Inference on the Mean of a Population - Variance Known

Horng-Chyi HorngHorng-Chyi Horng Statistics IIStatistics II 11

Inference on the Mean of a PopulationInference on the Mean of a Population--Variance KnownVariance Known

HH00: : = = 00

HH11: : 00 , where , where 00 is a specified constant. is a specified constant.

Sample mean is the unbiased point estimator for population mean.Sample mean is the unbiased point estimator for population mean.

&4-4 (&8-2)

then),( trueis H if Therefore,

.,~ then , varianceand mean with

ondistributi a fromdrawn samples are ,,, If

The ReasoningThe Reasoning

For HFor H00 to be true, the value of Z to be true, the value of Z00 can not be too large or too can not be too large or too

small.small.

Recall that 68.3% of ZRecall that 68.3% of Z00 should fall within (-1, +1) should fall within (-1, +1)

95.4% of Z95.4% of Z00 should fall within (-2, +2) should fall within (-2, +2)

99.7% of Z99.7% of Z00 should fall within (-3, +3) should fall within (-3, +3)

What values of ZWhat values of Z00 should we reject H should we reject H00? (based on ? (based on value) value)

What values of ZWhat values of Z00 should we conclude that there is not should we conclude that there is not

enough evidence to reject Henough evidence to reject H00??

Example 8-2Example 8-2

Aircrew escape systems are powered by a solid propellant. The burning rate of this propellant is an important product characteristic. Specifications require that the mean burning rate must be 50 cm/s. We know that the standard deviation of burning rate is 2 cm/s. The experimenter decides to specify a type I error probability or significance level of α = 0.05. He selects a random sample of n = 25 and obtains a sample average of the burning rate of x = 51.3 cm/s. What conclusions should be drawn?

Hypothesis Testing on Hypothesis Testing on - Variance Known- Variance Known

H1 Test Statistic Reject H0 if 0 Z0 > z or Z0 < - z

> 0 Z0 > z

Z0 < -z

P-Values in Hypothesis Tests(I)P-Values in Hypothesis Tests(I)

Where ZWhere Z00 is the test statistic, and is the test statistic, and (z) is the standard (z) is the standard

normal cumulative function.normal cumulative function. In example 8-2, ZIn example 8-2, Z00 = 3.25, P-Value = 2[1- = 3.25, P-Value = 2[1-(3.25)] = (3.25)] =

0.00120.0012

P-Values of Hypothesis Testing on P-Values of Hypothesis Testing on - Variance Known- Variance Known

H1 Test Statistic P-Value

0 P = 2 * P(Z |Z0|)

> 0 P = P(Z Z0 )

P = P(Z Z0 )

P-Values in Hypothesis Tests(II)P-Values in Hypothesis Tests(II)

-value is the maximum type I error allowed, while P--value is the maximum type I error allowed, while P-value is the real type I error calculated from the sample.value is the real type I error calculated from the sample.

-value is preset, while P-value is calculated from the -value is preset, while P-value is calculated from the sample.sample.

When P-value is less than When P-value is less than -value, we can safely make the -value, we can safely make the conclusion “Reject Hconclusion “Reject H00”. By doing so, the error we are ”. By doing so, the error we are

subjected to (P-value) is less than the maximum error subjected to (P-value) is less than the maximum error allowed (allowed (-value). -value).

Type II ErrorType II Error- Fail to reject H- Fail to reject H00 while H while H00 is false is false

How to calculate Type II Error? (I)How to calculate Type II Error? (I)(H(H00: : = = 00 Vs. H Vs. H11: : 00))

Under the circumstance of type II error, HUnder the circumstance of type II error, H00 is false. Supposed is false. Supposed

that the true value of the mean is that the true value of the mean is = = 00 + + , where , where > 0. The > 0. The

distribution of Zdistribution of Z00 is: is:

1 ,~ Therefore,

How to calculate Type II Error? (II)How to calculate Type II Error? (II) - refer to section &4.3 (&8.1) - refer to section &4.3 (&8.1)

