copyright © 2013, 2010 and 2007 pearson education, inc. section inference about two means:...

Copyright © 2013, 2010 and 2007 Pearson Education, Inc.

Section

Inference about Two Means: Independent Samples

11.3


Objectives

1. Test hypotheses regarding the difference of two independent means

2. Construct and interpret confidence intervals regarding the difference of two independent means

11-2


Suppose that a simple random sample of size n1 is taken from a population with unknown mean μ1 and unknown standard deviation σ1. In addition, a simple random sample of size n2 is taken from a population with unknown mean μ2 and unknown standard deviation σ2. If the two populations are normally distributed or the sample sizes are sufficiently large(n1 ≥ 30, n2 ≥ 30) , then

Sampling Distribution of the Difference of Two Means: Independent Samples with Population

Standard Deviations Unknown (Welch’s t)

11-3


approximately follows Student’s t-distribution with the smaller of n1-1 or n2-1 degrees of freedom where is the sample mean and si is the sample standard deviation from population i.

Sampling Distribution of the Difference of Two Means: Independent Samples with Population

Standard Deviations Unknown (Welch’s t)

xi

t x

1 x

2 1

2 s

12

n1

s

22

n2

11-4


Objective 1

• Test Hypotheses Regarding the Difference of Two Independent Means

11-5


To test hypotheses regarding two population means, μ1 and μ2, with unknown population standard deviations, we can use the following steps, provided that:

the samples are obtained using simple random sampling;

the samples are independent; the populations from which the samples are

drawn are normally distributed or the sample sizes are large (n1 ≥ 30, n2 ≥ 30);

For each sample, the sample size is no more than 5% of the population size.

Testing Hypotheses Regarding the Difference of Two Means

11-6


Step 1: Determine the null and alternative hypotheses. The hypotheses are structured in one of three ways:

11-7

Step 2: Select a level of significance, α, based on the seriousness of making a Type I error.

𝑡 0=(𝑥1−𝑥2 ) − (𝜇1−𝜇2 )

√ 𝑠12

𝑛1

+𝑠2

2

𝑛2

Step 3: Compute the test statistic

Use Table VI to determine the critical value using the smaller of n1 – 1 or n2 – 1 degrees of freedom.

Step 4: Find the critical value(s).


Compare the critical value with the test statistic:

Classical Approach

11-8


Use Table VI to determine the P-value using the smaller of n1 – 1 or n2 – 1 degrees of freedom.

P-Value Approach

11-9

Technology: Use a statistical spreadsheet or calculator with statistical capabilities to obtain the P-value. The directions for obtaining the P-value using the TI-83/84 Plus graphing calculator, Excel, MINITAB, and StatCrunch are in the Technology Step-by-Step in the text.

Finally: State the conclusion.


These procedures are robust, which means that minor departures from normality will not adversely affect the results. However, if the data have outliers, the procedure should not be used.

11-10


A researcher wanted to know whether “state” quarters had a weight that is more than “traditional” quarters. He randomly selected 18 “state” quarters and 16 “traditional” quarters, weighed each of them and obtained the following data.

Parallel Example 1: Testing Hypotheses Regarding Two Means

11-11


Test the claim that “state” quarters have a mean weight that is more than “traditional” quarters at the α = 0.05 level of significance.

NOTE: A normal probability plot of “state” quarters indicates the population could be normal. A normal probability plot of “traditional” quarters indicates the population could be normal

11-13


No outliers.

11-14


Step 1: We want to determine whether state quarters weigh more than traditional quarters:

H0: μ1 = μ2 versus H1: μ1 > μ2

Step 2: The level of significance is α = 0.05.

Step 3: The test statistic is

Solution

t0 5.7022 5.6494

0.04972

18 0.06892

16

2.53.

11-15


Solution: Classical Approach

This is a right-tailed test with α = 0.05. Since n1 – 1 = 17 and n2 – 1 = 15, we will use 15 degrees of freedom. The corresponding critical value is t0.05=1.753.

11-16


Step 4: Since the test statistic, t0 = 2.53 is greater than the critical value t.05 = 1.753, we reject the null hypothesis.


