aluminum oxide
μa = Ashley Trails percentage of aluminum oxide
μd = μn - μa = Difference in aluminum oxide levels
The true mean difference in aluminum oxide levels between the New Forrest and Ashley Trails.

32. A:
SRS:
Says randomly selected
Normality:
Says population is approx normal
Independence:
It is safe to assume that there are more than 50 samples available N:
Matched Pairs t-interval

33. I:
df =
5 – 1 = 4

35. I:
df = 20 – 1 = 19

36. C:
I am 95% confident the true mean difference in aluminum oxide levels between the New Forrest and Ashley Trails is between –1.754 and
Can you say there is no difference between the sites?
zero is in the confidence interval, so it is safe to say there is no difference.
Yes,

37. Example #5
The National Endowment for the Humanities sponsors summer institutes to improve the skills of high school language teachers. One institute hosted 20 Spanish teachers for four weeks. At the beginning of the period, the teachers took the Modern Language Association’s listening test of understanding of spoken Spanish. After four weeks of immersion in Spanish in and out of class, they took the listening test again. (The actual spoken Spanish in the two tests was different, so that simply taking the first test should not improve the score on the second test.) Below is the pretest and posttest scores. Give a 90% confidence interval for the mean increase in listening score due to attending the summer institute. Can you say the program was successful?

38. Subject
Pretest
Posttest 1 30 29 11 32 2 28 12 3 31 13 34 4 26 14 5 20 16 15 6 25 27 7 17 8 18 9 33 19 10

39. P:
μB = Pretest score
μA = Posttest score
μd = μB - μA = Difference in test scores
The true mean difference in test scores between the Pretest and Posttest

40. A:
SRS:
We must assume the 20 teachers are randomly selected
Normality:

42. A:
SRS:
We must assume the 20 teachers are randomly selected
Normality:
15<n<30 and distribution is approximately normal, so safe to assume
Independence:
It is safe to assume that there are more than 200 Spanish teachers N:
Matched Pairs t-interval

43. I: 19
df =
20 – 1 =

45. I:
df = 20 – 1 = 19

46. C:
I am 90% confident the true mean difference in test scores between the Pretest and Posttest
is between –2.689 and –
Can you say the program was successful?
zero is not in the confidence interval, so the pretest score is lower than the posttest score.
Yes,