Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Chapter Hypothesis Test of a Claim...

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

Chapter

Hypothesis Test of a Claim

Regarding a Population Parameter

( p , μ )

10

Chap 2 2


Section

The Language of Hypothesis Testing

10.1

10-4

An hypothesis is a statement regarding a characteristic of one or more populations.

From the Greek, an hypothesis is a statement, or claim, yet to be proved.

In statistics, hypothesis testing is the use of sampling and probability theory to evaluate a

claim that someone has made regarding a characteristic of the population.

10-5

1. In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher claims that this percentage is different

today.

2. According to a study published in 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher claims

that call length has increased since then.

3. Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 oz. With new equipment,

the quality control manager claims the standard deviation has decreased.

10-6

We test these types of hypotheses (claims) using sample data because it is usually

impossible/impractical to gain access to the entire population database.

If we have access to population data, obviously we have no need for inferential statistics to

test these claims because we can just look at the facts.

10-7

1. Note a claim someone has made regarding the population. Generate two mutually exclusive math hypotheses including the claim.

2. Collect evidence (sample data) pertinent to the claim. Make a decision as to which hypothesis is more likely.

3. Analyze the decision to assess the plausibility of the claim.

Steps in Hypothesis Testing

10-8

The null hypothesis, denoted H0, is a statement to be tested.

The null hypothesis is a statement which is assumed true until/unless evidence

(sample) indicates otherwise.

(Innocent until proven Guilty)

10-9

The alternative hypothesis, denoted H1, is a statement that contradicts the H0 .

We look at sample data (evidence) and decide to support/reject one of these

mutually exclusive hypotheses.We use that decision to make a comment about the plausibility of the claim we have

been given to evaluate.

10-10

“In Other Words”

The null hypothesis is a statement of status quo or no difference and always contains a

statement of equality. The null hypothesis is initially assumed to be

true until/unless we have evidence to the contrary.

We look to see if there is evidence that rejects the null and therefore supports the

alternative hypothesis.

10-11

“On Trial”

null hypothesis: man = innocent and is assumed true until/unless we find evidence to

the contrary. alternative hypothesis: man ≠ innocent

Evaluate evidence (sample data) and decide to reject or fail-to-reject the null.

The evidence will cause us to support or reject the initial claim of innocence.

in the courtroom…in the courtroom…

We are trying Karl for impersonating a Math We are trying Karl for impersonating a Math Professor. He looks guilty, but we must Professor. He looks guilty, but we must assume he is innocent anyway….assume he is innocent anyway….

We ask 100 of his students whether Karl We ask 100 of his students whether Karl knows anything about math. For a response. knows anything about math. For a response. we use a scale of 0 to 10, with 10 meaning we use a scale of 0 to 10, with 10 meaning he’s a math genius, and 0 meaning he’s a he’s a math genius, and 0 meaning he’s a math idiot (not a total idiot).math idiot (not a total idiot).

The mean of our sample turns out to be The mean of our sample turns out to be 1.0445 on a scale of 0 to 101.0445 on a scale of 0 to 10

in the courtroom…in the courtroom…

We conclude that it would be such a rare We conclude that it would be such a rare event (exceptionally small probability) for event (exceptionally small probability) for Karl to actually be a Math Professor and get Karl to actually be a Math Professor and get such a pathetically low sample score of such a pathetically low sample score of 1.0445, that our initial assumption of 1.0445, that our initial assumption of innocence was probably wrong, so we reject innocence was probably wrong, so we reject it…it…

(and we send him to jail where he belongs, (and we send him to jail where he belongs, for impersonating a Math Professor.)for impersonating a Math Professor.)

10-14

For each of the following claims, determine the null and alternative hypotheses. State whether the test is

two-tailed, left-tailed or right-tailed.In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher claims that this percentage is different today. H0: p=0.62According to a study published in 2006, the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher claims that the mean length of a call has increased since then. H0: μ = 3.25.Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager claims the standard deviation has decreased.