11-17


Because this is a right-tailed test, the P-value is the area under the t-distribution to the right of the test statistict0 = 2.53. That is, P-value = P(t > 2.53) ≈ 0.01.

Solution: P-Value Approach

11-18


Step 4: Since the P-value is less than the level of significance α = 0.05, we reject the null hypothesis.


11-19


Step 5: There is sufficient evidence at the α = 0.05 level to conclude that the state quarters weigh more than the traditional quarters.

Solution

11-20


NOTE: The degrees of freedom used to determine the critical value in the last example are conservative. Results that are more accurate can be obtained by using the following degrees of freedom:

df

s12

n1

s2

2

n2

2

s12

n1

2

n1 1

s22

n2

2

n2 1

11-21


Objective 2

• Construct and Interpret Confidence Intervals Regarding the Difference of Two Independent Means

11-22


A simple random sample of size n1 is taken from a population with unknown mean μ1 and unknown standard deviation σ1. Also, a simple random sample of size n2 is taken from a population with unknown mean μ2 and unknown standard deviation σ2. If the two populations are normally distributed or the sample sizes are sufficiently large (n1 ≥ 30 and n2 ≥ 30), a(1 – α)•100% confidence interval about μ1 – μ2 is given by . . .

Constructing a (1 – α)•100% ConfidenceInterval for the Difference of Two Means

11-23


Lower bound:

and

Upper bound:

where tα/2 is computed using the smaller of

n1 – 1 or n2 – 1 degrees of freedom or Formula

(2).

Constructing a (1 – α)•100% ConfidenceInterval for the Difference of Two Means

x

1 x

2 t2

s

12

n1

s

22

n2

x

1 x

2 t2

s

12

n1

s

22

n2

11-24


Construct a 95% confidence interval about the difference between the population mean weight of a “state” quarter versus the population mean weight of a “traditional” quarter.

Parallel Example 3: Constructing a Confidence Interval for the Difference of Two Means

11-25


• We have already verified that the populations are approximately normal and that there are no outliers.

• Recall = 5.702, s1 = 0.0497, =5.6494 and s2 = 0.0689.

• From Table VI with α = 0.05 and 15 degrees of freedom, we find tα/2 = 2.131.

Solution

x 1

x 2

11-26


Thus,

• Lower bound =

• Upper bound =

Solution

5.702 5.649 2.1310.04972

18

0.06892

160.0086

5.702 5.649 2.1310.04972

18

0.06892

160.0974

11-27


We are 95% confident that the mean weight of the “state” quarters is between 0.0086 and 0.0974 ounces more than the mean weight of the “traditional” quarters. Since the confidence interval does not contain 0, we conclude that the “state” quarters weigh more than the “traditional” quarters.

Solution

11-28


When the population variances are assumed to be equal, the pooled t-statistic can be used to test for a difference in means for two independent samples. The pooled t-statistic is computed by finding a weighted average of the sample variances and using this average in the computation of the test statistic.

The advantage to this test statistic is that it exactly follows Student’s t-distribution with n1+n2-2 degrees of freedom.

The disadvantage to this test statistic is that it requires that the population variances be equal.

11-29


Section

Inference about Two Population Standard Deviations

11.4


Objectives

1. Find critical values of the F-distribution

2. Test hypotheses regarding two population standard deviations

11-31


Objective 1

• Find Critical Values of the F-distribution

11-32


Requirements for Testing Claims Regarding Two Population Standard

Deviations

1. The samples are independent simple random samples.

11-33


Requirements for Testing Claims Regarding Two Population Standard

Deviations

1. The samples are independent simple random samples.2. The populations from which the samples are drawn are normally distributed.

11-34


CAUTION!

If the populations from which the samples are drawnare not normal, do not use the inferential proceduresdiscussed in this section.

11-35


Notation Used When Comparing Two Population Standard Deviations

12

: Variance for population 1

: Variance for population 2

: Sample variance for population 1

: Sample variance for population 2

n1 : Sample size for population 1

n2 : Sample size for population 2

22

s12

s22

11-36


Fisher's F-distribution

If and and are sample variances fromindependent simple random samples of size n1 andn2, respectively, drawn from normal populations, then

follows the F-distribution with n1 – 1 degrees offreedom in the numerator and n2 – 1 degrees of freedom in the denominator.