Type II error occurred when (fail to reject HType II error occurred when (fail to reject H00 while H while H00 is is

false) false)

~ where 02

Therefore, Therefore,

The Sample Size (I)The Sample Size (I)

Given values of Given values of and and , find the required sample size n to , find the required sample size n to achieve a particular level of achieve a particular level of ....

re whe

/ Then,

0 when /

// Since

The Sample Size (II)The Sample Size (II)

Two-sided Hypothesis TestingTwo-sided Hypothesis Testing

One-sided Hypothesis TestingOne-sided Hypothesis Testing

The Operating Characteristic CurvesThe Operating Characteristic Curves- Normal test (z-test)- Normal test (z-test)

Use to performing sample size or type II error Use to performing sample size or type II error calculations.calculations.

The parameter d is defined as:The parameter d is defined as:

so that it can be used for all problems regardless of the so that it can be used for all problems regardless of the values of values of 00 and and ..

Chart VI a,b,c,d are for Z-test.Chart VI a,b,c,d are for Z-test.

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Large Sample TestLarge Sample Test

If n If n 30, then the sample variance s 30, then the sample variance s22 will be close to will be close to 22 for most samples.for most samples.

Therefore, if population variance Therefore, if population variance 22 is unknown but n is unknown but n 30, we can substitute 30, we can substitute with s in the test procedure with with s in the test procedure with little harmful effect.little harmful effect.

Large Sample Hypothesis Testing on Large Sample Hypothesis Testing on - Variance Unknown but n - Variance Unknown but n 30 30

H1 Test Statistic Reject H0 if 0 Z0 > z or Z0 < - z

> 0 Z0 > z

Z0 < -z

Statistical Vs. Practical SignificanceStatistical Vs. Practical Significance

Practical Significance = 50.5-50 = 0.5Practical Significance = 50.5-50 = 0.5 Statistical Significance P-Value for each sample size n.Statistical Significance P-Value for each sample size n.

NotesNotes

be careful when interpreting the results from hypothesis tebe careful when interpreting the results from hypothesis testing when the sample size is large, sting when the sample size is large, because any small depbecause any small departure from the hypothesized value arture from the hypothesized value 00 will probably be det will probably be det

ected, even when the difference is of little or no practical siected, even when the difference is of little or no practical significance.gnificance.

In general, two types of conclusion can be drawn:In general, two types of conclusion can be drawn:

1. At 1. At = 0.**, we have enough evidence to reject H= 0.**, we have enough evidence to reject H00..

2. At 2. At = 0.**, we do not have enough evidence to reject = 0.**, we do not have enough evidence to reject HH00..

Confidence Interval on the Mean (I)Confidence Interval on the Mean (I)

Point Vs. Interval EstimationPoint Vs. Interval Estimation

The general form of interval estimate is The general form of interval estimate is

L L U U

in which we always attach a possible error in which we always attach a possible error such that such that

P(L P(L U) = 1- U) = 1-That is, we have 1-That is, we have 1- confidence that the true value of confidence that the true value of will fall within [L, U].will fall within [L, U].

Interval Estimate is also called Confidence Interval (C.I.).Interval Estimate is also called Confidence Interval (C.I.).

Confidence Interval on the Mean (II)Confidence Interval on the Mean (II)

L is called the lower-confidence limit andL is called the lower-confidence limit and

U is the upper-confidence limit.U is the upper-confidence limit.

Two-sided C.I. Vs. One-sided C.I.Two-sided C.I. Vs. One-sided C.I.

Construction of the C.I.Construction of the C.I.

From Central Limit Theory, From Central Limit Theory,

. ,~ ,25 and ,~ 22

nNXnXIf Use standardization and the properties of Z, Use standardization and the properties of Z,

nzXnzXP

Formula for C.I. on the Mean with Variance KnownFormula for C.I. on the Mean with Variance Known

Used whenUsed when

1. Variance known1. Variance known

2. n 2. n 30, use s to estimate 30, use s to estimate ..