H1: σ < 0.23, a left-tailed test

Forming Hypotheses

Larson & Farber, Elementary Statistics: Picturing the World, 3e 15

Types of Errors

Actual Truth of H0

H0 is true H0 is false

Fail to reject H0

Reject H0

Correct Decision

Correct Decision

Type II Error

Type I Error

Decision

10-16

α = P(Type I Error) = P(rejecting H0 when H0 is true)

β = P(Type II Error) = P(not rejecting H0 when H1 is true)

10-17

For each of the following claims, explain what it would mean to make a Type I error. What would it mean to make a Type II error?

a)In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher claims that this percentage is different today.

b)According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher claims the mean length of a call has increased since then.

Type I and Type II Errors


Level of Significance

In a hypothesis test, the level of significance is your maximum allowable probability of making a type I error. It is denoted by , the Greek letter “Alpha”.

The probability of making a type II error is denoted by , the Greek letter “Beta”.

By setting the level of significance at a small value, you are saying that you want the probability of rejecting a true null hypothesis (Type IError) to be small.

Commonly used levels of significance: = 0.10 = 0.05 = 0.01

Hypothesis tests are based on .

10-19

The probability of making a Type I error, α, is chosen by the researcher before the sample data is collected.

The level of significance (LOS), α, is the probability of making a

Type I error.

As the probability of a Type I error increases, the probability of a Type II error decreases, and vice-versa.

10-20

We assume the null hypothesis is true.We look at the evidence (sample) and then decide to reject H0 or fail to reject

H0 ,the null hypothesis.

This is just like the court system where we assume the defendant is innocent, and then either find him “not guilty”

(FTR H0 ) or “guilty” (Reject H0 ).

10-21

According to a study published in 2006, the mean length of a phone call on a cell phone was

3.25 minutes. A researcher claims that the mean length of a call has increased since then.

a)Suppose the sample evidence indicates that the null hypothesis should be rejected (Reject H0 ). State the wording of the conclusion.

b)Suppose the sample evidence indicates that the null hypothesis should not be rejected (Fail to Reject H0 ). State the wording of the conclusion.

Stating the Conclusion


Statistical Tests

The statistic that is compared with the parameter in the null hypothesis is called the test statistic.

After stating the null and alternative hypotheses and specifying the level of significance, a random sample is taken from the population and sample statistics are calculated.

X2s22

zpt (n < 30)z (n 30)μ

Standardized test statistic

Test statistic

Population parameter

p̂

x

Hypothesis Testing:Hypothesis Testing:Quick review of two final steps:Quick review of two final steps:

6. 6. DecisionDecision: Reject/FTR: Reject/FTR

77. . InterpretInterpret Decision Decision::Suf / Insuf evidence to Suf / Insuf evidence to SupportSupport Claim Claim

(Claim is the )(Claim is the )Suf / Insuf evidence to Suf / Insuf evidence to RejectReject Claim Claim (Claim is the )(Claim is the )

0H

1H

0H


Interpreting a Decision

ClaimClaim is H0 Claim is H1

Do not reject H0

Reject H0There is enough evidence to reject the claim.

Decision

There is not enough evidence to reject the claim.

There is enough evidence to support the claim.

There is not enough evidence to support the claim.


Section

Hypothesis Tests for a Population

Proportion

( P )

10.2

10-26

A researcher assumes (based on her prior research) that the population proportion of people who are in favor of the

banning cell phone use while driving is 0.5

She obtains a random sample of 1000 people and finds that 534 are in favor of the banning cell phone use while

driving.

so = 534/1000 = 0.534

Would it be unusual to obtain a sample proportion of 0.534 or higher from a population whose proportion is

0.5?

What is convincing, or statistically significant, evidence?

ˆ p

10-27

When observed results are unlikely compared to null hypothesis (H0 ), we say the result is statistically

significant and we reject the null hypothesis (which we always

assume to be true at the start).

10-28

To determine if a sample proportion of 0.534 is statistically significant, we build a probability model.

1. npq = 100(0.5)(0.5) = 250 ≥ 10

2. Sample n = 1000 is < 5% population size

We can use the normal model to describe the distribution.

The mean of the distribution

and the standard deviation is

öp0.5

öp

0.5 1 0.5

10000.016.

