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Chapter 17

Inferences:

Continuous Response

Introduction

• This chapter covers random sampling evaluations from

a population that has a continuous response. • An example of a continuous response is the amount of

tire tread that exists after 40,000 km of automobile

usage. One tire might, for example, have 6.0 mm of

remaining tread while another tire might measure 5.5

mm.

• In this chapter the estimation of population mean and

standard deviation from sampled data is discussed in

conjunction with probability plotting.

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17.1 Summarizing Sample Data

• The classical analysis of sampled data taken from a

continuous response population has focused on determining

a sample mean (𝑥 ) and standard deviation (𝑠), along with

perhaps confidence interval statements.

• Other information with probabilities may be more meaningful.

• For example, an experiment might be able to indicate 90% of

the automobiles using certain type of tire will have at least 4.9

mm of tire tread after 40,000 km. Such a statement can be

more informative than a statement that only relates to the

mean tire tread after 40,000 km

17.2 Sample Size:

Hypothesis Test of a Mean

• Given the producer’s risk 𝛼, consumer’s risk 𝛽, and the

acceptable amount of uncertainty 𝛿, Diamond’s (1989)

equation can be used to determine the sample size (n)

necessary to evaluate a hypothesis test criterion:

𝑛 = (𝑈𝛼 + 𝑈𝛽)2𝜎2

𝛿2

• 𝑈𝛽 is determined from single sided Table B in Appendix E.

• If the alternative hypothesis is single-sided (𝜇 < criterion), 𝑈𝛼 is also determined from Table B, but if the alternative

hypothesis is double-sided (𝜇 <> criterion), 𝑈𝛼 is determined

from Table C.

• If 𝜎 is unknown, 𝛿 can be expressed in terms of 𝜎

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17.2 Sample Size:

Hypothesis Test of a Mean

• If the standard deviation is not known, the sample size

should be adjusted to using (Diamond 1989):

𝑛 = (𝑡𝛼 + 𝑡𝛽)2𝑠2

𝛿2

• 𝑡𝛽 is determined from single sided Table D in Appendix E.

• If the alternative hypothesis is single sided (μ < criterion),

𝑡𝛼 is also determined from Table D, but if the alternative

hypothesis is double sided (μ <> criterion), 𝑡𝛼 is determined

from Table E.

• A stereo amplifier output level is desired to be on the average

at least 100 watts (W) per channel. Determine the sample

size that is needed to verify this criterion given the following:

𝛼 = 0.1, which from Table B yield 𝑈𝛼 = 1.282

𝛽 = 0.05, which from Table B yield 𝑈𝛽 = 1.645 𝛿 = 0.5𝜎

𝑛 = (𝑈𝛼 +𝑈𝛽)2𝜎2

𝛿2= (1.282 + 1.645)2

𝜎2

0.5𝜎 2= 34.26~35

17.3 Example 17.1

Sample Size Determination for a Mean

Criterion Test

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17.3 Example 17.1

Sample Size Determination for a Mean

Criterion Test • If the standard deviation is not known, this sample need to

be adjusted. Given the number degrees of freedom for the

t-table value equals 34 (i.e. 35 – 1) interpolation in Table D

yields to t0.1;34 = 1.307 and t0.05;34 = 1.692

𝑛 = (𝑡𝛼 + 𝑡𝛽)2𝑠2

𝛿2= (1.307 + 1.692)2

𝑠2

0.5𝑠 2= 35.95~36

17.4 Confidence Intervals on the Mean

and Hypothesis Test Criteria Alternatives:

Table 17.1

Single-Sided Double-Sided

σ Known

𝜇 ≤ 𝑥 +𝑈𝛼𝜎

𝑛 𝑜𝑟

𝜇 ≥ 𝑥 −𝑈𝛼𝜎

𝑛

𝑥 −𝑈𝛼𝜎

𝑛≤ 𝜇 ≤ 𝑥 +

𝑈𝛼𝜎

𝑛

σ Unkown

𝜇 ≤ 𝑥 +𝑡𝛼𝑠

𝑛 𝑜𝑟

𝜇 ≥ 𝑥 −𝑡𝛼𝑠

𝑛

𝑥 −𝑡𝛼𝑠

𝑛≤ 𝜇 ≤ 𝑥 +

𝑡𝛼𝑠

𝑛

Reference

tables

𝑈𝛼: Table B

𝑡𝛼: Table D (𝜈 = 𝑛 − 1) 𝑈𝛼 :Table C

𝑡𝛼: Table E (𝜈 = 𝑛 − 1)

