chapter 10 statistical inferences based on two samples statistics for business (env) 1

Chapter 10Statistical Inferences Based on

Two Samples

Statistics for Business(Env)

1

Statistical Inferences Based on Two Samples

10.1 Comparing Two Population Means by Using Independent Samples: Variances Known

10.2 Comparing Two Population Means by Using Independent Samples: Variances Unknown

10.3 Paired Difference Experiments10.4 Comparing Two Population Proportions by Using

Large, Independent Samples

2

Comparing Two Population Means by Using Independent Samples: Variances Known

• Suppose a random sample has been taken from each of two different populations

• Suppose that the populations are independent of each other– Then the random samples are independent of

each other• Then the sampling distribution of the difference in

sample means is normally distributed

3

4

Do the achievement scores for children taught by method A differ from the scores for children taught by method B?

5

A research design that uses a separate sample for each treatment condition (or for each population) is called an independent-measures research design or a between-subjects design.The goal of an independent-measures research study is to evaluate the mean difference between two populations (or between two treatment conditions).

Sampling Distribution of theDifference of Two Sample Means #1

• Suppose population 1 has mean µ1 and variance σ12

– From population 1, a random sample of size n1 is selected which has mean and variance s1

2

• Suppose population 2 has mean µ2 and variance σ22

– From population 2, a random sample of size n2 is selected which has mean and variance s2

2

• Then the sample distribution of the difference of two sample means…

6


• Is normal, if each of the sampled populations is normal– Approximately normal if the sample sizes n1 and n2

are large

• Has mean = µ1 – µ2

• Has standard deviation2

22

1

21

21 nnxx

7

8

µ1µ2

If you select one score from each of these two populations, the closest two values are X1 =50 and X2 =30. The two values that are farthest apart are X1 =70 and X2 =20.


9

z-Based Confidence Interval for the Difference in Means (Variances Known)

• Let be the mean of a sample of size n1 that has been randomly selected from a population with mean 1 and standard deviation 1

• Let be the mean of a sample of size n2 that has been randomly selected from a population with 2 and 2

• Suppose each sampled population is normally distributed or that the samples sizes n1 and n2 are large

• Suppose the samples are independent of each other, then …

10

z-Based Confidence Interval for the Difference in Means Continued

• A 100(1 – ) percent confidence interval for the difference in populations µ1–µ2 is

2

22

1

21

221 nnzxx

11

Example 10.1 The Bank Customer Waiting Time Case #1

• A random sample of size 100 waiting times observed under the current system of serving customers has a sample mean of 8.79– Call this population 1– Assume population 1 is normal or sample size is large– The variance is 4.7

• A random sample of size 100 waiting times observed under the new system of time of 5.14– Call this population 2– Assume population 2 is normal or sample size is large– The variance is 1.9

• Then if the samples are independent …

12

Example 10.1 The Bank Customer Waiting Time Case #2

• At 95% confidence, z/2 = z0.025 = 1.96, and

• According to the calculated interval, the bank manager can be 95% confident that the new system reduces the mean waiting time by between 3.15 and 4.15 minutes

154153

50350653

100

91

100

74961145798

2

22

1

21

221

.,.

..

.....

nnzxx

13

z-Based Test About the Difference in Means (Variances Known)

• Test the null hypothesis aboutH0: µ1 – µ2 = D0

– D0 = µ1 – µ2 is the claimed difference between the population means

– D0 is a number whose value varies depending on the situation

– Often D0 = 0, and the null means that there is no difference between the population means

14

z-Based Test About the Difference in Means (Variances Known)

• Use the notation from the confidence interval statement on a prior slide

• Assume that each sampled population is normal or that the samples sizes n1 and n2 are large

15

Test Statistic (Variances Known)

• The test statistic is

• The sampling distribution of this statistic is a standard normal distribution

• If the populations are normal and the samples are independent ...

2

22

1

21

021

nn

Dxxz

16

z-Based Test About the Difference inMeans (Variances Known)

• Reject H0: µ1 – µ2 = D0 in favor of a particular alternative hypothesis at a level of significance if the appropriate rejection point rule holds (i.e. calculated z is in the rejection region).