10-29

Sampling distribution of the sample proportion

10-30

Recall that our simple random sample yielded a sample proportion of 0.534, so

Our sample was 2.15 std deviations above the hypothesized proportion of 0.5 which is an

“unlikely” result.

Therefore, we reject the null hypothesis (H0 ) that the population proportion was 0.50.

z öp öp

öp

0.534 0.5

0.5 1 0.5 1000

2.15

10-31

Why does it make sense to reject the null hypothesis if the sample proportion is more than 2 standard deviations

away from the hypothesized proportion?

The area under the standard normal curve to the right ofz = 2 is 0.0228, or only 2.3% of the time would you expect to get a sample of 0.532

10-32

If the sample proportion Z-score is too many standard deviations (generally 2 or more) from the

proportion stated in the null hypothesis H0 (assumed true), then we delete that assumption

based on evidence (our sample) and “reject the H0”

Hypothesis Testing

10-33

The best point estimate of p, the proportion of the population is given

by

where x is the number of “success” individuals in the sample and n is the sample size.

Testing Hypotheses Regarding a Population Proportion, p

ˆ p x

n

10-34

The sampling distribution of is approximately normal, with mean

and standard deviation

provided that the following requirements are satisfied:

1. The sample is a simple random sample.2. np(1-p) ≥ 10 and n < 5% N3. The sampled values are independent of

each other.

ˆ p

ˆ p p

ˆ p p(1 p)

n

10-35

A. Critical Value Test orClassical Approach

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

10-36


Step 2: Select a level of significance, α, based on the max allowable probability of making a

Type I error.

10-37


Step 3: Compute the standardized test statistic

Note: p0 refers to the value of p in the H0

hypothesis

z0 ˆ p p0

p0(1 p0)

n

10-38

Use TI-84 or Table V to determine the critical value.

Left-Tailed

(critical value)

10-39


Right-Tailed

(critical value)

10-40

Two-Tailed

(critical value)


10-41

Step 4: Compare the Critical Z Value (based on α) with the “Standardized Test Statistic or STS”

(which is the Z-score of your sample).

If the STS is outside (more extreme than) the Critical Z, then your sample falls in the Rejection Region and

your Decision is to Reject the null (H0 ).

If the STS is not in the Rejection Region, then your Decision is to Fail to Reject the null (H0 ).


10-42

The P-Value Approach has the same objective as the Critical Value Approach, which is to decide whether to Reject or Fail-To-Reject (FTR) the null H0 based on the

LOS ( α )of the problem

The BIG advantage is that the TI-84 does all the work of calculating the STS (below) and the std dev of the

sample.

B. P-Value Approach

z0 ˆ p p0

p0(1 p0)

n

10-43

If the P-value < α, the Z of sample is in the Rejection Region, so your Decision is :

Reject the null hypothesis H0.

If P ≥ α , then the Z of your sample is not in the Rejection Region, so your Decision is:

Fail-To-Reject the H0

Now, finally “Interpret the Decision” you just made by making a statement about the Claim…

B. P-Value Approach

10-44

Testing a Hypothesis about a Population Proportion: Large Sample Size

In 1997, 46% of Americans said they did not trust the media “when it comes to reporting the news fully,

accurately and fairly”.

In a 2007 poll of 1010 adults nationwide, 525 stated they did not trust the media.

At the α = 0.05 level of significance (LOS), is there evidence to support the claim that the percentage of Americans that do not trust the media has increased

since 1997?

10-45

We want to evaluate the claim that p > 0.46.

First, we must verify that we can perform the hypothesis test using z-scores. For the sampling

distribution of to be approximately normal, so we require npq be at least 10.

1. This is a simple random sample.

2. np0q0 = 1010(0.46)(0.54) = 251 > 10

3. The sample size is less than 5% of the population size, so the assumption of independence is met.

ˆ p

10-46

Step 1: H0: p ≤ 0.46 H1: p > 0.46 CLAIM

Step 2: LOS = α = 0.05

Step 3: The sample proportion is .

