AP Statistics Chap 101 Confidence Intervals. AP Statistics Chap 102 Confidence Intervals...

Upload
herbertsharp 
Category
Documents

view
222 
download
0
Embed Size (px)
Transcript of AP Statistics Chap 101 Confidence Intervals. AP Statistics Chap 102 Confidence Intervals...
AP Statistics Chap 101
Confidence Intervals
AP Statistics Chap 102
Confidence Intervals
Population Mean
σ Unknown(Lock 6.5)
ConfidenceIntervals
PopulationProportion(Lock 6.1)
σ Known(Lock 5.2)
Central Limit Theorem
AP Statistics Chap 103
Normal if:Asymmetric: 30+Symmetric: 15+Normal: >1
AP Statistics Chap 104
Confidence Interval for μ(σ Known)
Assumptions Population standard deviation σ is known Population is normally distributed If population is not normal, use large sample
Confidence interval estimate
n
σzx α/2
AP Statistics Chap 105
Common Levels of Confidence
Commonly used confidence levels are 90%, 95%, and 99%
Confidence Level
Confidence Coefficient,
z value,
1.28
1.645
1.96
2.33
2.58
3.08
3.27
.80
.90
.95
.98
.99
.998
.999
80%
90%
95%
98%
99%
99.8%
99.9%
1 /2z
AP Statistics Chap 106
Finding the Critical Value
Consider a 95% confidence interval:
z.025= 1.96 z.025= 1.96
.951
.0252
α .025
2
α
Point EstimateLower Confidence Limit
UpperConfidence Limit
z units:
x units: Point Estimate
0
1.96zα/2
Confidence Intervals
90% Samples
95% Samples
99% Samples
+1.65x +2.58x
x_
X
+1.96x
2.58x 1.65x
1.96x
X= ± Zx
AP Statistics Chap 108
μμx
Interval and Level of Confidence
Confidence Intervals
Intervals extend from
to
100(1)%of intervals constructed contain μ;
100% do not.
Sampling Distribution of the Mean
n
σzx /2
n
σzx /2
x
x1
x2
/2 /21
AP Statistics Chap 109
Margin of Error(MoE)
Margin of Error (e): the amount added and subtracted to the point estimate to form the confidence interval
n
σzx /2
n
σze /2
Example: Margin of error for estimating μ, σ known:
AP Statistics Chap 1010
Factors Affecting MoE
Data variation, σ : e as σ
Sample size, n : e as n
Level of confidence, 1  : e if 1 
n
σze /2
AP Statistics Chap 1011
Example 1
A sample of 11 circuits from a large normal population has a mean resistance of 2.20 ohms. We know from past testing that the population standard deviation is .35 ohms.
Determine a 95% confidence interval for the true mean resistance of the population.
AP Statistics Chap 1012
Example 1
A sample of 11 circuits from a large normal population has a mean resistance of 2.20 ohms. We know from past testing that the population standard deviation is .35 ohms.
(continued)
n
σ /2zx Solution:
2.20±1.96(.35/sqrt(11)
2.20±.2068
[1.99, 2.4068]
ZInterval w/ Calculator
AP Statistics Chap 1013
AP Statistics Chap 1014
Interpretation
We are 95% confident that the true mean resistance is between 1.9932 and 2.4068 ohms
Although the true mean may or may not be in this interval, 95% of intervals formed in this manner will contain the true mean
An incorrect interpretation is that there is 95% probability that this
interval contains the true population mean.
(This interval either does or does not contain the true mean, there is no probability for a single interval)
Thinking Challenge
You’re a Q/C inspector for Gallo. The for 2liter bottles is .05 liters. A random sample of 100 bottles showsX = 1.99 liters. What is the 90% confidence interval estimate of the true mean amount in 2liter bottles?
2 liter
© 19841994 T/Maker Co.