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17.4 Confidence Intervals on the Mean

and Hypothesis Test Criteria Alternatives

• If sample size is calculated before conducting the

experiment using desired values of 𝛼, 𝛽, and 𝛿, the null

hypothesis is not rejected if the criterion is contained within

the appropriate confidence interval for 𝜇. This decision is

made with the 𝛽 risk of error that was used in calculating

the sample size.

• However, if the criterion is not contained within the interval,

the null hypothesis is rejected. This decision is made with

the 𝛼 risk of error.

• Other methods can be used when setting up a hypothesis

test criterion. Consider 𝐻𝑎: 𝜇 > 𝜇𝑎, where 𝜇𝑎 is a product

specification vriterion, it can be determined that

𝑥 𝑐𝑟𝑖𝑡𝑒𝑟𝑖𝑜𝑛 = 𝜇𝑎 +𝑡𝛼𝑠

𝑛

When 𝑥 is greater than the test 𝑥 𝑐𝑟𝑖𝑡𝑒𝑟𝑖𝑜𝑛, 𝐻0 is rejected.

When 𝑥 is less than 𝑥 𝑐𝑟𝑖𝑡𝑒𝑟𝑖𝑜𝑛, 𝐻0 is not rejected.

• Alternative approach to this problem is to use this equation

𝑡0 =(𝑥 − 𝜇𝑎)

𝑠/ 𝑛

Where the null hypothesis is rejected if 𝑡0 > 𝑡𝛼

17.4 Confidence Intervals on the Mean

and Hypothesis Test Criteria Alternatives

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17.5 Example 17.2:

Confidence Intervals on the Mean:

𝜎 is known

• 𝑥 = 5.77, 𝜎 = 2, 𝛼 = 0.10

One-Sample Z

The assumed standard deviation = 2

90% Upper

N Mean SE Mean Bound

16 5.770 0.500 6.411

Minitab:

Stat

Basic Statistics

1-Sample Z-test

90% Lower

N Mean SE Mean Bound

16 5.770 0.500 5.129

N Mean SE Mean 90% CI

16 5.770 0.500 (4.948, 6.592)

17.5 Example 17.2:

Confidence Intervals on the Mean:

𝜎 is unknown

• 𝑥 = 5.77, 𝑠 = 2.41,𝛼 = 0.10

Minitab:

Stat

Basic Statistics

1-Sample t-test

One-Sample T

90% Upper

N Mean StDev SE Mean Bound

16 5.770 2.410 0.603 6.578

90% Lower

N Mean StDev SE Mean Bound

16 5.770 2.410 0.603 4.962

N Mean StDev SE Mean 90% CI

16 5.770 2.410 0.603 (4.714, 6.826)

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17.6 Example 17.3:

Sample Size -- An Alternative Approach

• An alternative approach to determine the sample size is to fix

the width (𝑤) of the margin of error in the confidence intervals.

𝑤 =𝑈𝛼𝜎

𝑛 , 𝑠𝑜 𝑛 =

𝑈𝛼2𝜎2

𝑤2

17.7 Standard Deviation

Confidence Interval

• When a sample of size n is taken from a population that is

normally distributed, the double-sided confidence interval

equation for the population’s standard deviation is

(𝑛−1)𝑠2

𝜒2𝛼2;𝜈

12

≤ 𝜎 ≤ (𝑛−1)𝑠2

𝜒21−𝛼2;𝜈

12

• where 𝑠 is the standard deviation of the sample 𝜒2 values

are taken from Table G with 𝛼/2 risk and 𝜈 degrees of

freedom equal to the sample size minus 1.

• This relationship is not robust for data from non-normal

distribution.

