FPP 23

Confidence Interval of a Mean

Confidence intervals for proportion reviewGeneric formula for a confidence interval

estimate ± multiplier*SERecall the multiplier depends on the level of

confidence

For a population proportion we have

The multiplier here is found using the normal distribution €

ˆ p ± multiplier *ˆ p (1− ˆ p )

n

Confidence interval for a meanGeneric formula

estimate ± multiplier*SEAn estimate for a population mean μis the sample mean

(typically denoted by ) SE is given by σ/√nMultiplier found using the normal distribution

But we don’t know σ. So what do we do?Use the sample standard deviation

Thus

But since we use s instead of σ we must use a t-distribution with n – 1 degrees of freedom (d.f.) instead of a normal distribution to find the multiplier

€

s = s2 = (x i − x )2 /(n −1)i=1

n

∑

€

SE = s / n

€

x

t-table

CI of a mean recapEquation for a confidence interval of a mean

sample mean ± multiplier*SE

The multiplier comes from the t-distribution with n – 1 d.f., s is the sample standard deviation, n is the sample size

All the ideas of confidence intervals for a proportion carry over to means.InterpretationsThe meaning of statistical confidence.

€

x ± multiplier * s/ n

Application of CI’s: Mercury levels in NC rivers

Rivers in North Carolina contain small concentrations of mercury which can accumulate in fish over their lifetimes. Because mercury cannot be excreted from the body it builds up in the tissues. The concentration of mercury in fish tissues can be obtained at considerable expense by catching fish and sending samples to a lab for analysis. Directly measuring the mercury concentration in the water is impossible since it is almost always below detectable limits

A study was recently conducted by researchers at the Nicholas School of the Environment at Duke in the Wacamaw and Lumber Rivers to investigate mercury levels in tissues of large mouth bass. At several stations along each river, a group of fish were caught, weighted and measured. In addition a filet from each fish caught was sent to the lab so that the tissue concentration of mercury (in parts per million) could be determined for each fish.

Mercury in concentrations greater than 1 part per million are considered unsafe for humans to ingest. Are fish in the Lumber and Wacamaw Rivers too contaminated to eat?

EDA for mercuryThe distribution of

mercury is right-skewed in both rivers. There are a few outliers in Lumber River, but the large sample size should allow us to use the Central Limit Theorem for CI’s.

The sample average mercury level for both rivers is above 1.0 ppm.

95% CI’s for population average mercury levels in two rivers:

0 .5 1 1.5 2 2.5 3 3.5 4

Mean

Std Dev

Std Err Mean

upper 95% Mean

lower 95% Mean

N

1.0780822

0.648611

0.0759142

1.2294143

0.92675

73

Moments

mercury

Distributions

river=lumber

0 .5 1 1.5 2 2.5 3 3.5 4

Mean

Std Dev

Std Err Mean

upper 95% Mean

lower 95% Mean

N

1.2764286

0.8291484

0.0837566

1.4426623

1.1101948

98

Moments

mercury

Distributions

river=wacamaw

Conclusions based on CI’sWe are 95% confident that the population

average mercury level in fish in the Lumber River is between .93 and 1.23 ppm. Since 1.0 ppm is inside the CI, we do not feel confident that the average level is below or above the danger level. More study is needed.

We are 95% confident that the population average mercury level in fish in the Wacamaw River is between 1.11 and 1.44 ppm. It is likely that the average mercury level is beyond 1.0 ppm and therefore unsafe. Don’t eat Wacamaw bass!

Interpretation of CI’s for averagesWrong:

“95% of all fish in Wacamaw river have mercury levels between 1.11 and 1.44 pm”

Right“We are 95% confident that the average

mercury level of fish in the Wacamaw river is between 1.11 and 1.44ppm”

Special consideration for CI’s of averagesBeware of outliers

Outliers can dramatically inflate estimates of the SE. This could lead to CI’s so wide they aren’t useful.

What to do when you have outliers:

1.Check for data entry errors2.Do analyses with and without outliers. When

results differ substantially, report both analyses. Otherwise, report original analyses only.

Example 1Suppose Brent Matthews, manager of a

Sam’s Club, wants to know how much milk he should stock daily. Brent checked the sales records for random sample of 16 days and found the mean number of gallons sold is 150 gallons per day, the sample standard deviation is 12 gallons. Determine the number of gallons that Brent should stock daily with a 95% confidence interval.

Example 2It is important for airlines to follow the

published scheduled departure times of flights. Suppose that one airline that recently sampled the records of 246 flights originating in Orlando found that 10 flights were delayed for severe weather, 4 flights were delayed for maintenance concerns, and all the other flights were on time. Determine the percentage of on-time departures using a 95% confidence interval.