Example 8-6Example 8-6 Consider the rocket propellant problem in Example 8-2. Consider the rocket propellant problem in Example 8-2. Find a 95% C.I. on the mean burning rate?Find a 95% C.I. on the mean burning rate?

95% C.I => 95% C.I => = 0.05, = 0.05,

zz/2 /2 = z= z0.0250.025 = 1.96 = 1.96

Notes - C.I. Notes - C.I.

Relationship between Hypothesis Testing and C.I.sRelationship between Hypothesis Testing and C.I.s

Confidence level (1-Confidence level (1-) and precision of estimation (C.I. * ) and precision of estimation (C.I. * 1/2)1/2)

Sample size and C.I.sSample size and C.I.s

Choice of Sample Size to Achieve Precision of EstimationChoice of Sample Size to Achieve Precision of Estimation

One-Sided C.I.s on the MeanOne-Sided C.I.s on the Mean

Inference on the Mean of a PopulationInference on the Mean of a Population--Variance UnknownVariance Unknown

HH00: : = = 00

HH11: : 00 , where , where 00 is a specified constant. is a specified constant.

Variance unknown, therefore, use s instead of Variance unknown, therefore, use s instead of in the test in the test statistic.statistic.

&4-5 (&8-3)

If n is large enough (If n is large enough ( 30), we can use the test procedure 30), we can use the test procedure in &4-4 (&8-2). However, n is usually small. In this case, in &4-4 (&8-2). However, n is usually small. In this case, T0 will not follow the standard normal distribution.T0 will not follow the standard normal distribution.

Inference on the Mean of a PopulationInference on the Mean of a Population--Variance UnknownVariance Unknown

Let XLet X11, X, X22, …, X, …, Xnn be a random sample for a normal distribu be a random sample for a normal distribu

tion with unknown mean tion with unknown mean and unknown variance and unknown variance 22. The . The quantityquantity

has a t distribution with n - 1 degrees of freedom.has a t distribution with n - 1 degrees of freedom.

freedom. of degrees ofnumber theis

:ondistributi t of pdf

The ReasoningThe Reasoning

For HFor H00 to be true, the value of T to be true, the value of T00 can not be too large or too can not be too large or too

small.small.

What values of TWhat values of T00 should we reject H should we reject H00? (based on ? (based on value) value)

What values of TWhat values of T00 should we conclude that there is not should we conclude that there is not

enough evidence to reject Henough evidence to reject H00??

Although when n Although when n 30, we can use Z 30, we can use Z00 in section &8-2 to in section &8-2 to

perform the testing instead. We prefer using Tperform the testing instead. We prefer using T00 to more to more

accurately reflect the real testing result if t-table is available.accurately reflect the real testing result if t-table is available.

Testing for Normality (Example 8-8)Testing for Normality (Example 8-8)- t-test assumes that the data are a random sample from a - t-test assumes that the data are a random sample from a normal populationnormal population

(1) Box Plot(1) Box Plot (2) Normality Probability Plot(2) Normality Probability Plot

Hypothesis Testing on Hypothesis Testing on - Variance Unknown- Variance Unknown

H1 Test Statistic Reject H0 if

0 T0 > t or T0 < - t

> 0 T0 > t

T0 < -t

Finding P-ValuesFinding P-Values

Steps:Steps:

1. Find the degrees of freedom (k = n-1)in the t-table.1. Find the degrees of freedom (k = n-1)in the t-table.

2. Compare T2. Compare T00 to the values in that row and find the to the values in that row and find the

closest one.closest one.

3. Look the 3. Look the value associated with the one you pick. The value associated with the one you pick. The p-value of your test is equal to this p-value of your test is equal to this value. value.

