steps to construct any confidence interval: “panic” p:parameter of interest (what are you...

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 #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

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

ˆ ˆ1ˆ *

p pp Z

n

N: One Sample Proportion Interval

I:

Z* = ?

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

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

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

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:

*(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!

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


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


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.

(Z-Interval)

estimate margin of error

Confidence Interval for a Population Mean ( known)

x *Zn

estimate critical value standard error

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

Calculator Tip: Z-Interval

Stat – Tests – ZInterval

Data: If given actual values

Stats: If given summary of values

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:

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

(CLT or population approx normal)

(problem should say)

10N n

(Population 10x sample size)

2. Normality

3. Independence

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 #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.

A: SRS:

Normality:

Independence:

N: One sample Z-Interval


Approximately normal by the CLT (n 30)

I am assuming that Dr. Oswick has 300 patients or more.

I: *x Zn

13.450.67 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


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.


c. What two things could you do to decrease your margin of error?

Increase n

Lower confidence level

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

A: SRS:

Normality:

Independence:

N: One sample Z-Interval


Approximately normal because the population is approximately normal

I am assuming that Snickershas 80 bars or more in the 1-oz size

I: *x Zn

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 mn

2*Z

nm

Note: Always round up! You can’t have part of a person! Ex: 163.2 rounds up to 164.

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

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

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

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.

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)

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 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

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


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



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



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



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

Calculator Tip: Finding P(t)

2nd – Dist – tcdf( lower bound, upper bound, degrees of freedom)

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

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

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 #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.

A: SRS:

Normality:

Independence:

N: One Sample t-interval


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.

I: 1*ns

x tn

df = 20 – 1 = 19

I:

1.24.3 2.861

20

4.3 0.7677

3.5323, 5.0677

1*ns

x tn

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.

steps to construct any confidence interval: “panic” p:parameter of interest (what are you...

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