Confidence Intervals

Mrs. Medina

Why would we want to estimate a parameter and/or be confident about your estimate?

• Estimate? – We want to estimate because we cannot test the

entire population. It might be too large or inaccessible.

• Confident?– Because anyone can spout a predication, but is the

real value within a reasonable distance to their prediction.

Population vs. Sample

• Identify the population and the sample:– a) A survey of 1353 American households found

that 18% of the households own a computer.– b) A recent survey of 2625 elementary school

children found that 28% of the children could be classified obese.

– c) The average weight of every sixth person entering the mall within 3 hour period was 146 lb.

Parameter vs. Statistic

• A recent survey by the alumni of a major university indicated that the average salary of 10,000 of its 300,000 graduates was 125,000.

• The average salary of all assembly-line employees at a certain car manufacturer is $33,000.

• The average late fee for 360 credit card holders was found to be $56.75.

identify the population, sample, population parameters, and sample statistics:

• a) In a USA Today Internet poll, readers responded voluntarily to the question “Do you consume at least one caffeinated beverage every day?”

• b) Astronomers typically determine the distance to galaxy (a galaxy is a huge collection of billions of stars) by measuring the distances to just a few stars within it and taking the mean (average) of these distance measurements.

Determine whether the data are qualitative or quantitative

• a) the colors of automobiles on a used car lot• b) the numbers on the shirts of a girl’s soccer

team• c) the number of seats in a movie theater• d) a list of house numbers on your street• e) the ages of a sample of 350 employees of a

large hospital

Population Distribution Probability vs. Sampling Distribution Probability

• Assume that cholesterol levels for women ages 20-34 are approximately normally distributed with µ = 185 and σ = 39.

– What is the probability that a woman age 20-34 has a cholesterol level below 240 mg/dL?

– What is the probability that the average cholesterol level of 20 women between the age of 20-34 is below 240 mg/dL?

Population Distribution Probability vs. Sampling Distribution Probability

• Assume that the weights of 5-year old boys for a normal distribution with µ = 43 lbs and σ = 5 lbs. What is more likely?

– That a randomly selected 5-year old will weigh more than 49 lbs?

– Or, that the average of a 8 5-year old boys will weigh more than 46 lbs?

Objective

• The objective of estimation is…– to estimate the unknown value of a population

parameter, like the mean , or a proportion p, on the basis of a sample statistic (sampling distribution) calculated from sample data by giving a good solid range.

We make point estimates

• Proportions, p– What percent of the students at LHS vote for Jane Doe

for class president?– Ex: 150 out 350 voted for Jane Doe.p = 150/350 = .429 or 42.9%

• Means, m– What is the average GPA of the students at LHS?– If 100 students are sampled (n=100), the mean of the

GPAs is 2.98.

How confident are you about these

statistics?

Confidence Interval

• A confidence interval is a range (or an interval) of values used to estimate the unknown value of a population parameter .

p

.95

ˆ95% of the time p will be in this interval

Use Goldgar’s animation on confidence intervals

When constructing confidence intervals…

• Check conditions– Is it a proportion? What is the sample size, n? Which

equations do you use?– Is it a mean? What is the sample size, n? Which

equations do you use?• Calculate the CI = confidence interval

– Follow through the calculations using the equations• Interpret the results

– Write a statement that shows your understanding of the Confidence interval.

Calculate confidence intervals for proportions

pzpCI ˆ%% ˆ

n

ppp

)1(ˆ

size sample

countˆ p

CL/2 + 0.5

Make sure CL (confidence level) is in decimal form

proportionStandard Error

Margin of error

Medication side effects (confidence interval for p)

Arthritis is a painful, chronic inflammation of the joints.

An experiment on the side effects of pain relievers

examined arthritis patients to find the proportion of

patients who suffer side effects.

What are some side effects of ibuprofen?

Serious side effects (seek medical attention immediately):Allergic reaction (difficulty breathing, swelling, or hives), Muscle cramps, numbness, or tingling, Ulcers (open sores) in the mouth, Rapid weight gain (fluid retention), Seizures, Black, bloody, or tarry stools, Blood in your urine or vomit, Decreased hearing or ringing in the ears, Jaundice (yellowing of the skin or eyes), or Abdominal cramping, indigestion, or heartburn,

Less serious side effects (discuss with your doctor):Dizziness or headache, Nausea, gaseousness, diarrhea, or constipation, Depression, Fatigue or weakness, Dry mouth, or Irregular menstrual periods

Calculate a 90% confidence interval for the population proportion p of arthritis patients who suffer some “adverse symptoms.”

INTERPRETATION OF THE CI: We are 90% confident that the interval

(.034, .070) contains the true proportion of arthritis patients that experience

some adverse symptoms when taking ibuprofen.

440 subjects with chronic arthritis were given ibuprofen for pain relief; 23 subjects suffered from adverse side effects.

052.440

23

size sample

countˆ p 011.

440

)052.1(052.)1(ˆ

n

ppp

65.1 ;95.0@

95.05.02/9.0

%90 zp

018.052.

)011)(.65.1(052.

CI

CI

pzpCI ˆ%% ˆ

Confidence Intervals for sample means

xx zCI %%

nx

CL/2 + 0.5

Make sure CL (confidence level) is in decimal form

Sample meanStandard Error

Margin of error

You are interested in the mean annual salary for registered nurses in the U.S. Assume that s = $1,700. You test random samples of 35 nurses from hospitals and find the mean to be $38,000. – Construct a 90% confidence interval.– Construct a 95% confidence interval.– Construct a 99% confidence interval.

CL =90%: 0.9/2 + 0.5 = .95@p=0.95, Z90%=1.65

CL= 95%: 0.95/2+ 0.5 = .975 @p=.975; z95% = 1.96

CL=99%: 0.99/2 + 0.5 = .995@p=.995, z99% = 2.58

474000,38

)35

1700)(65.1(000,38

%90

%90

CI

CI

563000,38

)35

1700)(96.1(000,38

%95

%95

CI

CI

741000,38

)35

1700)(58.2(000,38

%99

%99

CI

CI

The higher the confidence level, the wider the interval

xx zCI %%

nx

Write 1 interpretation

What sample size of nurses should you use to have a confidence interval of 99% with only $500 from the mean.

• This is a backwards approach.CL=99%: 0.99/2 + 0.5 = .995@p=.995, z = 2.58

19458.2/500

58.2500

x

S

7.76

sides)both (square 76.8194/1700

1700194

1700194

n

n

n

n

The sample size tested has to be of 77 nurses in order to be 99% confident that the mean salary of nurses is 38,000±500

Margin of error

xx zCI %%

nx