# SEU Statistics Minitab Problems

Applied Statistics and Probabilityfor Engineers

Seventh Edition

Douglas C. Montgomery

George C. Runger

Chapter 6

Descriptive Statistics

Chapter 6 Title Slide

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1

6

The Role of Statistics in Engineering

CHAPTER OUTLINE

6.1 Numerical Summaries of

Data

6.2 Stem-and-Leaf Diagrams

6.3 Frequency Distributions

and Histograms

Chapter 6 Contents

6.4

6.5

6.6

6.7

Box Plots

Time Sequence Plots

Scatter Diagrams

Probability Plots

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Learning Objectives for Chapter 6

After careful study of this chapter, you should be able to do the following:

1. Compute and interpret the sample mean, variance, standard deviation, median, and range

2. Explain the concepts of sample mean, variance, population mean, and population variance

3. Construct and interpret visual data displays, including stem-and-leaf display, histogram, and

the box plot

4. Explain the concept of random sampling

5. Construct and interpret normal probability plots

6. Explain how to use box plots and other data displays to visually compare two or more samples

of data

7. Know how to use simple time series plots to visually display the important features of timeoriented data

8. Know how to construct and interpret scatter diagrams of two or more variables

Chapter 6 Learning Objectives

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Numerical Summaries of Data

• Data summaries and displays are essential to good statistical thinking

• It is useful to describe data features numerically

• Characterizing the location or central tendency in the data is an example

of a numerical summary

• Data are often a sample of observations that have been selected from

some larger population of observations

• Many statistical software are available: Minitab, SPSS, Stata, Statistica,

etc.

• Many Excel add-ins: Analysis ToolPak, MegaStat, PHStat, etc.

Sec 6.1 Numerical Summaries of Data

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Sample Mean

The location or central tendency in the data can be characterized by the arithmetic

average or sample mean.

For a finite population with 𝑁 equally likely values, the probability mass function is

𝑓(𝑥𝑖 ) = 1/𝑁 and the mean is

Sec 6.1 Numerical Summaries of Data

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Example 6.1 | Sample Mean

Consider 8 observations (𝑥𝑖) of pull-off force from engine

connectors as shown in the table.

8

x = average =

x

i =1

i

8

12.6 + 12.9 + … + 13.1

=

8

104

=

= 13.0 pounds

8

xi

12.6

12.9

13.4

12.3

13.6

13.5

12.6

13.1

13.00

= AVERAGE($B2:$B9)

i

1

2

3

4

5

6

7

8

Figure 6.1 The sample mean is the balance point.

Sec 6.1 Numerical Summaries of Data

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Sample Variance and Standard Deviation

The variability or scatter in the data may be described by the sample variance or

the sample standard deviation.

The units of measurement for the sample variance are the square of the original

units of the variable, while the standard deviation measures variability in the original

units.

Sec 6.1 Numerical Summaries of Data

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Example 6.2 | Sample Variance

The table displays the quantities needed for calculating

the sample variance and sample standard deviation.

The numerator of 𝑠 2 is

Sec 6.1 Numerical Summaries of Data

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𝟐

Computation of 𝒔

The prior calculation is definitional and tedious. A shortcut is derived here and

involves just 2 sums.

Sec 6.1 Numerical Summaries of Data

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Example 6.3 | Shortcut Calculation for

For Example 6.2, we calculate the sample variance and standard deviation using the

shortcut method.

Sec 6.1 Numerical Summaries of Data

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The meaning of 𝒏 − 𝟏 in the denominator

• The population variance is calculated with 𝑁, the population size. Why

isn’t the sample variance calculated with 𝑛, the sample size?

• The true variance is based on data deviations from the true mean, 𝜇.

• The sample calculation is based on the data deviations from 𝑥,ҧ not 𝜇.

• 𝑥ҧ is an estimator of 𝜇; close but not the same.

• So the 𝑛 − 1 divisor is used to compensate for the error in the mean

estimation.

Sec 6.1 Numerical Summaries of Data

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Degrees of Freedom

• When the sample variance is calculated with the quantity 𝑛 − 1 in the

denominator, the quantity 𝑛 − 1 is called the degrees of freedom

Sec 6.1 Numerical Summaries of Data

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Sample Range

In addition to the sample variance and sample standard deviation, the

sample range is a useful measure of variability.

For Example 6.3 (pull-off force data), the sample range is 𝑟 = 13.6 − 12.3 =

1.3.

Sec 6.1 Numerical Summaries of Data

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Stem-and-Leaf Diagrams

Sec 6.2 Stem-and-Leaf Diagrams

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Example 6.4a | Alloy Strength

• Consider the data below. We select as stem values the numbers 7, 8, 9, …, 24.

Sec 6.2 Stem-and-Leaf Diagrams

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Example 6.4b | Alloy Strength

• The resulting stem-and-leaf diagram is shown.

• Inspection of the diagram reveals that most of the

comprehensive strengths lie between 110 and

200 psi and that a central value is somewhere

between 150 and 160 psi.

• The strengths are distributed approximately

symmetrically about the central value

Sec 6.2 Stem-and-Leaf Diagrams

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Frequency Distributions and Histograms

• A frequency distribution is a more compact summary of data than a stem

– and – leaf diagram. It shows distribution of the data

• To construct, we must divide the range of the data into intervals, which are

usually called class intervals, cells, or bins

• Choosing number of bins approximately equal to the square root of the

number of observations often works well in practice

• Bin width(max-min)/number of bins

• Using table 6-2, 80 = 9…So we can use 9 bins

• Bin width=(245-76)/9..approximately 20, we start below minimum(76) like

70 and end wit more than maximum like 250

Sec 6.3 Frequency Distributions and Histograms

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Frequency Distribution Table

Relative frequency=frequency/n(number of data)

Cumulative frequency: is the total number of observations

that are less than or equal to the upper limit of the bin.