The standardized test statistic (STS) is:

ˆ p 525

10100.52

0

0.52 0.463.83

0.46(0.54)1010

z

10-47

A. Critical Value Approach

Step 4: This is a right-tailed test, and the critical value at the α = 0.05 level is:

z0.05 = 1.645 (See Table V)

Step 5: The STS is z0 = 3.83, which is

greater than the critical value 1.645.

Therefore, the sample lies in the Rejection Region, so we decide to:

Reject the null hypothesis.

10-48

B. P-Value Approach

Step 4: Since this is a right-tailed test, the P- value is the area to the right of (more extreme) the STS test statistic z0=3.83.

P-value = normcdf(3.83, 1E99) = 6.4E-5 ≈ 0.

Step 5: Since the P-value is less than α = 0.05,

we decide to: Reject the H0

10-49

“Interpret the Decision”

Step 6: Based on our sample, there is sufficient evidence at the α = 0.05 level of significance to support the claim that the percentage of Americans that do not trust the media has increased since 1997.

TI-84 TI-84 (Proportion)(Proportion)

STAT:TESTS:5:1-PropZTestSTAT:TESTS:5:1-PropZTestpo = p used in Hopo = p used in Hox = number of successes in the sample x = number of successes in the sample

(must be an Integer)(must be an Integer)n from your samplen from your sampleFor prop line: use the Ha symbolFor prop line: use the Ha symbol

DrawDraw: will show the shaded part of the : will show the shaded part of the curve and also show z and P-value.curve and also show z and P-value.

TI-84 TI-84 (Proportion)(Proportion)

STAT:TESTS:5:1-PropZTestSTAT:TESTS:5:1-PropZTestpo = 0.46po = 0.46x = 525 x = 525 (must be an Integer)(must be an Integer)n = 1010n = 1010For prop line: > po (Right-tailed Test)For prop line: > po (Right-tailed Test)

DrawDraw: shows z = 3.8133 and p=1E-4: shows z = 3.8133 and p=1E-4

Note: The TI-84 does Note: The TI-84 does notnot know LOS, so know LOS, so it does not know where the Reject it does not know where the Reject Region is.Region is.


Section


Mean

( μ )

10.3

10-53

To test hypotheses regarding the population mean assuming the population standard

deviation is unknown, we use the t-distribution rather than the Z-distribution. When we replace

σ with s,

follows Student’s t-distribution with n –1 degrees of freedom.

t0

x 0

s

n

10-54

1. The t-distribution is different for different degrees of freedom.

2. The t-distribution is centered at 0 and is symmetric about 0.

3. The area in the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution because using s

as an estimate of σ introduces more variability to the t-statistic.

Properties of the t-Distribution: Table VI

10-55

4. As the sample size n increases, the density curve of t gets closer to the standard normal z

density curve.

5. This result occurs because as the sample size increases, the values of “s” get closer to the values of “σ” by the Law of Large Numbers.

Properties of the t-Distribution

10-56

To test hypotheses regarding the population mean, we require that:

1.The sample is obtained using simple random sampling.

2.The sample has no outliers, and the population from which the sample is drawn is normally

distributed OR the sample size is large (n ≥ 30).3.The sampled values are independent of each

other.

Testing Hypotheses Regarding aPopulation Mean

10-57

Step 1: Determine the null and alternative hypotheses. They can be structured in

one of three ways:

10-58

Classical Approach

Step 2: Compute the standardized test statistic

which follows the Student’s t-distribution with n – 1 degrees of freedom.

Step 3. Use Table VI to determine the critical t value.

t0

x 0

s

n

10-59

P-Value Approach

Step 1: Compute the STS:

Use Table VI or the TI-84 to approximate the P-value.

t0

x 0

s

n

10-60

P-Value ApproachLeft-Tailed

10-61

P-Value ApproachRight-Tailed

10-62

P-Value ApproachTwo-Tailed

10-63

If the P-value < α, the “t” of sample is in the Rejection Region, so your Decision is :

Reject the null hypothesis H0.