2 liter
Confidence Interval Solution
AP Statistics Chap 1017
Computer Simulation
AP Statistics Chap 1018
Example 2
Tim Kelley weighs himself once a week for several years. Last month he weighed himself 4 times with an average of 190.5. Examination of Tim’s past data reveals that over relatively short periods of time, his weight measurements are approximately normal with a standard deviation of about 3. Find a 90% confidence interval for his mean weight for last month. Then, find a 99% confidence interval.
97.192,03.1884
3645.15.190
CI
37.194,63.1864
358.25.190
CI
AP Statistics Chap 1019
Example 2
Suppose Tim had only weighed himself once last month and that his one observation was x=190.5 (the same as the mean before). Estimate µ with 90% confidence.
(continued)
24.198,76.1821
3645.15.190
CI
AP Statistics Chap 1020
Example 2
Tim wants to have a margin of error of only 2 pounds with 95% confidence. How many times must he weigh himself to achieve this goal?
(continued)
93
96.12 nn
AP Statistics Chap 1021
Confidence Intervals
with
Unknown Population
Standard Deviation
AP Statistics Chap 1022
Confidence Intervals
Population Mean
σ Unknown
ConfidenceIntervals
PopulationProportion
σ Known
AP Statistics Chap 1023
If the population standard deviation σ is unknown, we can substitute the sample standard deviation, s
This introduces extra uncertainty, since s is variable from sample to sample
So we use the t distribution instead of the normal distribution
Confidence Interval for μ(σ Unknown)
AP Statistics Chap 1024
Assumptions Population standard deviation is unknown Population is normally distributed If population is not normal, use large sample
Use Student’s t Distribution Confidence Interval Estimate
Confidence Interval for μ(σ Unknown)
n
stx /2
(continued)
AP Statistics Chap 1027
Student’s t Distribution
t0
t (df = 5)
t (df = 13)tdistributions are bellshaped and symmetric, but have ‘fatter’ tails than the normal
Standard Normal
(t with df = )
Note: t z as n increases
AP Statistics Chap 1028
Student’s t Table
Upper Tail Area
df .25 .10 .05
1 1.000 3.078 6.314
2 0.817 1.886 2.920
3 0.765 1.638 2.353
t0 2.920The body of the table contains t values, not probabilities
Let: n = 3 df = n  1 = 2 = .10 /2 =.05
/2 = .05
AP Statistics Chap 1029
t distribution values
With comparison to the z value
Confidence t t t z Level (10 d.f.) (20 d.f.) (30 d.f.) ____
.80 1.372 1.325 1.310 1.28
.90 1.812 1.725 1.697 1.64
.95 2.228 2.086 2.042 1.96
.99 3.169 2.845 2.750 2.58
Note: t z as n increases
AP Statistics Chap 1030
Example
A random sample of n = 25 has x = 50 and s = 8. Form a 95% confidence interval for μ
2.0639t so, 24, 1– n d.f. .025,241n,/2 t
25
8(2.0639)50
n
stx /2
CI = 46.698, 53.302
The confidence interval is:
Thinking Challenge
You’re a time study analyst in manufacturing. You’ve recorded the following task times (min.): 3.6, 4.2, 4.0, 3.5, 3.8, 3.1.What is the 90% confidence interval estimate of the population mean task time?