• Rules are on the next slide…

17

Hypothesis Tests forTwo Population Means

Lower-tail test:

H0: μ1 μ2

H1: μ1 < μ2

i.e.,

H0: μ1 – μ2 0H1: μ1 – μ2 < 0

Upper-tail test:

H0: μ1 ≤ μ2

H1: μ1 > μ2

i.e.,

H0: μ1 – μ2 ≤ 0H1: μ1 – μ2 > 0

Two-tail test:

H0: μ1 = μ2

H1: μ1 ≠ μ2

i.e.,

H0: μ1 – μ2 = 0H1: μ1 – μ2 ≠ 0

Two Population Means, Known Population Variances

18

Two Population Means, Known Population Variances

Lower-tail test:

H0: μ1 – μ2 0H1: μ1 – μ2 < 0

Upper-tail test:

H0: μ1 – μ2 ≤ 0H1: μ1 – μ2 > 0

Two-tail test:

H0: μ1 – μ2 = 0H1: μ1 – μ2 ≠ 0

/2 /2

-z -z/2z z/2

Reject H0 if Z < -Z Reject H0 if Z > Z Reject H0 if Z < -Z/2

or Z > Z/2

Hypothesis tests for μ1 – μ2

19

EXAMPLE

The mean income in Kingston is $35,000 for a sample of 35 households. The population s.d. is known to be $7,000.

Two cities, Boston and Kingston are both in Massachusetts.

The mean household income in Boston is $38,000. The population s.d. is known to be $6,000 for a sample of 40 households.

At the .01 significance level can we conclude the mean income in Boston is more?

Step 2 Select the level of significance.

The .01 significance level is stated in the problem.

Step 3 Find the appropriate test

statistic. Since both samples are more than 30, we can use

z as the test statistic.

Step 1 State the null and

alternate hypotheses.H0: µB < µK

H1: µB > µK

Step 4 State the decision rule.The null hypothesis is

rejected if t is greater than 2.326 or p < .01.

EXAMPLE

21

98.1

35

)000,7($

40

)000,6($

000,35$000,38$t

22

Because the computed Z of 1.98 < critical Z of 2.26, the p-value of .0239 > .01 (), the decision is not to reject the H0. We cannot conclude that the mean household income in Boston is larger.

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

EXAMPLE

22

Comparing Two Population Means by Using Independent Samples: Variances Unknown

• In general, the true values of the population variances σ1

2 and σ22 are not known

• They have to be estimated from the sample variances s1

2 and s22, respectively

23

Comparing Two Population Means by Using Independent Samples: Variances Unknown #2

• Also need to estimate the standard deviation of the sampling distribution of the difference between sample means

• Two approaches:1. If it can be assumed that σ1

2 = σ22 = σ2, then

calculate the “pooled estimate” of σ2

2. If σ12 ≠ σ2

2, then use approximate methods

24

Pooled Estimate of σ2

• Assume that σ12 = σ2

2 = σ2

• The pooled estimate of σ2 is the weighted averages of the two sample variances, s1

2 and s22

• The pooled estimate of σ2 is denoted by sp2

• The estimate of the population standard deviation of the sampling distribution is

2

11

21

222

2112

nn

snsns p

21

2 1121 nn

s pxx

25

26

One sample compared with 2 samples statistics

df2

SS2

Assume that σ12 = σ2

2 = σ2

Mean

t-Based Confidence Interval for the Difference in Means (Variances Unknown)

• Select independent random samples from two normal populations with equal variances

• A 100(1 – ) percent confidence interval for the difference in populations µ1 – µ2 is

• where

• and t/2 is based on (n1+n2-2) degrees of freedom (df)

21

2221

11

nnstxx p

2

11

21

222

2112

nn

snsns p

27

2

)1()1(

21

222

2112

nn

snsns p

21p

21

n

1

n

1s

XXt

Step Two: Determine the value of t from the following formula.

Step One: Pool the sample standard deviations.

