Confidence Intervals – Review and Basics

The admissions director at Big City University proposes using the IQ scores of current students as a marketing tool. He administers IQ tests to a SRS of 50 of the 5000 freshmen. The mean IQ score for the sample is x bar = 112. What can the director say about the mean score (mu) of all 5000 freshmen? How would the sample mean x bar vary if we took many samples of 50 freshmen from the same population?

-Because of law of large numbers, mean for mu should be close to 112.

-Mean is the same as mu, need to find standard deviation. I will tell you we know that the IQ scores have a standard deviation of 15. So we can find our standard deviation of our samples by using sigma/square root n = 2.1

-This is normal because 50 > 30.

-Since it is Normal, we can use what rule?

-The true value of mu should be between 95% of the data

-We can say that we are 95% confident that the mean IQ for all Big City University freshmen is between 107.8 and 116.2

Confidence Interval and Confidence Level

-A confidence interval calculated from the data usually of the form estimate ± margin of error

-A confidence level C, which gives the probability that the interval will capture the true parameter value in repeated samples.

Why do we need a margin of error? The ± shows how accurate we believe our guess is based on variability. 95% confident means it catches all 95% of samples. The ± 4.2 means give or take 4.2 (that is our margin of error).

Example:

In late 2009, the Pew Internet and American Life Project asked a random sample of 2253 US adults, “Do you ever…use Twitter or another service to share updates about yourself or to see updates about others?” Of the sample, 19% said “Yes”. According to Pew, the resulting 95% confidence interval is (.167, .213).

Confidence interval: We are 95% confident that the proportion of people that use Twitter or other services to update about themselves is from .167 to .213

Confidence level: In 95% of all possible samples of 2253 US Adults, the resulting confidence interval would capture the actual population proportion of US adults who use Twitter or another service.

According to the Gallup poll on 8/13/10, the 95% confidence interval of the true proportion of Americans who approved of the job Obama was doing as president was .44 ± .03.

Interpret the confidence interval and the confidence level.

Interval – We are 95% confident that the proportion of Americans who approve of the job of the president is between the interval of .41 to .47.

Level – In 95% of all possible samples of the same size of Americans who approve of the job the president is doing the interval would capture the actual population proportion.

Example: How much does the fat content of Brand X hot dogs vary? To find out, researchers measured the fat content in grams of a random sample of 10 Brand X hot dogs. A 95% confidence interval for the population standard deviation σ is 2.84 to 7.55.

TRUE OR FALSE: The interval from 2.84 to 7.55 has a 95% chance of containing the actual population σ.

FALSE – Probability is either 1 or 0 …. Its there or its not. *******

Conditions for constructing a confidence interval for µ

-Data must come from a SRS from the population

-Sampling distribution of x bar is approximately Normal

  • CLT …. n ≥ 30

-Individual observations are independent

  • N ≥ 10n

*** You MUST show that the conditions exist BEFORE you construct a valid confidence interval in order to get full credit*** EVEN IF IT TELLS YOU TO CONSTRUCT AN INTERVAL, THIS MUST BE DONE!

Z* =

90% confidence z* = 1.64595% confidence z* = 1.9699% confidence z*=2.576

Confidence Interval for a population mean (if you know σ)

n = 70, σ = 15 they gave me 70 values….x bar = 306.3, find and interpret a 90% confidence interval for mean tension of all the screens produced on this day.

Let’s try this – Page 624 #1, 2, 5

Page 632 #9, 10, 12 ---

you do not have to draw scatterplot

READ the EXCLAMATION POINTS on page 636

y(hat) = 27.64 + .527x

Margin of error = z*

Margin of error = m µ

We can use the equation above to figure out how large of a sample we need for a specific

margin of error.

EXAMPLE: Researchers would like to estimate the mean cholesterol level µ of a particular variety of monkey that is often used in lab experiments. They would like their estimate to be within 1 milligram per deciliter of the true value at a 95% confidence level. A previous study involving this variety of monkey suggests that the standard deviation is σ = 5 mg/dl. How many monkeys do I need?

So z* ≤ mz* ≤ 11.96 n ≥ 97. We would want 97 monkeys.

Make sure you read all the cautions on 636.

ESTIMATING A POPULATION MEAN

How can we construct a confidence interval for an unknown population mean when we don’t know the population standard deviation?????

Must still verify the three conditions:

-SRS

-NORMALITY

-INDEPENDENCE

Mr. V’s class wants to construct a confidence interval for the proportion p of red beads in a container. A simple random sample they selected had 107 red beads and 144 white beads. There were 3,000 beads.

Are the conditions for a confidence interval met?

Is it a SRS?

How do you check normal for a proportion?

How do you check Independent?

Need to find a standard deviation (use formula from last chapter)…. This is your standard error in this problem. It is used as your Standard Deviation when you aren’t given one.

Can you find Z* on your own? Look for 90% in the back of your book.

95%?80%96%

Alcohol abuse has been described by college presidents as the number one problem on campus and it is an important cause of death in young adults. How common is it? A survey of 10,904 randomly selected US College students collected info on drinking behavior and alcohol related problems. The researchers defined frequent binge drinking as having 5 or more drinks in a row three or more times in the past 2 weeks. According to this definition 2486 students were classified as frequent binge drinkers.

-Identify population and parameters

-Check conditions for constructing confidence interval

-Find 99% confidence interval

-Interpret the interval

-Do it again for an 86% confidence interval

ASSIGNMENT

Page 669 #45 – 49

Page 672 #51 – 55