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17.8 Example 17.4: Standard

Deviation Confidence statement

• Consider again the 16 data points from sample 1 of table 3.3,

which had a mean of 5.77 and standard deviation of 2.41. Given

that a standard deviation was not known, the 90% confidence

interval for the standard deviation of the population would then be

(𝑛−1)𝑠2

𝜒2𝛼2;𝜈

12

≤ 𝜎 ≤ (𝑛−1)𝑠2

𝜒21−𝛼2;𝜈

12

(16−1)2.412

𝜒20.12 ;15

12

≤ 𝜎 ≤ (16−1)2.412

𝜒21−0.12 ;15

12

1.87 ≤ 𝜎 ≤ 3.46

17.9 Percentage of the Population

Assessments: Prediction Intervals, Tolerance

Intervals, and Probability Plots

• A confidence interval for a parameter is an interval that is

likely to contain the true value of the parameter.

• Prediction and tolerance intervals are concerned with the

population itself and with values that may be sampled

from it in the future.

• These intervals are only useful when the shape of the

population is known, here we assume the population is

known to be normal.

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Prediction Interval

• A prediction interval is an interval that is likely to contain

the value of an item that will be sampled from the

population at a future time.

• We “predict” that a value that is yet to be sampled from

the population will fall within the predication interval.

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100(1 – α)% Prediction Interval

• Let 𝑥1, … , 𝑥𝑛 be a random sample from a normal population. Let 𝑦 be another item to be sampled from this population, whose value has not yet been observed. The

100(1 – 𝛼)% prediction interval for 𝑦 is

𝑥 ± 𝑡𝛼/2,𝑛−1 𝑠 1 +1

𝑛

• The probability is 1 – 𝛼 that the value of 𝑦 will be contained in this interval.

• One sided intervals may also be constructed.

𝑥 + 𝑡𝛼,𝑛−1 𝑠 1 +1

𝑛 or 𝑥 − 𝑡𝛼,𝑛−1𝑠 1 +

1

𝑛

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Example of Prediction Intervals

• A sample of 10 concrete blocks manufactured by a

certain process has a mean compressive strength of

1312 MPa, with standard deviation of 25 MPa. Find a

95% prediction interval for the strength of a block that has

not yet been measured.

𝑥 ± 𝑡𝛼/2,𝑛−1 𝑠 1 +1

𝑛

1312± 𝑡.025,9(25) 1 +1

10

1312± 2.262 25 1.1 𝑜𝑟 (1253,1371)

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Comparing CI and PI

• The formula for the PI is similar to the formula for the CI of a mean of normal population.

• The prediction interval has a small adjustment to the standard error with the additional + 1 under the square root.

• This reflects the random variation in the value of the sampled item that is to be predicted.

• Prediction intervals are sensitive to the assumption that the population is normal.

• If the shape of the population differs much from the normal curve, the prediction interval may be misleading.

• Large samples do not help, if the population is not normal then the prediction interval is invalid.

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Tolerance Intervals

• A tolerance interval is an interval that is likely to contain a specified proportion of the population.

• First assume that we have a normal population whose mean 𝜇 and standard deviation 𝜎 are known.

• To find an interval that contains 90% of the population, we have 𝜇 ± 1.645𝜎.

• In general, the interval 𝜇 ± 𝑧𝛾 2 𝜎 will contain

100(1 – 𝛾)% of the population.

• In practice, we do not know 𝜇 or 𝜎. Instead we use the sample mean and sample standard deviation.

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Consequences

• Since we are estimating the mean and standard

deviation from the sample,

– We must make the interval wider than it would be if 𝜇 and 𝜎 were known.

– We cannot be 100% confident that the interval actually

contains the required proportion of the population.

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Construction of Interval

• We must specify the proportion 100(1 – 𝛾)% of the population that we wish the interval to contain.

• We must also specify the confidence 100(1 – 𝛼)% that the interval actually contains the specified proportion.

