Definition. Statistics

Содержание

Слайд 2

Ex. 1a: Independent and Dependent Samples Classify each pair of samples

Ex. 1a: Independent and Dependent Samples

Classify each pair of samples as

independent or dependent:
Sample 1: Resting heart rates of 35 individuals before drinking coffee.
Sample 2: Resting heart rates of the same individuals after drinking two cups of coffee.
Слайд 3

Ex. 1: Independent and Dependent Samples Sample 1: Resting heart rates

Ex. 1: Independent and Dependent Samples

Sample 1: Resting heart rates of

35 individuals before drinking coffee.
Sample 2: Resting heart rates of the same individuals after drinking two cups of coffee.
These samples are dependent. Because the resting heart rates of the same individuals were taken, the samples are related. The samples can be paired with respect to each individual.
Слайд 4

Ex. 1b: Independent and Dependent Samples Classify each pair of samples

Ex. 1b: Independent and Dependent Samples

Classify each pair of samples as

independent or dependent:
Sample 1: Test scores for 35 statistics students
Sample 2: Test scores for 42 biology students who do not study statistics
Слайд 5

Ex. 1b: Independent and Dependent Samples Sample 1: Test scores for

Ex. 1b: Independent and Dependent Samples

Sample 1: Test scores for 35

statistics students
Sample 2: Test scores for 42 biology students who do not study statistics
These samples are independent. It is not possible to form a pairing between the members of samples—the sample sizes are different and the data represent test scores for different individuals.
Слайд 6

Note: Dependent samples often involve identical twins, before and after results

Note:

Dependent samples often involve identical twins, before and after results for

the same person or object, or results of individuals matched for specific characteristics.
Слайд 7

The t-Test for the Difference Between Means To perform a two-sample

The t-Test for the Difference Between Means

To perform a two-sample hypothesis

test with dependent samples, you will use a different technique. You will first find the difference for each data pair,
. The test statistic is the mean of these differences,
Слайд 8

To conduct the test, the following conditions are required: The samples

To conduct the test, the following conditions are required:

The samples must

be dependent (paired) and randomly selected.
Both populations must be normally distributed.
If these two requirements are met, then the sampling distribution for , the mean of the differences of the paired data entries in the dependent samples,
Слайд 9

To conduct the test, the following conditions are required: has a

To conduct the test, the following conditions are required:

has a

t-distribution with n – 1 degrees of freedom, where n is the number of data pairs.
Слайд 10

The following symbols are used for the t-test for μd. Although

The following symbols are used for the t-test for μd.

Although formulas

are given for the mean and standard deviation of differences, we suggest you use a technology tool to calculate these statistics.
Слайд 11

Because the sampling distribution for is a t-distribution, you can use

Because the sampling distribution for is a t-distribution, you can use

a t-test to test a claim about the mean of the differences for a population of paired data.

STUDY TIP: If n > 29, use the last row (∞) in the t-distribution table.

Слайд 12

Слайд 13

Ex. 2: The t-Test for the Difference Between Means A golf

Ex. 2: The t-Test for the Difference Between Means

A golf club

manufacturer claims that golfers can lower their score by using the manufacturer’s newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are again asked to give their most recent scores. The scores for each golfer are given in the next slide. Assuming the golf scores are normally distributed, is there enough evidence to support the manufacturer’s claim at α = 0.10?
Слайд 14

The claim is that “golfers can lower their scores.” In other

The claim is that “golfers can lower their scores.” In other

words, the manufacturer claims that the score using the old clubs will be greater than the score using the new clubs. Each difference is given by:
d = (old score) – (new score)
The null and alternative hypotheses are
Ho: μd ≤ 0 and Ha: μd > 0 (claim)
Слайд 15

Because the test is a right-tailed test, α = 0.10, and

Because the test is a right-tailed test, α = 0.10, and

d.f. = 8 – 1 = 7, the critical value for t is 1.415. The rejection region is t > 1.415. Using the table below, you can calculate and sd as follows:
Слайд 16

Using the t-test, the standardized test statistic is: The graph below

Using the t-test, the standardized test statistic is:

The graph below shows

the location of the rejection region and the standardized test statistic, t. Because t is in the rejection region, you should decide to reject the null hypothesis. There is not enough evidence to support the golf manufacturer’s claim at the 10% level The results of this test indicate that after using the new clubs, golf scores were significantly lower.
Слайд 17

Ex. 3: The t-Test for the Difference Between Means A state

Ex. 3: The t-Test for the Difference Between Means

A state legislator

wants to determine whether her voter’s performance rating (0-100) has changed from last year to this year. The following table shows the legislator’s performance rating for the same 16 randomly selected voters for last year and this year. At α = 0.01, is there enough evidence to conclude that the legislator’s performance rating has changed? Assume the performance ratings are normally distributed.
Слайд 18

If there is a change in the legislator’s rating, there will

If there is a change in the legislator’s rating, there will

be a difference between “this year’s” ratings and “last year’s) ratings. Because the legislator wants to see if there is a difference, the null and alternative hypotheses are:
Ho: μd = 0 and Ha: μd ≠ 0 (claim)
Слайд 19

Because the test is a tw0-tailed test, α = 0.01, and

Because the test is a tw0-tailed test, α = 0.01, and

d.f. = 16 – 1 = 15, the critical values for t are 2.947. The rejection region are t < -2.947 and t > 2.947.
Слайд 20

Using the t-test, the standardized test statistic is: The graph shows

Using the t-test, the standardized test statistic is:

The graph shows the

location of the rejection region and the standardized test statistic, t. Because t is not in the rejection region, you should fail to reject the null hypothesis at the 1% level. There is not enough evidence to conclude that the legislator’s approval rating has changed.
Слайд 21

Using Technology If you prefer to use a technology tool for

Using Technology

If you prefer to use a technology tool for this

type of test, enter the data in two columns and form a third column in which you calculate the difference for each pair. You can now perform a one-sample t-test on the difference column as shown in Chapter 7.

Stat|Edit|enter data
Subtract L1 – L2 = in L3.
STAT|Tests|t-test
Data
μ = 0
List: L3
Freq: 1
μ ≠ 0
Calculate