Download Estimating with t-distribution notes for Oct 22

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 _________)
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
A: SRS: Says randomly selected
Normality:
1200
p̂ 

1500
0.80
n 1  pˆ   10
npˆ  10
1500  0.8  10
1500 1  0.8  10
1200  10
300  10
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
N: One Sample Proportion Interval
I:
pˆ  Z *
Z* = ?
pˆ 1  pˆ 
n
Confidence Level (C)
Upper tail prob.
1  C 1  0.92

 0.04
2
2
92%
0.04
0.04
0.92
Z=?
Z=?
Z* Value
Confidence Level (C)
Upper tail prob.
1  C 1  0.92

 0.04
2
2
92%
0.04
0.04
0.92
Z=?
Z=?
Z* Value
 1.75
N: One Sample Proportion Interval
I:
pˆ  Z *
0.80  1.75
pˆ 1  pˆ 
n
0.8 1  0.8 
0.8  0.0181
 0.7819,
0.8181
1500
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!
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
p̂
Note: If p is unknown use the most conservative
value of p = 0.5. Since n is the sample size, it must
be a whole number!!! Round up!
p *(1  p*)
Z*
m
n
n
p *(1  p*)
 m 


 Z *
2
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.
p *(1  p*) 0.19(1  0.19)
.1539
n



2
2
.000318877551
 .035 
 m 




1.96


Z
*


482.6304  483
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.
n
p *(1  p*)
 m 


 Z *
2

0.5(1  0.5)
 .035 


 1.96 
2
.25


.000318877551
784
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.
483
784
Using 0.5 makes the sample size very large, ensuring
that enough people will be surveyed.
Confidence Interval for a Population Mean ( known)
(Z-Interval)
estimate  margin of error
estimate  critical value  standard error
x

  
Z *

 n
Properties of Confidence Intervals for Population Mean
  
x  Z *

 n
• 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
Calculator Tip: Z-Interval
Stat – Tests – ZInterval
Data: If given actual values
Stats: If given summary of values
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
CAUTION!
Never Say:
The interval will capture the true mean C% of the time.
It either does or does not!
Conditions for a Z-Interval:
1. SRS (problem should say)
2. Normality (CLT or population approx normal)
3. Independence (Population 10x sample size)
 N  10n
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 _________)
Example #1
Serum Cholesterol-Dr. Paul Oswick wants to estimate the true
mean serum HDL cholesterol for all of his 20-29 year old female
patients. He randomly selects 30 patients and computes the
sample mean to be 50.67. Assume from past records, the
population standard deviation for the serum HDL cholesterol for
20-29 year old female patients is =13.4.
a. Construct a 95% confidence interval for the mean
serum HDL cholesterol for all of Dr. Oswick’s 20-29
year old female patients.
P: The true mean serum HDL cholesterol for all of
Dr. Oswick’s 20-29 year old female patients.
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
I:
  
x  Z *

 n
 13.4 
50.67  1.96 

 30 
50.67  4.795
 45.875, 55.465
C: I am 95% confident the true mean serum HDL
cholesterol for all of Dr. Oswick’s 20-29 year old
female patients is between 45.875 and 55.465
Example #1
Serum Cholesterol-Dr. Paul Oswick wants to estimate the true
mean serum HDL cholesterol for all of his 20-29 year old female
patients. He randomly selects 30 patients and computes the
sample mean to be 50.67. Assume from past records, the
population standard deviation for the serum HDL cholesterol for
20-29 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 20-29 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.
Example #1
Serum Cholesterol-Dr. Paul Oswick wants to estimate the true
mean serum HDL cholesterol for all of his 20-29 year old female
patients. He randomly selects 30 patients and computes the
sample mean to be 50.67. Assume from past records, the
population standard deviation for the serum HDL cholesterol for
20-29 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
Example #2
Suppose your class is investigating the weights of Snickers 1ounce 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 = 0.005 ounces. Several
candy bars are randomly selected and weighed with sensitive
balances borrowed from the physics lab. The weights are
0.95
1.02
0.98 0.97 1.05 1.01 0.98 1.00
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
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
I:
  
x  Z *

 n
 0.005 
.995  1.645 

 8 
.995  0.0029
.9921, .9979
C: I am 90% confident the true mean weight of
Snickers 1-oz Fun-size candy bars is between
.9921 and .9979 ounces. I am not confident that
the candy bars weigh as advertised at the 90%
level.
Choosing a Sample Size for a specific margin of error
  
Z *
m
 n
 Z * 
n

 m 
2
Note: Always round up! You can’t have part of a
person! Ex: 163.2 rounds up to 164.
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.
a. What is the sample mean income?
  
x  Z *

 n
18, 201  21, 799
x
 $20, 000
2
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?
x  20, 000
  
x  Z *

 n
m
m = 21,799 – 20,000
m = 1,799
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.
 Z * 
n

 m 
2
 2.576 19, 700 
n

1000


n   50.7472
2
n  2575.2783
2
n  2576
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.
Standard Error: When the standard deviation of a
statistic is estimated from the data
s
n
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)
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
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
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
Example #2
Practice finding t*
n
Degrees of
Freedom (n-1)
Confidence
Interval
n = 10
9
99% CI
n = 20
90% CI
n = 40
95% CI
n = 30
99% CI
t*
Example #2
Practice finding t*
n
Degrees of
Freedom
Confidence
Interval
n = 10
9
99% CI
n = 20
19
90% CI
n = 40
95% CI
n = 30
99% CI
t*
3.250
Example #2
Practice finding t*
n
Degrees of
Freedom
Confidence
Interval
n = 10
9
99% CI
3.250
n = 20
19
90% CI
1.729
n = 40
39
95% CI
n = 30
99% CI
t*
Example #2
Practice finding t*
n
Degrees of
Freedom
Confidence
Interval
t*
n = 10
9
99% CI
3.250
n = 20
19
90% CI
1.729
n = 40
39
95% CI
2.042
n = 30
29
99% CI
Example #2
Practice finding t*
n
Degrees of
Freedom
Confidence
Interval
t*
n = 10
9
99% CI
3.250
n = 20
19
90% CI
1.729
n = 40
39
95% CI
2.042
n = 30
29
99% CI
2.756
Calculator Tip: Finding P(t)
2nd – Dist – tcdf( lower bound, upper
bound, degrees of freedom)
One-Sample t-interval:
 s 
x  t *n1 

 n
Calculator Tip:
One sample t-Interval
Go to: Stat – Tests – TInterval
Data: If given actual values
Stats: If given summary of values
Conditions for a t-interval:
1. SRS (problem should say)
2. Normality (population approx normal and n<15, or
moderate size (15≤ n < 30) with moderate
skewness or outliers, or large sample size
n ≥ 30)
3. Independence (Population 10x sample size)
 N  10n
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.
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.
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
I:
 s 
x  t *n1 

 n
df = 20 – 1 = 19
I:
 s 
x  t *n1 

 n
 1.2 
4.3  2.861

 20 
4.3  0.7677
3.5323, 5.0677
df = 20 – 1 = 19
C: I am 99% confident the true mean waste
generated per person per day is between 3.5323
and 5.0677 pounds.