If P ≥ α , then the “t” of your sample is not in the Rejection Region, so your Decision is:

Fail-To-Reject the H0

Now, finally “Interpret the Decision” you just made by making a statement about the Claim…

P-Value Approach

10-64

The procedure is robust, which means that minor departures from normality will not

adversely affect the results of the test. However, for small samples, if the data have outliers, or multiple modes, this procedure

should not be used because the distribution will not be approximately normal.

10-65

Testing a Hypothesis about a Population Mean, Large Sample

Assume the resting metabolic rate (RMR) of healthy males in complete silence is 5710. Researchers

measured the RMR of 45 healthy males who were listening to calm classical music and found their mean

RMR to be 5500 with a standard deviation of 992.

At the α = 0.05 level of significance, is there evidence to support the researcher’s claim that the mean RMR of males listening to calm classical music is not 5710?

10-66

Solution

Step 1: H0: μ = 5710 versus H1: μ ≠ 5710 Claim

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

Step 3: The sample mean is = 5500 and the sample standard deviation is s = 992. The STS is

0

5500 57101.4201

992 45t

x

10-67

Solution: Classical Approach

Step 4: Since this is a two-tailed test, the critical values at the α = 0.05 level with (n –1) = 44 to be

t0.025 = ± 2.021

Step 5: Since the test statistic, t0 = – 1.4201, is between

the critical values, (not in the Reject Region) we fail to reject the null hypothesis.

TI-84 TI-84 (n<30)(n<30)

STAT:TESTS:2:STAT:TESTS:2:T-TestT-Test

oo = 5710 ( Ho )= 5710 ( Ho )X-bar = 5500 n = 45X-bar = 5500 n = 45Sx = 992 from sampleSx = 992 from sampleFor For line: ≠ line: ≠ oo (H1 symbol)(H1 symbol)

DrawDraw: t = - 1.4201 p = 0.1626: t = - 1.4201 p = 0.1626

10-69

Solution: P-Value Approach

Step 5: Since the P-value is greater than the LOS

(0.1626 > 0. 05), our Decision is to Fail to Reject H0

Step 6: There is insufficient evidence at the α = 0.05 level of significance to support the claim that

the mean RMR of males listening to calm classical music is not 5710.

10-70

Testing a Hypothesis about a Population Mean,

Small Sample

According to the US Mint, standard quarters weigh 5.67 grams (28g/oz). A researcher suspects the new “US State” quarters have a weight that is heavier than the normal 5.67 grams. He randomly selects 18 “state” quarters, weighs them and obtains the following data.

5.70 5.67 5.73 5.61 5.70 5.67

5.65 5.62 5.73 5.65 5.79 5.73

5.77 5.71 5.70 5.76 5.73 5.72

At the α = 0.05 level of significance, is there evidence to support the claim that “State” quarters weigh more than 5.67 grams?

10-71

Solution

Step 1: H0: μ ≤ 5.67 versus H1: μ > 5.67 Claim


Step 3: From the data, the sample mean is calculated to be 5.7022 and the sample standard deviation is s = 0.0497. The STS statistic is

t0 5.7022 5.67

.0497 182.75

TI-84 TI-84 (n<30)(n<30)

STAT:TESTS:2:STAT:TESTS:2:T-TestT-Test

oo = 5.67 ( Ho )= 5.67 ( Ho )X-bar = 5.7022 n = 18X-bar = 5.7022 n = 18Sx = 0.0497 from sampleSx = 0.0497 from sampleFor For line: > line: > oo (H1 symbol)(H1 symbol)

DrawDraw: t = 2.7488 p = 0.0069: t = 2.7488 p = 0.0069

10-73

Solution: P-Value Approach

Step 5: Since the P-value is less than the level of significance (0.0069 < 0.05), our Decision is to

Reject the H0 .

Step 6: There is sufficient evidence at the α = 0.05 level of significance to support the claim that the mean weight of the “State” quarters is greater than 5.67 grams.

Note: t–crit = 1.740 and STS = 2.7488, so sample is in the Reject Region (Critical Value Method).


Section


Standard Deviation

( σ )

10.4

10-75

If a random sample of size “n” is obtained from a normally distributed population with mean “μ”

and standard deviation “σ”, then

where s2 is a sample variance has a chi-square distribution with n – 1 degrees of freedom.