12 2

2

s12

s22

F s1

2

s22

11-37


Characteristics of the F-distribution

1. It is not symmetric. The F-distribution is skewed right.

11-38



1. It is not symmetric. The F-distribution is skewed right.2. The shape of the F-distribution depends upon the degrees of freedom in the numerator and denominator. This is similar to the χ2 distribution and Student’s t-distribution, whose shapes depend upon their degrees of freedom.

11-39



1. It is not symmetric. The F-distribution is skewed right.2. The shape of the F-distribution depends upon the degrees of freedom in the numerator and denominator. This is similar to the χ2 distribution and Student’s t-distribution, whose shape depends upon their degrees of freedom.3. The total area under the curve is 1.

11-40



1. It is not symmetric. The F-distribution is skewed right.2. The shape of the F-distribution depends upon the degrees of freedom in the numerator and denominator. This is similar to the χ2 distribution and Student’s t-distribution, whose shape depends upon their degrees of freedom.3. The total area under the curve is 1.4. The values of F are always greater than or equal to zero.

11-41


is the critical F with n1 – 1 degrees of freedom in the numerator and n2 – 1 degrees of freedom in the denominator and an area of α to the right of the critical F.

F ,n1 1,n2 1

11-43


To find the critical F with an area of α to the left, use the following:

F1 ,n1 1,n2 1 1

F ,n2 1,n1 1

11-44


Find the critical F-value:

a) for a right-tailed test with α = 0.1, degrees of freedom in the numerator = 8 and degrees of freedom in the denominator = 4.

b) for a two-tailed test with α = 0.05, degrees of freedom in the numerator = 20 and degrees of freedom in the denominator = 15.

Parallel Example 1: Finding Critical Values for the F-distribution

11-45


a) F0.1,8,4 = 3.95

b) F.025,20,15 = 2.76

Solution

F

.975,20,15

1

F.025,15,20

1

2.570.39

11-46


NOTE:

If the number of degrees of freedom is not found in the table, we follow the practice of choosing the degrees of freedom closest to that desired. If the degrees of freedom is exactly between two values, find the mean of the values.

11-47


Objective 2

• Test Hypotheses Regarding Two Population Standard Deviations

11-48


To test hypotheses regarding two population standard deviations, σ1 and σ2, we can use the following steps, provided that

1. the samples are obtained using simple random sampling,

2. the sample data are independent, and

3. the populations from which the samples are drawn are normally distributed.

Test Hypotheses Regarding Two Population Standard Deviations

11-49


Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:

Two-Tailed Left-Tailed Right-Tailed

H0: σ1 = σ2 H0: σ1 = σ2 H0: σ1 = σ2

H1: σ1 ≠ σ2 H1: σ1 < σ2 H1: σ1 > σ2

Note: σ1 is the population standard deviation for population 1 and σ2 is the population standard deviation for population 2.

11-50


Step 2: Select a level of significance, α, based on the seriousness of making a Type I error.

11-51


Step 3: Compute the test statistic

which follows Fisher’s F-distribution with n1 – 1 degrees of freedom in the numerator and n2 – 1 degrees of freedom in the denominator.

21

0 22

sF

s

11-52


Use Table VIII to determine the critical value(s) using n1 – 1 degrees of freedom in the numerator and n2 – 1 degrees of freedom in the denominator.

Classical Approach

11-53


Classical Approach

(critical value)

Two-Tailed

11-54


Classical Approach

(critical value)

Left-Tailed

11-55


Classical Approach

(critical value)

Right-Tailed

11-56


Step 4: Compare the critical value with the test statistic:

Classical Approach

Two-Tailed Left-Tailed Right-Tailed

If or , reject the null hypothesis.

If , reject the null hypothesis.

If , reject the null hypothesis.

