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Created by Bethany Stubbe and Stephan Kogitz

Two-Sample Tests of Hypothesis

GOALS

When you have completed this chapter, you will be able to:

ONE

Conduct a test of hypothesis about the difference between two independent populations means.

TWO

Conduct a test of hypothesis about the difference between two population proportions.

Two Sample Tests of Hypothesis

GOALS

When you have completed this chapter, you will be able to:

THREE

Conduct a test of a hypothesis about the mean difference between paired or dependent observations.

FOUR

Understand the difference between dependent and independent samples.

Comparing two populations

- We wish to know whether the distribution of the differences in sample means has a mean of 0.

If both samples contain at least 30 observations we use the z distribution as the test statistic.

Comparing Two Population Means

- No assumptions about the shape of the populations are required.
- The samples are from independent populations.

Comparing Two Population Means

- The formula for computing the value of z is:

EXAMPLE 1

Two cities, Bradford and Kane, are separated only by the Conewango River. There is competition between the two cities. The local paper recently reported that the mean household income in Bradford is $38,000 with a standard deviation of $6,000 for a sample of 40 households. The same article reported the mean income in Kane is $35,000 with a standard deviation of $7,000 for a sample of 35 households. At the .01 significance level, can we conclude the mean income in Bradford is more?

EXAMPLE 1 continued

- Step 1:State the null and alternate hypotheses.
- H0: µB≤ µK ; H1: µB> µK

Step 2: State the level of significance. The .01 significance level is stated in the problem.

EXAMPLE 1 continued

Step 3:Find the appropriate test statistic.Because both samples are more than 30, we can use z as the test statistic.

Step 4:State the decision rule. The null hypothesis is rejected if z is greater than 2.33.

Example 1 continued

Step 5: Compute the value of z and make a decision.

Example 1 continued

- The decision is to not reject the null hypothesis. We cannot conclude that the mean household income in Bradford is larger.
- The p-value is:
- P(z > 1.98) = .5000 - .4761 = .0239

Two Sample Tests of Proportions

We investigate whether two samples came from populations with an equal proportion of successes.

Two Sample Tests of Proportions

The two samples are pooled using the following formula.

where X1 and X2 refer to the number of successes in the respective samples of n1 and n2.

Two Sample Tests of Proportions

The value of the test statistic is computed from the following formula.

Example 2

- Are unmarried workers more likely to be absent from work than married workers? A sample of 250 married workers showed 22 missed more than 5 days last year, while a sample of 300 unmarried workers showed 35 missed more than five days. Use a .05 significance level.

Example 2 continued

- The null hypothesis is rejected if the computed value of z is greater than 1.65.
- The pooled proportion is

Example 2 continued

- The value of the test statistic is

Example 2 continued

- The null hypothesis is not rejected. We cannot conclude that a higher proportion of unmarried workers miss more days in a year than the married workers.

The p-value is:

P(z > 1.10) = .5000 - .3643 = .1457

Small Sample Tests of Means

- The t distribution is used as the test statistic if one or more of the samples have less than 30 observations.

Small Sample Tests of Means

The required assumptions are:

1. Both populations must follow the normal distribution.

2. The populations must have equal standard deviations.

3. The samples are from independent populations.

Small sample test of means continued

- Finding the value of the test statistic requires two steps.
- 1. Pool the sample standard deviations.

Small sample test of means continued

2. Determine the value of t from the following formula.

EXAMPLE 3

- A recent consumer study compared the battery life of two brands of portable CD players. A sample of 15 Brand A players revealed a mean of 33.7 hours with a standard deviation of 2.4 hours. A sample of 12 Brand B players revealed a mean of 35.7 hours with a standard deviation of 3.9. At the .05 significance level can we conclude that the battery life is higher in Brand B players?

Example 3 continued

- Step 1:State the null and alternate hypotheses.
- H0: µD≥ µI ; H1: µD< µI

Step 2: State the level of significance.The .05 significance level is stated in the problem.

Example 3 continued

Step 3:Find the appropriate test statistic.Both samples are less than 30, so we use the t distribution.

Step 4:The decision rule is to reject H0 if

t<-1.708. There are 25 degrees of freedom.

EXAMPLE 3 continued

- Step 5:We compute the pooled variance:

Example 3continued

- We compute the value of t as follows.

Example 3 continued

- H0 is not rejected. There is insufficient sample evidence to claim a higher battery life in Brand B CD players.

Hypothesis Testing Involving Paired Observations

Dependent samples are samples that are paired or related in some fashion. For example:

- If you wished to buy a car you would look at the same car at two (or more) different dealerships and compare the prices.
- If you wished to measure the effectiveness of a new diet you would weigh the dieters at the start and at the finish of the program.

Hypothesis Testing Involving Paired Observations

- Use the following test when the samples are dependent:
- where is the mean of the differences, sd is the standard deviation of the differences, and n is the number of pairs (differences)

EXAMPLE 4

- An independent testing agency is comparing the daily rental cost for renting a compact car from Hertz and Avis. A random sample of eight cities revealed the following information. At the .05 significance level can the testing agency conclude that there is a difference in the rental charged?

EXAMPLE 4 continued

- The null hypothesis and the alternate hypothesis are:Step 1:

H0 is rejected if t < -2.365 or t > 2.365. We use the t distribution with 7 degrees of freedom.

Example 4 continued

- City Hertz Avis d d2
- Calgary 42 40 2 4
- Charlottetown 56 52 4 16
- Fredericton 45 43 2 4
- Montreal 48 48 0 0
- Regina 37 32 5 25
- Toronto 45 48 -3 9
- Saint John 41 39 2 4
- Vancouver 46 50 -4 16

Example 4 continued

- Because 0.894 is less than the critical value, do not reject the null hypothesis. There is no difference in the mean amount charged by Hertz and Avis.

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