Finding the value of the test statistic requires two steps:

28

Two Population Means, Unknown Population Variances

Lower-tail test:

H0: μ1 – μ2 0H1: μ1 – μ2 < 0

Upper-tail test:

H0: μ1 – μ2 ≤ 0H1: μ1 – μ2 > 0

Two-tail test:

H0: μ1 – μ2 = 0H1: μ1 – μ2 ≠ 0

/2 /2

-t -t/2t t/2

Reject H0 if t < -t Reject H0 if t > t Reject H0 if t < -t/2

or t > t/2

Hypothesis tests for μ1 – μ2

29

A recent EPA study compared the highway fuel economy of domestic and imported passenger cars. A sample of 15 domestic cars revealed a mean of 33.7 mpg with a sample standard deviation of 2.4 mpg.

A sample of 12 imported cars revealed a mean of 35.7 mpg with a sample standard deviation of 3.9. At the .05 significance level can the EPA conclude that the mpg is higher on the imported cars?

Example:

30

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.

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

Example: (continued)

31

918.921215

)9.3)(112()4.2)(115(

2

))(1())(1(

22

21

222

2112

nn

snsns p

Step 4

The decision rule is to reject H0 if t<-1.708. There are n1 + n2 – 2 or 25 degrees of freedom.

Step 5 We compute the pooled variance.


32

640.1

121

151

918.9

7.357.33

11

21

2

21

nns

XXt

p

We compute the value of t as follows.


33

Since a computed z of –1.64 > critical z of –1.71, H0 can not be rejected. There is insufficient sample evidence to claim a higher mpg on the imported cars.


34

-1.71 -1.64

35

To show if boys are heavier than girls of the same age, a survey is conducted in which a sample of 15 boys shows a mean weight of 41Kg and a standard deviation of 3Kg. A group of 10 girls of the same age shows a mean weight of 38Kg and a standard deviation of 2Kg. Assuming both the weights of boys and girls follow the normal distribution. At the level of significant 0.05, test if the average weight of boys is greater than the average weight of girls of the same age.

Example: Comparing Mean weights


hypotheses. H0: µg > µb H1: µg < µb




36

Step 4

The decision rule is to reject H0 if t > t0.05 =1.714. There are n1 + n2 – 2 or 23 degrees of freedom.

37


Step 5 Compute the pooled variance and t. S2

p = [(15-1)*32 + (10-1)*22]/ (15+10-2) = 7.04Sp = 2.65 t = (41-38) / sqrt(7.04*(1/15 + 1/10)) = 2.77

Since t =2.77 > t0.05 =1.714, we reject H0 .So the mean weight of boys is larger than the mean weight of girls of the same age.

=1.714

38

Example: Directed reading activities in the classroom

A class of 21 third-graders participates in these activities for 8 weeks

while a control classroom of 23 third-graders follows the same

curriculum without the activities. After the 8 weeks, all children take a

reading test (scores in table). At a level of significance 0.05, can we

conclude directed reading activities help improve reading ability?

Step 1 State the null and

alternate hypotheses. H0: µ1 = µ2

H1: µ1 = µ2

39



Step 4

The decision rule is to reject H0 if t > t0.025 =1.97 or t < -t0.025 . There are n1 + n2 – 2 or 42 degrees of freedom.

Example: Directed reading activities (continued)

40



p = [(21-1)*11.012 + (23-1)*17.152]/ (21+23-2) = 211.79t = (51.48-41.52) / sqrt(211.79*(1/21 + 1/23)) = 9.96/4.39=2.27

Since t =2.27 > t0.025 =1.97, we reject H0 .So there are significant difference between the 2 group.

41


hypotheses. H0: µ2 > µ1 H1: µ2 < µ1


p = [(21-1)*11.012 + (23-1)*17.152]/ (21+23-2) = 211.79t = (51.48-41.52) / sqrt(211.79*(1/21 + 1/23)) = 9.96/4.39=2.27

There are n1 + n2 – 2 or 42 degrees of freedom. The rule is to reject H0 if t > t0.05

=1.65.