F0 F1 2,n1 1,n2 1

F0 F 2,n1 1,n2 1

F0 F1 ,n1 1,n2 1

F0 F,n1 1,n2 1

11-57


Step 3: Use technology to determine the P-value.

P-Value Approach

11-58

Step 4: If P-value < α, reject the null hypothesis.

Step 5: State the conclusion.


CAUTION!

The procedures just presented are not robust,minor departures from normality will adversely affect the results of the test. Therefore, the test should be used only when the requirement of normality has been verified.

11-59


A researcher wanted to know whether “state” quarters had a standard deviation weight that is less than “traditional” quarters. He randomly selected 18 “state” quarters and 16 “traditional” quarters, weighed each of them and obtained the data on the next slide. A normal probability plot indicates that the sample data could come from a population that is normal. Test the researcher’s claim at the α = 0.05 level of significance.

Parallel Example 2: Testing Hypotheses Regarding Two Population Standard Deviations

11-60


Step 1: The researcher wants to know if “state” quarters have a standard deviation weight that is less than “traditional” quarters. Thus

H0: σ1 = σ2 versus H1: σ1 < σ2

This is a left-tailed test.

Step 2: The level of significance is α = 0.05.

Solution

11-62


Step 3: The standard deviation of “state” quarters was found to be 0.0497 and the standard deviation of “traditional” quarters was found to be 0.0689. The test statistic is then

Solution

F0 0.04972

0.068920.52

11-63


Since this is a left-tailed test, we determine the critical value at the 1 – α = 1 – 0.05 = 0.95 level of significance with n1 – 1=18 – 1=17 degrees of freedom in the numerator and n2 – 1 = 16 – 1 = 15 degrees of freedom in the denominator. Thus,

Note: we used the table value F0.05,15,15 for the above calculation since this is the closest to the required degrees of freedom available from Table VIII.


F

.95,17,15

1

F.05,15,17

1

2.400.42.

11-64


Step 4: Since the test statistic F0= 0.52 is greater than the critical value F0.95,17,15=0.42, we fail to reject the null hypothesis.


11-65


Step 3: Using technology, we find that the P-value is 0.097. If the statement in the null hypothesis were true, we would expect to get the results obtained about 10 out of 100 times. This is not very unusual.


11-66


Step 4: Since the P-value is greater than the level of significance, α = 0.05, we fail to reject the null hypothesis.


11-67


Step 5: There is not enough evidence to conclude that the standard deviation of weight is less for “state” quarters than it is for “traditional” quarters at the α = 0.05 level of significance.

Solution

11-68


Section

Putting It Together: Which Method Do I Use?

11.5


Objective

1. Determine the appropriate hypothesis test to perform

11-70


Objective 1

• Determine the Appropriate Hypothesis Test to Perform

11-71


What parameter is addressed in the hypothesis?

• Proportion, p• σ or σ2

• Mean, μ

11-72


Proportion, p

Is the sampling Dependent or Independent?

11-73


Proportion, p

Dependent samples:Provided the samples are obtained randomly and the total number of observations where the outcomes differ is at least 10, use the normal distribution with

z0 f12 f21 1

f12 f21

11-74


Proportion, p

Independent samples:

Provided for each sample and the sample size is no more than 5% of the populationsize, use the normal distribution with

where

nˆ p 1 ˆ p 10

z0 ˆ p 1 ˆ p 2

ˆ p 1 ˆ p 1

n1

1

n2

ˆ p x1 x2

n1 n2

11-75


σ or σ2

Provided the data are normally distributed, use the F-distribution with

F0 s1

2

s22

11-76


Mean, μ

Is the sampling Dependent or Independent?

11-77


Mean, μ

Dependent samples:Provided each sample size is greater than 30 or the differences come from a population that is normally distributed, use Student’s t-distributionwith n-1 degrees of freedom with

t0 d d

sd

n

11-78


Mean, μ

Independent samples:Provided each sample size is greater than 30 or eachpopulation is normally distributed, use Student’st-distribution

t0 x 1 x 2 1 2

s12

n1

s22

n2

11-79

copyright © 2013, 2010 and 2007 pearson education, inc. section inference about two means:...

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