In example 8-8, TIn example 8-8, T00 = 4.90, k = n-1 = 21, P-Value < 0.0005 = 4.90, k = n-1 = 21, P-Value < 0.0005

because the t value associated with (k = 21, because the t value associated with (k = 21, = 0.0005) is = 0.0005) is 3.819.3.819.

P-Values of Hypothesis Testing on P-Values of Hypothesis Testing on - Variance Unknown- Variance Unknown

H1 Test Statistic P-Value

0 P = 2 * P(tn-1 |T0|)

> 0 P = P(tn-1 T0 )

P = P(tn-1 T0 )

The Operating Characteristic CurvesThe Operating Characteristic Curves- t-test- t-test

Use to performing sample size or type II error calculations.Use to performing sample size or type II error calculations. The parameter d is defined as:The parameter d is defined as:

so that it can be used for all problems regardless of the valso that it can be used for all problems regardless of the values of ues of 00 and and ..

Chart VI e,f,g,h are used in t-test. (pp. A14-A15)Chart VI e,f,g,h are used in t-test. (pp. A14-A15)

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In example 8-8, if the mean load at failure differs from 10 In example 8-8, if the mean load at failure differs from 10 MPa by as much as 1 MPa, is the sample size n = 22 adeqMPa by as much as 1 MPa, is the sample size n = 22 adequate to ensure that Huate to ensure that H00 will be rejected with probability at l will be rejected with probability at l

east 0.8?east 0.8?

s = 3.55, therefore, d = 1.0/3.55 = 0.28.s = 3.55, therefore, d = 1.0/3.55 = 0.28.

Appendix Chart VI g, for d = 0.28, n = 22 => Appendix Chart VI g, for d = 0.28, n = 22 => = 0.68 = 0.68

The probability of rejecting HThe probability of rejecting H00: : = 10 if the true mean exceeds this by 1.0 MP = 10 if the true mean exceeds this by 1.0 MP

a (reject Ha (reject H00 while H while H00 is false) is approximately 1 - is false) is approximately 1 - = 0.32, which is too smal = 0.32, which is too smal

l. Therefore n = 22 is not enough. l. Therefore n = 22 is not enough.

At the same chart, d = 0.28, At the same chart, d = 0.28, = 0.2 (1- = 0.2 (1-=0.8) => n = 75=0.8) => n = 75

Construction of the C.I. on the MeanConstruction of the C.I. on the Mean - Variance Unknown - Variance Unknown

In general, the distribution of In general, the distribution of

is t with n-1 d.f. is t with n-1 d.f.

Use the properties of t with n-1 d.f., Use the properties of t with n-1 d.f.,

1,2/1,2/

nstXnstXP

Formula for C.I. on the Mean with Variance UnknownFormula for C.I. on the Mean with Variance Unknown

Example 8-10Example 8-10 Reconsider the tensile adhesive problem in Example 8-8. Reconsider the tensile adhesive problem in Example 8-8. Find a 95% C.I. on the mean?Find a 95% C.I. on the mean?

N = 22, sample mean = 13.71, s = 3.55, tN = 22, sample mean = 13.71, s = 3.55, t/2,n-1 /2,n-1 = t= t0.025,21 0.025,21 = 2.= 2.

080080

13.71 - 2.080 (3.55) / 13.71 - 2.080 (3.55) / 22 22 13.71 + 2.080 (3.55) / 13.71 + 2.080 (3.55) / 2222

13.71 - 1.57 13.71 - 1.57 13.71 + 1.57 13.71 + 1.57

12.14 12.14 15.28 15.28

The 95% C.I. On the mean is [12.14, 15.28] The 95% C.I. On the mean is [12.14, 15.28]

nstXnstX nn // 1,2/1,2/

Final Note for the Inference on the MeanFinal Note for the Inference on the Mean

Variance Sample Size Population Type Testing Method

N > 30 All Z

Normal Z

N < 30

Non-normal Chebyshev’s

Inequality

N > 30 All T or Z

Normal T

Unknown

N < 30

Non-normal Non-parametric

Inference on the Mean of a Population - Variance Known

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