CH.9 Tests of Hypotheses for a Single Sample · Test of a Hypothesis •Hypotheses are statements...

1

CH.9 Tests of Hypotheses for a Single Sample

• Hypotheses testing• Tests on the mean of a normal distribution-

variance known• Tests on the mean of a normal distribution-

variance unknown• Tests on the variance and standard deviation of a

normal distribution• Tests on a population proportion• Testing for goodness of fit

2

9-1 Hypothesis Testing9-1.1 Statistical Hypotheses

Statistical hypothesis testing and confidence interval estimation of parameters are the fundamental methods used at the data analysisstage of a comparative experiment, in which the engineer is interested, for example, in comparing the mean of a population to a specified value.

Statistical inference may be divided into two major areas:• Parameter estimation• Hypothesis testing

3


For example, suppose that we are interested in the burning rate of a solid propellant used to power aircrew escape systems.

• Now burning rate is a random variable that can be described by a probability distribution.

• Suppose that our interest focuses on the mean burning rate (a parameter of this distribution).

• Specifically, we are interested in deciding whether or not the mean burning rate is 50 centimeters per second.

4


null hypothesis

alternative hypothesis

One-sided Alternative Hypotheses

Two-sided Alternative Hypothesis

5


Test of a Hypothesis•Hypotheses are statements about population or distribution under study.

•A procedure leading to a decision about a particular hypothesis

• Hypothesis-testing procedures rely on using the information in a random sample from the population of interest.

• If this information is consistent with the hypothesis, then we will conclude that the hypothesis is true (fail to reject hypothesis); if this information is inconsistent with the hypothesis, we will conclude that the hypothesis is false (reject hypothesis).

6

9-1 Hypothesis Testing9-1.2 Tests of Statistical Hypotheses

Decision criteria for testing H0:μ = 50 cm/s versus H1:μ ≠ 50 cm/s.

7


Decision criteria for testing H0:μ = 50 cm/s versus H1:μ ≠ 50 cm/s.

Acceptance RegionCritical Region Critical Region

critical values

8

9-1 Hypothesis Testing9-1.2 Tests of Statistical HypothesesTwo wrong conclusions are possible:

Type I Error

Type II Error

9


Sometimes the type I error probability is called the significance level, or the α-error, or the size of the test.

10

9-1 Hypothesis Testing

2.5 0.7910X n

σσ = = =

9-1.2 Tests of Statistical Hypotheses

Ex: σ =2.5 n=10 Standard deviation of the sample mean:

xμ

Xσ

11


X

Z1.90-1.90 0

Probability distribution of

Correspondingprobability distribution of Z

Critical Region for 0 1: 50 : 50H Hμ μ= ≠

12


How can we reduce α ?- by widening acceptance region

(if take critical values 48 and 52, α =0.0114, Verify !)- by increasing sample size

(if take n=16, α =0.0164 Verify !)

13


The probability of type II error when μ = 52 and n = 10.

14


15


The probability of type II error when μ = 50.5 and n = 10.

16


17


The probability of type II error when μ = 52 and n = 16.

18


19


• Generally α is controlable when critical values are selected.• Thus, rejection of null hypothesis H0 is a strong conclusion.• β is not constant but depends on the true value of the parameter and

sample size.• Accepting H0 is a weak conclusion unless β is acceptably small.• Prefer the terminology “fail to reject H0” rather than “accept H0”• Fail to reject H0

– implies we have not found sufficient evidence to reject H0.– does not necessarily mean there is a high probability that H0 is true.– means more data are required to reach a strong conclusion.

20


Power

• The power is computed as 1 - β, and power can be interpreted as the probability of correctly rejecting a false null hypothesis. We often compare statistical tests by comparing their power properties.

• For example, consider the propellant burning rate problem whenwe are testing H 0 : μ = 50 cm/s against H 1 : μ not equal 50 cm/s . Suppose that the true value of the mean is μ = 52. When n = 10, we found that β = 0.2643, so the power of this test is 1 - β = 1 - 0.2643 = 0.7357 when μ = 52.

21


9-1.3 One-Sided and Two-Sided HypothesesTwo-Sided Test:

One-Sided Tests:

22


Example 9-1

• Suppose if the propellant burning rate is less than 50 cm/s• Want to show this with a strong conclusion CLAIM• Hypotheses should be stated as

H0: μ = 50 cm/sH1: μ < 50 cm/s

• Since the rejection of H0 is always a strong conclusion, thisstatement of the hypotheses will produce the desired outcome ifH0 is rejected.

• Although H0 is stated with an equal sign, it is understood toinclude any value of μ not specified by H1.

• Failing to reject H0 does not mean μ = 50 cm/s exactly.• Failing to reject H0 means we do not have strong evidence in

support of H1.

23


The bottler wants to be sure that the bottles meet the specification on mean internal pressure or bursting strength, which for 10-ounce bottles is a minimum strength of 200 psi.

