AP STATISTICS: CHAPTER 23

Inferences About Means

HYPOTHESIS TESTING FOR MEANS WITH A KNOWN POPULATION STANDARD DEVIATION ( )

1. A school administrator has developed an individualized reading-comprehension program for eight grade students. To evaluate this new program, a random sample of 45 eighth-grade students was selected. These students participated in the new reading program for one semester and then took a standard reading-comprehension examination. The mean test score for the population of students who had taken this test in the past was 76 with a standard deviation of = 8. The sample results for the 45 students provided a mean of 79. Is there significant evidence at the .05 level of significance that scores have improved with the new program?

HYPOTHESIS TESTING FOR MEANS WITH A KNOWN POPULATION STANDARD DEVIATION ( )

2. The pain reliever currently used in a hospital is known to bring relief to patients in a mean time of 3.9 minutes with a standard deviation of 1.14 minutes. To compare a new pain reliever with the current one, the new drug is administered to a random sample of 40 patients. The mean time to provide relief for the sample of patients is 3.5 minutes. Do the data provide sufficient evidence to conclude that the new drug was effective in reducing the mean time until a patient receives relief from pain? Use a .01 level of significance.

If our data come from a simple random sample (SRS) and the sample size is sufficiently large, then we know that the sampling distribution of the sample means is approximately normal with mean

μ and standard deviation .

n

PROBLEM:If Ç is unknown, then we cannot calculate

the standard deviation for the sampling model.

We must estimate the value of Ç in order to use the methods of inference that we have learned.

 SOLUTION:We will use s (the standard deviation of

the sample) to estimate Ç.

Then the standard error of the sample mean ˜ is .

In order to standardize ˜, we subtract its mean and divide by its standard deviation.

has the normal distribution N( 0, 1).

PROBLEM:If we replace Ç with s, then the statistic has

more variation and no longer has a normal distribution so we cannot call it z. It has a new distribution called the t distribution.

n

s

n

xz

t is a standardized value. Like z, t tells us how many standardized units ˜ is from the mean Ã.

When we describe a t-distribution we must identify its degrees of freedom because there is a different t-statistic for each sample size. The degrees of freedom for the one-sample t statistic is n – 1 .

The t distribution is symmetric about zero and is bell-shaped, but there is more variation so the spread is greater.

As the degrees of freedom increase, the t distribution gets closer to the normal distribution, since s gets closer to σ .

Normal Distribution

t- Distribution

We can construct a confidence interval using the t-distribution in the same way we constructed confidence intervals for the z distribution.

Remember, the t Table uses the area to the right of t*.

n

stx df*

One-sample t procedures are exactly correct only when the population is normal . It must be reasonable to assume that the population is approximately normal in order to justify the use of t procedures.

The t procedures are strongly influenced by outliers . Always check the data first! If there are outliers and the sample size is small , the results will not be reliable.

When to use t procedures: If the sample size is less than 15 , only

use t procedures if the data are close to Normal .

If the sample size is at least 15 but less than 40 , only use t procedures if the data is unimodal and reasonably symmetric .

If the sample size is at least 40 , you may use t procedures, even if the data is skewed.

EXAMPLE: A coffee vending machine dispenses coffee

into a paper cup. You’re supposed to get 10 ounces of coffee, but the amount varies slightly from cup to cup. Here are the amounts measured in a random sample of 20 cups. Is there evidence that the machine is shortchanging customers?

Use PHANTOMS!!

9.9 9.7 10.0 10.1 9.9 9.6 9.8 9.8 10.0 9.5

9.7 10.1 9.9 9.6 10.2 9.8 10.0 9.9 9.5 9.9

EXAMPLE: A company has set a goal of developing a

battery that lasts five hours (300 minutes) in continuous use. In a first test of these batteries, the following lifespans (in minutes) were measured: 321, 295, 332, 351, 336, 311, 253, 270, 326, 311, and 288.

Find a 90% confidence interval for the mean lifespan of this type of battery.

Use PANIC!!

If we wish to conduct another trial, how many batteries must we test to be 95% sure of estimating the mean lifespan to within 15 minutes? To within 5 minutes?