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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 _________)
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Example #1A 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
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A: SRS:
Normality:
Independence:
Says randomly selected
It is safe to assume that there are more than 15,000 people who file their taxes
ˆ1 10n p ˆ 10np
p̂ 1200
1500 0.80
1500 0.8 10 1500 1 0.8 10 1200 10 300 10
Yes, safe to assume an approximately normally distribution
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ˆ ˆ1ˆ *
p pp Z
n
N: One Sample Proportion Interval
I:
Z* = ?
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Confidence Level (C) Upper tail prob. Z* Value
92%
Z=? Z=?
0.040.04
1
2
C
1 0.92
2
0.04
0.92
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Confidence Level (C) Upper tail prob. Z* Value
92%
Z=? Z=?
0.040.04
1
2
C
1 0.92
2
0.04 1.75
0.92
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ˆ ˆ1ˆ *
p pp Z
n
N: One Sample Proportion Interval
I:
0.8 1 0.80.80 1.75
1500
0.8 0.0181
0.7819, 0.8181
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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!
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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:
*(1 *)*
p pZ m
n
n
2
*(1 *)
*
p p
mZ
p* = An estimate for p̂Note: 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!
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Example #2You 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.
482.6304
2
*(1 *)
*
p pn
mZ
2
0.19(1 0.19)
.0351.96
.1539
.000318877551
483
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Example #2You 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.
784
2
*(1 *)
*
p pn
mZ
2
0.5(1 0.5)
.0351.96
.25
.000318877551
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Example #2You 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.
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(Z-Interval)
estimate margin of error
Confidence Interval for a Population Mean ( known)
x *Zn
estimate critical value standard error
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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
*x Zn
• If you increase n, then the margin of error decreases
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Calculator Tip: Z-Interval
Stat – Tests – ZInterval
Data: If given actual values
Stats: If given summary of values
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Interpreting a Confidence Interval:
I am C% confident that the true parameter is captured in the interval _____ to ______
What you will say:
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
What it means:
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CAUTION!
Never Say:
The interval will capture the true mean C% of the time.
It either does or does not!
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Conditions for a Z-Interval:
1. SRS
(CLT or population approx normal)
(problem should say)
10N n
(Population 10x sample size)
2. Normality
3. Independence
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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 _________)
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Example #1Serum 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.
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A: SRS:
Normality:
Independence:
N: One sample Z-Interval
Says randomly selected
Approximately normal by the CLT (n 30)
I am assuming that Dr. Oswick has 300 patients or more.
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I: *x Zn
13.450.67 1.96
30
50.67 4.795
45.875, 55.465
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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
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Example #1Serum 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.
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Example #1Serum 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
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Example #2Suppose 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 = 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 1.00ounces. 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
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A: SRS:
Normality:
Independence:
N: One sample Z-Interval
Says randomly selected
Approximately normal because the population is approximately normal
I am assuming that Snickershas 80 bars or more in the 1-oz size
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I: *x Zn
0.005.995 1.645
8
.995 0.0029
.9921, .9979
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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.
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Choosing a Sample Size for a specific margin of error
*Z mn
2*Z
nm
Note: Always round up! You can’t have part of a person! Ex: 163.2 rounds up to 164.
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Example #3A 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.
18,201 21,799
2x
a. What is the sample mean income?
$20,000
*x Zn
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Example #3A 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?
20,000x
m = 21,799 – 20,000
m = 1,799
*x Zn
m
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Example #4A 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.
22.576 19,700
1000n
2*Z
nm
250.7472n
2575.2783n
2576n
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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.
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When the standard deviation of a statistic is estimated from the data
Standard Error:
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)
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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
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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
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Use invT on calculator:Go to 2nd – VARS - #4 invTType 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 = 9Calculate: invT((1+.90)/2, 9) = 1.833t* = 1.833
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Example #2Practice 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 99% CI
9
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Example #2Practice finding t*
n Degrees of Freedom
Confidence Interval
t*
n = 10 99% CI
n = 20 90% CI
n = 40 95% CI
n = 30 99% CI
9 3.250
19
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Example #2Practice finding t*
n Degrees of Freedom
Confidence Interval
t*
n = 10 99% CI
n = 20 90% CI
n = 40 95% CI
n = 30 99% CI
9 3.250
19 1.729
39
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Example #2Practice finding t*
n Degrees of Freedom
Confidence Interval
t*
n = 10 99% CI
n = 20 90% CI
n = 40 95% CI
n = 30 99% CI
9 3.250
19 1.729
39 2.042
29
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Example #2Practice finding t*
n Degrees of Freedom
Confidence Interval
t*
n = 10 99% CI
n = 20 90% CI
n = 40 95% CI
n = 30 99% CI
9 3.250
19 1.729
39 2.042
29 2.756
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Calculator Tip: Finding P(t)
2nd – Dist – tcdf( lower bound, upper bound, degrees of freedom)
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One-Sample t-interval:
1*ns
x tn
Calculator Tip:
One sample t-Interval Go to: Stat – Tests – TInterval
Data: If given actual valuesStats: If given summary of values
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Conditions for a t-interval:
1. SRS
(population approx normal and n<15, or moderate size (15≤ n < 30) with moderate skewness or outliers, or large sample size n ≥ 30)
(problem should say)
10N n
(Population 10x sample size)
2. Normality
3. Independence
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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.
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Example #3As 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.
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A: SRS:
Normality:
Independence:
N: One Sample t-interval
Says randomly selected
15<n<30. We must assume the population doesn’t have strong skewness. Proceeding with caution!
It is safe to assume that there are more than 200 Americans that create waste.
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I: 1*ns
x tn
df = 20 – 1 = 19
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I:
1.24.3 2.861
20
4.3 0.7677
3.5323, 5.0677
1*ns
x tn
df = 20 – 1 = 19
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C: I am 99% confident the true mean waste generated per person per day is between 3.5323 and 5.0677 pounds.