The bottler has decided to formulate the decision procedure for a specific lot of bottles as a hypothesis testing problem.

There are two possible formulations for this problem: either

or

Which is correct? Depends on the objective of the analysis.

24

9-1 Hypothesis Testing9-1.4 P-Values in Hypothesis Tests•When H0 is rejected at a specified α level, this gives no idea aboutwhether the computed value of the test statistic

• is just barely in the rejection region

• or it is very far into this region.

•Thus, P-value has been adopted widely in practice

25

9-1 Hypothesis Testing9-1.4 P-Values in Hypothesis TestsConsider the two-sided hypothesis test for burning rate:

H0 : μ = 50 cm/sH1 : μ ≠ 50 cm/s

n=16, σ=2.5,

P-value?

x 51.3=

26


x 51.3=

9-1.5 Connection between Hypothesis Tests and Confidence IntervalsClose relation between hypothesis tests and confidence intervals

H0 : μ = 50 cm/s n=16, σ=2.5, α=0.05, H1 : μ ≠ 50 cm/s

Critical z values are zα/2=z0.025=1.96 and –z0.025=-1.96 which corresponds to

Critical values = [48.775 ; 51.225]2.550 1.9616

±

x 51.3= is not in the acceptance region [48.775 ; 51.225]. So reject null hypothesis.

Confidence interval for μ at α=0.05 is

That is

2.551.3 1.9616

±

50.075 52.525≤ μ ≤

μ=50 is not in the confidence interval [50.075 ; 52.525]. So reject null hypothesis.

same conclusion !

27

9-1 Hypothesis Testing9-1.5 Connection between Hypothesis Tests and Confidence Intervals

28

1. From the problem context, identify the parameter of interest.2. State the null hypothesis, H0 .3. Specify an appropriate alternative hypothesis, H1 .4. Choose a significance level, α.5. Determine an appropriate test statistic.6. State the rejection region for the statistic.7. Compute any necessary sample quantities, substitute these

into the equation for the test statistic, and compute that value.8. Decide whether or not H0 should be rejected and report that in

the problem context.


9-1.6 General Procedure for Hypothesis Tests

29

9-2 Tests on the Mean of a Normal Distribution, Variance Known

9-2.1 Hypothesis Tests on the Mean

We wish to test:

The test statistic is:

30



Reject H0 if the observed value of the test statistic z0 is either:

z0 > zα/2 or z0 < -zα/2

Fail to reject H0 if -zα/2 < z0 < zα/2

31


Example 9-2

32


Example 9-2

33


Example 9-2

34



35


9-2.1 Hypothesis Tests on the Mean (Continued)

36


37


9-2.1 Hypothesis Tests on the Mean (Continued)

38


P-Values in Hypothesis Tests

39


9-2.2 Type II Error and Choice of Sample SizeFinding the Probability of Type II Error β

40



The distribution of Z0 under H0 and H1

41



0 if δ>0

Let zβ be the 100β upper percentile of the standard normal distr. β = Φ(-zβ)

/ 2nz zβ α

δσ

− −

42


9-2.2 Type II Error and Choice of Sample SizeSample Size Formulas

For a two-sided alternative hypothesis:

43


9-2.2 Type II Error and Choice of Sample SizeSample Size Formulas

For a one-sided alternative hypothesis:

44


Example 9-3

504949.216 50.784

Under H1: µ≠50 Under H0: µ=50H0: µ=50H1: µ≠50σ=2α=0.05n=25If true µ=49, β=?

z0.025 =1.96

Critical points are:50 ± z0.025σ/√n

= 50 ± 1.96*2/549.216 and 50.784

45


Example 9-3

5049 49.216 50.784

Under H1: µ≠50 Under H0: µ=50

H0: µ=50H1: µ≠50σ=2α=0.05n=25If true µ=49, β=?

( )

(49.216 50.784 when =49)49.216 49 50.784 49

2 / 25 2 / 250.54 4.46 0.295

P X

P Z

P Z

β μ

β

β

= ≤ ≤

− −⎛ ⎞= ≤ ≤⎜ ⎟⎝ ⎠

= ≤ ≤ =0.295 0.54z zβ = =

46


Example 9-3

N(-2.5,1) N(0,1)

0-2.5 -1.96 1.96

Under H1: µ≠50 Under H0: µ=50

Standardize thenormal graphs

( )

( 1.96 1.96 when = -2.5)1.96 2.5 1.96 2.5

1 10.54 4.46 0.295

P X

P Z

P Z

β μ

β

β

= − ≤ ≤

− + +⎛ ⎞= ≤ ≤⎜ ⎟⎝ ⎠

= ≤ ≤ =

With standard normal graph:

1* 25 1* 251.96 1.962 2

(4.46) (0.54) 0.295

β

β

⎛ ⎞ ⎛ ⎞− −=Φ − −Φ − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠=Φ −Φ =

With the formula:

1

1* 25 2.52

n

δ

δσ

= −

−= = −

47


Example 9-3

48


9-2.2 Type II Error and Choice of Sample SizeUsing Operating Characteristic Curves

I I

49


Example 9-4

50


9-2.3 Large Sample Test

If the distribution of the population is not known, but n>40sample standard deviation “s” can be substituted for “σ” andtest procedures in Section 9.2 are valid.