Chap 9-42

Pooled Variance t Test: Example

You are a financial analyst for a brokerage firm. Is there a difference in dividend yield between stocks listed on the NYSE & NASDAQ? You collect the following data: NYSE NASDAQNumber 21 25Sample mean 3.27 2.53Sample std dev 1.30 1.16

Assuming both populations are approximately normal with equal variances, isthere a difference in average yield ( = 0.05)?

Calculating the Test Statistic

1.5021

1)25(1)-(21

1.161251.30121

1)n()1(n

S1nS1nS

22

21

222

2112

p

2.040

251

211

5021.1

02.533.27

n1

n1

S

μμXXt

21

2p

2121

The test statistic is:

43

Solution

H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)

H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)

= 0.05df = 21 + 25 - 2 = 44Critical Values: t = ± 1.96

Test Statistic: Decision:

Conclusion:

Reject H0 at = 0.05

There is evidence of a difference in means.

t0 1.96-1.96

.025

Reject H0 Reject H0

.025

2.040

2.040

251

211

5021.1

2.533.27t

44

Two kinds of studies

So far, we have studied :

two sets of sample data that come from two independent populations (e.g. women and men, or students from program A and from program B).

However, sometimes we want to study

two sets of sample data that come from related populations (e.g. “before treatment” and “after treatment”).

Independent samples

Paired samples

45

Paired/Dependent Samples

Dependent samples are samples that are paired or related in some fashion.

*The same subjects measured at two different points in time (repeated-measures).

*Matched or paired observations*Hypothesis test proceeds just as in the one

sample case.

Independent samples are samples that are not related in any way.

46

Existing System (1) New Software (2) Difference Di

9.98 Seconds 9.88 Seconds .109.88 9.86 .029.84 9.75 .099.99 9.80 .199.94 9.87 .079.84 9.84 .009.86 9.87 - .0110.12 9.98 .149.90 9.83 .079.91 9.86 .05

Paired-Sample t Test: ExampleAssume you work in the finance department. Is the new financial package faster (=0.05 level)? You collect the following processing times for same set of jobs:

2

.072

1 .06215

i

iD

DD

n

D DS

n

47

Paired-Sample t Test: Example

Is the new financial package faster (0.05 level)?

.072D =

H0: D H1: D

Test Statistic

Critical Value=1.8331 df = n - 1 = 9

Reject

1.8331

Decision: Reject H0

t Stat. in the rejection zone.

Conclusion: The new software package is faster.

3.66

t

.072 03.66

/ .06215/ 10D

D

DtS n

48

Suppose we collect 8 pairs of twins. The first twin in the pair is healthy; the second is not. For each twin, we measure grey matter density (gmd).Is grey matter density in the populations significantly different ?

Processed data from the 8 pairs is shown below (units not given).Consider the population differences, D = X1 - X2,

Paired-Sample: Example-twins

Hypothesis Testing Involving Paired Observations

where D is the mean of the differencessd is the (sample) s.d. of the differencesn is the number of pairs (differences)

If σD is unknown, we can estimate the unknown population standard deviation with a sample standard deviation:

1n

)D(DS

n

1i

2i

D

n

SμD

tD

D

The test statistic for D is now a t statistic, with n-1 d.f.

(continued)

50

The confidence interval for μD is

1n

)D(DS

n

1i

2i

D

n

StD D

1n

Confidence Interval of Paired Observations, σD Unknown

where SD is:

(continued)

51

Chap 9-52

Lower-tail test:

H0: μD 0H1: μD < 0

Upper-tail test:

H0: μD ≤ 0H1: μD > 0

Two-tail test:

H0: μD = 0H1: μD ≠ 0

Paired Samples

Hypothesis Testing for Mean Difference, σD Unknown

/2 /2

-t -t/2t t/2

Reject H0 if t < -t Reject H0 if t > t Reject H0 if t < -t or t > t Where t has n - 1 d.f.

• Assume you send your salespeople to a “customer service” training workshop. Has the training made a difference in the number of complaints? You collect the following data:

Paired Samples Example

Number of Complaints: (2) - (1)Salesperson Before (1) After (2) Difference, Di

Chen 6 4 - 2 Li 20 6 -14 Zhang 3 2 - 1 Wang 0 0 0 Wan 4 0 - 4 -21

D = Di

n

5.67

1n

)D(DS

2i

D

= -4.2

Chap 9-53

• Has the training made a difference in the number of complaints (at the 0.01 level)?