• It is then possible to find a number 𝑘𝑛, 𝛼, 𝛾 such that the interval

𝑥 ± 𝑘𝑛,𝛼,𝛾𝑠

will contain at least 100(1 – 𝛾)% of the population with

confidence 100(1 – 𝛼)%. Values of 𝑘𝑛, 𝛼, 𝛾 are presented in the following table.

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Tolerance Factors for the

Normal Distribution

24

n 90% 95% 99% 90% 95% 99%

2 32.0187 37.6746 48.4296 160.1940 188.4915 242.3004

3 8.3795 9.9158 12.8613 18.9304 22.4009 29.0553

4 5.3692 6.3699 8.2993 9.3984 11.1501 14.5274

5 4.2749 5.0787 6.6338 6.6118 7.8550 10.2602

6 3.7123 4.4140 5.7746 5.3366 6.3453 8.3013

7 3.3686 4.0074 5.2481 4.6129 5.4877 7.1868

8 3.1358 3.7317 4.8907 4.1473 4.9355 6.4683

9 2.9670 3.5317 4.6310 3.8223 4.5499 5.9660

10 2.8385 3.3794 4.4330 3.5821 4.2647 5.5943

11 2.7372 3.2592 4.2766 3.3970 4.0449 5.3075

12 2.6550 3.1617 4.1496 3.2497 3.8700 5.0792

13 2.5868 3.0808 4.0441 3.1295 3.7271 4.8926

14 2.5292 3.0124 3.9549 3.0294 3.6081 4.7371

15 2.4799 2.9538 3.8785 2.9446 3.5073 4.6053

95% Conf 99% Conf

Source: Principles of Statistics for Engineers and Scientists, by

William Navidi, 2010

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Tolerance Factors for the

Normal Distribution

25 Source: Principles of Statistics for Engineers and Scientists, by

William Navidi, 2010

n 90% 95% 99% 90% 95% 99%

16 2.4371 2.9029 3.8121 2.8717 3.4207 4.4920

17 2.3995 2.8583 3.7538 2.8084 3.3453 4.3934

18 2.3662 2.8188 3.7022 2.7527 3.2792 4.3068

19 2.3366 2.7835 3.6560 2.7034 3.2205 4.2300

20 2.3099 2.7518 3.6146 2.6594 3.1681 4.1614

25 2.2083 2.6310 3.4565 2.4941 2.9715 3.9039

30 2.1398 2.5494 3.3497 2.3848 2.8414 3.7333

35 2.0899 2.4900 3.2719 2.3063 2.7479 3.6107

40 2.0516 2.4445 3.2122 2.2468 2.6770 3.5177

45 2.0212 2.4083 3.1647 2.1998 2.6211 3.4443

50 1.9964 2.3787 3.1259 2.1616 2.5756 3.3846

60 1.9578 2.3328 3.0657 2.1029 2.5057 3.2929

70 1.9291 2.2987 3.0208 2.0596 2.4541 3.2251

80 1.9068 2.2720 2.9859 2.0260 2.4141 3.1725

90 1.8887 2.2506 2.9577 1.9990 2.3819 3.1303

100 1.8738 2.2328 2.9343 1.9768 2.3555 3.0955

95% Conf 99% Conf

Example of Tolerance Intervals

• The lengths of bolts manufactured by a certain process

are known to be normally distributed. In a sample of 30

bolts, the average length was 10.25 cm, with a standard

deviation of 0.20 cm. Find a tolerance interval that

includes 90% of the lengths of the bolts with 95%

confidence.

𝑥 ± 𝑘𝑛,𝛼,𝛾𝑠

10.25± 𝑘30,.05,.10(0.20)

10.25 ± 2.1398 0.20 𝑜𝑟 (9.82, 10.68)

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17.10 Example 17.5: Percentage of

the Population Statements

17.11 Statistical Tolerancing

• Consider that measurements for n components that are

each centered at mean (𝜇𝑖) of normal distribution with plus

or minus tolerance (𝑇𝑖) around this mean value.

• The worst-case overall tolerance (𝑇𝑤) for this situation is

simply the addition of these tolerances.