51

9-3 Tests on the Mean of a Normal Distribution, Variance Unknown

9-3.1 Hypothesis Tests on the MeanOne-Sample t-Test

52



The reference distribution for H0: μ = μ0 with critical region for (a) H1: μ ≠ μ0 , (b) H1: μ > μ0, and (c) H1: μ < μ0.

53


Example 9-6

54


Example 9-6

55


Example 9-6

Normal probability plot of the coefficient of restitution data from Example 9-6.

56


Example 9-6

57


9-3.2 P-value for a t-TestThe P-value for a t-test is just the smallest level of significance at which the null hypothesis would be rejected.

t0 = 2.72, this is between two tabulated values, 2.624 and 2.977. Therefore, the P-value must be between 0.01 and 0.005.

0.005<P-value<0.01.

Suppose t0 = 2.72 for a two-sided test, then

0.005*2<P-value<0.01*2 0.01<P-value<0.02

58


9-3.3 Type II Error and Choice of Sample Size

The type II error of the two-sided alternative would be

Where denotes the noncentral t random variable. 0T ′

59


Example 9-7

60

9-4 Hypothesis Tests on the Variance and Standard Deviation of a Normal Distribution9-4.1 Hypothesis Test on the Variance

61

9-4 Hypothesis Tests on the Variance and Standard Deviation of a Normal Distribution9-4.1 Hypothesis Test on the Variance

The same test statistic is used for one-sided alternative hypotheses. For the one-sided hypothesis

62

9-4 Hypothesis Tests on the Variance and Standard Deviation of a Normal Distribution

9-4.1 Hypothesis Test on the Variance

63


Example 9-8

64


Example 9-8

65



For the two-sided alternative hypothesis:

Operating characteristic curves are provided in Charts VIIi and VIIj.

66


Example 9-9

67

9-5 Tests on a Population Proportion

9-5.1 Large-Sample Tests on a Proportion

Many engineering decision problems include hypothesis testing about p.

An appropriate test statistic is

68


Example 9-10

69


Example 9-10

70


Another form of the test statistic Z0 is obtained bydividing the numerator and denominator by n

or

71

9-5 Tests on a Population Proportion9-5.2 Type II Error and Choice of Sample Size

For a two-sided alternative

If the alternative is p < p0

If the alternative is p > p0where p is true value of thepopulation proportion

72


1

(1 )H

p pn

σ −= 0

0 0(1 )H

p pn

σ −=

0 01 2 2 2o H o HC p z C p zα ασ σ= − = +

9-5.2 Type II Error and Choice of Sample SizeLess complex representation:

Critical values (C1 and C2 ) are:

1 1

1 1

1 1

1 1

1 2

1 2

2 1

ˆ( )

( )H H

H H

H H

H H

P C P CC C

P Z

C C

βμ μ

σ σ

μ μσ σ

= < <− −

= < <

⎛ ⎞ ⎛ ⎞− −= Φ −Φ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Type II error is:

73



For a two-sided alternative

For a one-sided alternative

74

9-5 Tests on a Population ProportionExample 9-11

01 0.05 1.645(0.0154) 0.0247o HC p zασ= − = − =

Critical value (C1 ) is:

1

0.03(0.97) 0.012200Hσ = = 0

0.05(0.95) 0.0154200Hσ = =

( )

1ˆ( )

0.0247 0.03( )0.012

1 0.44 0.67

P C P

P Z

β = <−

= <

= −Φ − =

75


Example 9-11

76

9-7 Testing for Goodness of Fit • The test is based on the chi-square distribution.

• Assume there is a sample of size n from a population whose probability distribution is unknown.

• Let Oi be the observed frequency in the ith class interval.

• Let Ei be the expected frequency in the ith class interval.

The test statistic

has approximately chi-square distribution with k-p-1 degrees of freedom.

p: number of parameters of the hypothesized distribution estimated bysample statistics.

77

Example 9-12

9-7 Testing for Goodness of Fit

78


Example 9-12

79


Example 9-12

80


Example 9-12

81


Example 9-12

82


Example 9-12

83

9-7 Testing for Goodness of Fit Example 9-13

84


X=5.04+Z*0.08

Z X

-1.15 4.95

-0.68 4.99

-0.32 5.01

0 5.04

0.32 5.07

0.68 5.09

1.15 5.13

85


86


Date post:	23-Aug-2020
Category:	Documents
Upload:	others
View:	3 times
Download:	0 times

CH.9 Tests of Hypotheses for a Single Sample · Test of a Hypothesis •Hypotheses are statements...

Documents