Chi-Square Distribution

2 (n 1)s2

2

10-76

1. It is not symmetric.2. The shape of the chi-square

distribution depends on the degrees of freedom, just as with Student’s t-distribution.

3. As the number of degrees of freedom increases, the chi-square distribution becomes more nearly symmetric.

4. The values of χ2 are always nonnegative (greater than or equal to 0).

Characteristics of the Chi-Square Distribution

10-78

To test hypotheses about the population variance or standard

deviation, we can use the following steps, provided that:

The sample is obtained using simple

random sampling.The population is normally distributed.

Testing Hypotheses about a Population Variance or Standard Deviation

10-79

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

10-80

Compute the standardized test statistic

Use Table VII to determine the critical value using n – 1 degrees of freedom.

02

(n 1)s2

02

10-81

Classical Approach

Left -Tailed

10-82

Classical Approach

Right-Tailed

10-83

Classical Approach

Two-Tailed

10-84

Testing a Hypothesis about a Population Standard Deviation

A can of soda-pop states that the can contains 355 ml (11.5 oz) of soda. The can-filling machine has a

specification of ± 3.2 ml max standard deviation, but a quality control engineer suspects the machine is not calibrated correctly. She wants to verify the machine is not

under- or over-filling the cans, so she randomly selects 9 cans of the soda and measures the contents. She obtains the

following volume data:

351 360 358 356 359 358 355 361 352

Test her claim that the soda standard deviation, σ, is greater than 3.2 ml at the α = 0.05 level of significance.

10-85

Solution

Step 1: H0: σ = 3.2 H1: σ > 3.2 Claim

This is a right-tailed test.


Step 3: From the data, the sample standard deviation is computed to be s = 3.464. The standardized test statistic is

02

(9 1)(3.464)2

3.22 9.374

10-86

Solution: Classical Approach

Step 4: Since this is a right-tailed test, we determine the critical value at the α = 0.05 with 8 degrees of freedom to be χ2

0.05= 15.507.

Step 5: Since STS is less than the critical value, 15.507, so we Fail to Reject H0

Step 6: There is insufficient evidence at α = 0.05 LOS to support the claim that the standard deviation of the soda can contents is greater than 3.2 ml.

02 9.374


Section

Which Method Do I Use?

10.5


Section

The Probability of a Type II Error

and

the Power of the Test

10.6

10-90

The Probability of a Type II Error

Step 1: Determine the sample mean that separates the rejection region from the non-rejection region.

10-91

Earlier, we tested the hypothesis that the mean trade volume of Apple stock was greater than 35.14 million shares, H0: μ = 35.14 versus H1: μ > 35.14, based upon a random sample of size n = 40 with the population standard deviation, σ, assumed to be 15.07 million shares at the α = 0.1 level of significance. Compute the probability of a type II error given that the population mean is μ = 40.62.

Computing the Probability of a Type II Error

10-92

Step 1: Since z0.9=1.28, we let z = 1.28, μ0 = 35.14, σ = 15.07, and n = 40

find the sample mean that separates the rejection region from the nonrejection region:

For any sample mean less than 38.19, we do not reject the null hypothesis.

Solution

1.28 x 35.14

15.07

40

x 38.19

10-93

Solution

10-94

Solution

10-95

Step 3: P(Type II error) = β = P(do not reject H0 given H1 is

true) = P( < 38.19 given that μ

=40.62)

=

β = 0.1539

Solution

x

P Z 38.19 40.62

15.07

40

P(Z 1.02)

10-96

Power of the Test

The probability of rejecting the null hypothesis when the alternative

hypothesis is true is (1 – β) which is referred to as the power of the test. The higher the power of the test, the more likely the test will reject the null

when the alternative hypothesis is true.

10-97

Recall that β = 0.1539.The power of the test is

1 – β = 1 – 0.1539 = 0.8461.

There is a 84.61% chance of rejecting the null hypothesis when the true

population mean is 40.62.

Chap 2 98

Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Chapter Hypothesis Test of a Claim...

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