Confidence Interval Solution*
X = 3.7 S = 3.8987 n = 6, df = n  1 = 6  1 = 5 S / n = 3.99 / 6 = 1.592
t.05,5 = 2.0150
3.7  (2.015)(1.592) 3.7 + (2.015)(1.592)
0.492 6.908
AP Statistics Chap 1033
Approximation for Large Samples
Since t approaches z as the sample size increases, an approximation is sometimes used when n 30:
n
stx /2
n
szx /2
Correct formula
Approximation for large n
AP Statistics Chap 1036
If σ is unknown
If unknown, σ can be estimated when using the required sample size formula
Use a value for σ that is expected to be at least as large as the true σ
Select a pilot sample and estimate σ with the sample standard deviation, s
AP Statistics Chap 1037
Chapter 10
Confidence Intervals
for
Population Proportions
AP Statistics Chap 1038
Confidence Intervals
Population Mean
σ Unknown
ConfidenceIntervals
PopulationProportion
σ Known
AP Statistics Chap 1039
Confidence Intervals for the Population Proportion, p
An interval estimate for the population proportion ( p ) can be calculated by adding an allowance for uncertainty to the sample proportion ( ) p̂
AP Statistics Chap 1040
Confidence Intervals for the
Population Proportion, p
Recall that the distribution of the sample proportion is approximately normal if the sample size is large, with standard deviation
We will estimate this with sample data:
(continued)
n
)p(1ps
p
n
p)p(1σp
AP Statistics Chap 1041
Confidence interval endpoints
Upper and lower confidence limits for the population proportion are calculated with the formula
where z is the standard normal value for the level of confidence desired p is the sample proportion n is the sample size
n
)p(pzp /2
1
AP Statistics Chap 1042
Example
A random sample of 100 people
shows that 25 are lefthanded.
Form a 95% confidence interval for
the true proportion of lefthanders
AP Statistics Chap 1043
Example A random sample of 100 people shows
that 25 are lefthanded. Form a 95% confidence interval for the true proportion of lefthanders.
1.
2.
3.
.0433 .25(.75)/n)/np(1pS
.2525/100 p
p
0.3349 . . . . . 0.1651
(.0433) 1.96 .25
(continued)
AP Statistics Chap 1044
Interpretation
We are 95% confident that the true percentage of lefthanders in the population is between
16.51% and 33.49%.
Although this range may or may not contain the true proportion, 95% of intervals formed from samples of size 100 in this manner will contain the true proportion.
AP Statistics Chap 1045
Changing the sample size
Increases in the sample size reduce the width of the confidence interval.
Example: If the sample size in the above example is
doubled to 200, and if 50 are lefthanded in the sample, then the interval is still centered at .25, but the width shrinks to
.19 …… .31
Estimation Example Proportion
A random sample of 400 graduates showed 32 went to grad school. Set up a 95% confidence interval estimate for p.
Estimation Example Proportion
A random sample of 400 graduates showed 32 went to grad school. Set up a 95% confidence interval estimate for p.
ˆ ˆ ˆ ˆˆ ˆ
α/2 α/2
p×( 1p) p×( 1p)pZ × ≤p≤p+Z ×
n n
.08×( 1.08) .08×( 1.08).081.96× ≤p≤.08+1.96×
400 400
.053 p .107
Thinking Challenge
You’re a production manager for a newspaper. You want to find the % defective. Of 200 newspapers, 35 had defects. What is the 90% confidence interval estimate of the population proportion defective?
Confidence Interval Solution*
ˆ ˆ ˆ ˆˆ ˆ
a/2 a/2
p×( 1p) p×( 1p)pz × ≤p≤p+z ×
n n
.175×( .825) .175×( .825).1751.645× ≤p≤.175+1.645×
200 200.1308≤p≤.2192
AP Statistics Chap 1050
Finding the Required Sample Size
for proportion problems
n
)p(pze /2
1
Solve for n:
Define the margin of error:
2/2
e
)p(pzn
12
p can be estimated with a pilot sample, if necessary (or conservatively use p = .50)
AP Statistics Chap 1051
What sample size...?
How large a sample would be necessary to estimate the true proportion defective in a large population within 3%, with 95% confidence?
(Assume a pilot sample yields p = .12)
AP Statistics Chap 1052
What sample size...?
Solution:For 95% confidence, use Z = 1.96E = .03p = .12, so use this to estimate p
So use n = 451
450.74(.03)
.12)(.12)(1(1.96)
e
)p(pzn
2
2/2
2
2 1
(continued)
AP Statistics Chap 1054
Confidence Intervals
End
of
Basic Confidence Intervals