𝑇𝑤 = ± 𝑇𝑖 = ±(𝑇1 +

𝑛

𝑖=1

𝑇2 +⋯+ 𝑇𝑛)

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17.11 Statistical Tolerancing

• The serial 3𝜎 combination of the component tolerances

yields an overall product 3𝜎 tolerance (𝑇3𝜎) of

𝑇3𝜎 = ± 𝑇𝑖2

𝑛

𝑖=1

1/2

= ±(𝑇12 + 𝑇2

2 +⋯+ 𝑇𝑛2)1/2

• The assumption that each component follows a normal

distribution will not be valid. For example, a ± 10%

tolerance resistor may follow a bimodal, truncated, and/or

skewed, distribution because the best parts can be sorted

out and sold at a higher price, with a ±1% or ±5% tolerance.

17.11 Statistical Tolerancing

• Another situation where the normality assumption may be

invalid is where a manufacturer initially produces a part at

one tolerance extreme, anticipating tool wear in the

manufacturing process.

• An alternative approach is to estimate a distribution shape

for each component. Then conduct a Monte Carlo

simulation to yield an overall expected output.

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17.12 Example 17.6:

Combining Analytical Data with

Statistical Tolerancing • An automatic sheet feed device is to load a sheet of paper

into a printer such that the first character printed on the

paper will be 1.261.26 mm from the edge of the paper.

• In this problem, there are 2 variabilities that need

consideration:

– Variability within machines

– Variability between machines

• Three alternatives:

– Worst-case analysis

– All sheets of paper on all printer manufactured

– Tolerance of the sheets produced on a worst-case machine

17.12 Example 17.6:

Combining Analytical Data with

Statistical Tolerancing • Alternative 1: Worst-case analysis

– Worst-case machine tolerances are combined with the 3𝜎 limits of machine repeatability:

𝑇𝐴 = 𝑇𝑤 + 3𝜎𝑟 = ± 𝑇𝑖 ± 3𝜎𝑟

𝑛

𝑖=1

= ±2.02

– All sheets of paper on all printer manufactured

– Tolerance of the sheets produced on a worst-case machine

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17.12 Example 17.6:

Combining Analytical Data with

Statistical Tolerancing • Alternative 1: Worst-case analysis

– Worst-case machine tolerances are combined with the 3𝜎 limits of machine repeatability:

𝑇𝐴 = 𝑇𝑤 + 3𝜎𝑟 = ± 𝑇𝑖 ± 3𝜎𝑟

𝑛

𝑖=1

= ±2.02

• Alternative 2: All sheets of paper on all printers

𝑇𝐵 = ± 𝑇𝑖2

𝑛

𝑖=1

+ (3𝜎𝑟)2

0.5

= ±0.622

• Tolerance of the sheets produced on a worst-case machine

17.12 Example 17.6:

Combining Analytical Data with

Statistical Tolerancing • Alternative 3: Tolerance of the sheets produced on a worst-

case machine

𝑇𝐶 = ± 𝑇𝑖2

𝑛

𝑖=1

1/2

± (3𝜎𝑟) = ±0.85

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17.13 Nonparametric Estimates:

Run Test For Randomization

• A run is defined as a group of consecutive

observations either all greater than or less than

some value. To assess whether data are in

random order using a runs test, data are

evaluated in terms of the number of runs above

and below the median. Within this

nonparametric test no assumption is required

about the population distribution.

17.14 Example 17.7: Nonparametric

Runs Test For Randomization

• Forty people are selected randomly. Each person is asked a question,

which has five possible answers that are coded 1-5. A gradual bias in the

question phrasing or a lack of randomization when selecting people would

cause non-randomization of the responses.

1 1 2 1 1 1 1 1 1

2 3 3 2 0 0 0 0 1

1 3 3 4 4 5 5 5 5

2 1 1 2 2 2 1 1 3

3 3 3 2

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17.14 Example 17.7: Nonparametric

Runs Test For Randomization

Runs Test: C1

Runs test for C1

Runs above and below K = 2.05

The observed number of runs = 7

The expected number of runs = 19.2

14 observations above K, 26 below

P-value = 0.000

Minitab:

Stat

Nonparametrics

Run Test