1 .6 655 .6 7 /

04 .2

n/S

μDt

D

D

H0: μD = 0H1: μD 0

Test Statistic:

Critical Value = ± 4.604

Reject

/2

- 4.604 4.604

Decision: Do not reject H0

(t stat is not in the reject region)

Conclusion: There is not a significant change in the number of complaints.

Paired Samples: Solution

Reject

/2

- 1.66

= .01 d.f. = n - 1 = 4

D = -4.2

Chap 9-54

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?

City Hertz ($)

Avis ($)

Atlanta 42 40

Chicago 56 52

Cleveland 45 43

Denver 48 48

Honolulu 37 32

Kansas City 45 48

Miami 41 39

Seattle 46 50

Step 4

H0 is rejected if

t < -2.365 or t > 2.365;

or if p-value < .05.

We use the t distribution with n-1 or 7 degrees of freedom.

Step 2 The stated

significance level is .05.

Step 3 The appropriate test

statistic is the paired t-test.

Step 1Ho: d = 0H1: d 0

Step 5Perform the calculations and

make a decision.

56

City Hertz Avis d d2

Atlanta 42 40 2 4

Chicago 56 52 4 16

Cleveland45 43 2 4

Denver 48 48 0 0

Honolulu 37 32 5 25

Kansas City 45 48 -3 9

Miami 41 39 2 4

Seattle 46 50 -4 16

57

00.18

0.8

n

dd

1623.3

188

878

1

222

n

n

dd

sd

894.081623.3

00.1

ns

dt

d

58

P(t>.894) = .20 for a one-tailed t-test at 7 degrees of freedom.

Because 0.894 is less than the critical value, the p-value of .20 > a of .05, do not reject the null hypothesis. There is no difference in the mean amount charged by Hertz and Avis.

59

Comparing Two Population Proportions

Goal: test a hypothesis or form a confidence interval for the difference between two population proportions (p1 – p2).

The point estimate for the difference is

Assumptions: n1p1 5 , n1(1-p1) 5

n2p2 5 , n2(1-p2) 5

21 ss pp

60

two independent samples from two populations

Two Population Proportions

21

21

nn

XXp

The pooled estimate for the

overall proportion is:

where X1 and X2 are the numbers from samples 1 and 2 with the characteristic of interest

Since we begin by assuming the null hypothesis is true, we assume p1 = p2 and pool the two ps estimates

61

Two Population Proportions

21

21ss

n1

n1

)p1(p

ppppZ 21

The test statistic for p1 – p2 is a Z statistic:

(continued)

2

2s

1

1s

21

21

n

Xp ,

n

Xp ,

nn

XXp

21

where

62

Confidence Interval forTwo Population Proportions

Population proportions

2

ss

1

ssss n

)p(1p

n

)p(1pZpp 2211

21

The confidence interval for

p1 – p2 is:

63

ExampleAre 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.

64

The null hypothesis is rejected if the computed value of z is greater than 1.65 or the p-value < .05.

The pooled proportion

250300

2235

cp = .1036

The null and the alternate hypothesesH0: U < M H1: U > M

65

10.1

250

)1036.1(1036.

300

)1036.1(1036.250

22

300

35

z

Because the calculated z of 1.10 < a critical z of 1.65 ( of .05), 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.

66

Chapter Ten

Two-Sample Tests of Two-Sample Tests of HypothesisHypothesis

TWO- Conduct a test of hypothesis regarding the difference in two population proportions with Known/ Unknown Variances

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

ONE- Conduct a test of hypothesis about the difference between two independent population means with Known/ Unknown Variances

THREE- Understand the difference between dependent and independent samples.

67

68

Table of the Standard Normal Distribution

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141 0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517 0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879 0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224 0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852 0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389 1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319 1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857 2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890 2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916 2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936 2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986 3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990

chapter 10 statistical inferences based on two samples statistics for business (env) 1

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