Sec 6.3 Frequency Distributions and Histograms

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Histograms

• A histogram is a visual display of the frequency distribution

• Provides a visual impression of the shape and distribution of the

measurements and information about the central tendency and scatter or

dispersion in the data

Sec 6.3 Frequency Distributions and Histograms

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Histograms

Sec 6.3 Frequency Distributions and Histograms

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Box Plots

• The box plot is a graphical display that simultaneously describes several

important features of a data set, such as center, spread, departure from

symmetry, and identification of unusual observations or outliers

• Sometimes called box – and – whisker plots

• Displays three quartiles

• A line, or whisker, extends from each end of the box

Sec 6.4 Box Plots

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Quartile formula

N=number of observations

For table 6-2, Q1=(80+1)*1/4=20.25th order =143.5 see excel 6-2 file

Q2=81*2/4=40.5th order=161.5

Similarly Q3=181

IQR=181-143.5=37.5

q3 +1.5IQR = 237.25.

q1 -1.5IQR= 87.25

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Box Plots

Sec 6.4 Box Plots

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Scatter Diagrams

• Multivariate: each observation consists of

measurements of several variables

• The scatter diagram is a useful way to

graphically display the potential relationship

between quality and one of the other qualities

• When two or more variables exist, the matrix of

scatter diagrams may be useful in looking at all

of the pairwise relationships between the

variables in the sample

• The sample correlation coefficient is a

quantitative measure of the strength of the

linear relationship between two random

variables x and y

Sec 6.6 Scatter Diagrams

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Probability Plots

• A probability plot is a graphical method for determining whether sample

data conform to a hypothesized distribution based on a subjective visual

examination of the data

• To construct a probability plot:

• Rank the data observations in the sample from smallest to largest: x(1), x(2),…, x(n).

• The observed value x(j) is plotted against the observed cumulative frequency (j – 0.5)/n.

• The paired numbers are plotted on the probability paper of the proposed distribution.

• If the plotted points deviate a straight line, then the hypothesized

distribution adequately describes the data.

Sec 6.7 Probability Plots

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Important Terms and Concepts

• Box plot

• Degrees of freedom

• Digidot plot

• Frequency distribution and

histogram

• Histogram

• Interquartile range

• Matrix of scatter plots

• Multivariate data

• Normal probability plot

• Outlier

• Pareto chart

Chapter 6 Important Terms and Concepts

• Percentile

• Population mean

• Population standard

deviation

• Population variance

• Probability plot

• Quartiles and percentiles

• Relative frequency

distribution

• Sample correlation

coefficient

• Sample mean

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• Sample median

• Sample mode

• Sample range

• Sample standard deviation

• Sample variance

• Scatter diagram

• Stem-and-leaf diagram

• Time series

26

Applied Statistics and Probability

for Engineers

Seventh Edition

Douglas C. Montgomery

George C. Runger

Chapters 7 –8

Point Estimation of Parameters, Sampling Distributions,

and Statistical Intervals for a Single Sample

Chapter 7 Title Slide

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1

Point Estimation

• A point estimate is a reasonable value of a population

parameter.

• Statistics such, 𝑥,ҧ 𝑠2, and 𝑝Ƹ are random variables.

• Statistics have their unique distributions which are called

sampling distributions.

Sec 7.1 Point Estimation

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Point Estimator

Sec 7.1 Point Estimation

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Some Parameters & Their Statistics

• There could be choices for the point

estimator of a parameter.

• To estimate the mean of a population,

we could choose the:

▪ Sample mean.

▪ Sample median.

▪ Average of the largest & smallest

observations in the sample.

Sec 7.1 Point Estimation

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Some Definitions

• The random variables X1, X2,…,Xn are a random sample of size

n if:

a) The Xi ‘s are independent random variables.

b) Every Xi has the same probability distribution.

•

•

A statistic is any function of the observations in a random

sample.

The probability distribution of a statistic is called a sampling

distribution.

Sec 7-2 Sampling Distributions and the Central Limit Theorem

5

Central Limit Theorem

Sec 7.2 Sampling Distributions and the Central Limit Theorem

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Example 7.2 | Central Limit Theorem

Sec 7.2 Sampling Distributions and the Central Limit Theorem

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Confidence Interval & Properties

A confidence interval estimate for is an interval of the form

where the end-points L and U are computed from the sample data.

• There is a probability of selecting a sample for which the CI will contain the true value of

𝜇.

• The endpoints or bounds L and U are called lower- and upper-confidence limits and 1 − 𝛼

is called the confidence coefficient or confidence level.

Sec 8.1.1 Development of the Confidence Interval and Its Basic Properties

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Confidence Interval on the Mean, Variance

Known

Sec 8.1.1 Development of the Confidence Interval and Its Basic Properties

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Example 8.1 | Metallic Material Transition

• Ten measurements of impact energy (J) on specimens of A238 steel cut at 60°C are as

follows: 64.1, 64.7, 64.5, 64.6, 64.5, 64.3, 64.6, 64.8, 64.2, and 64.3. The impact energy is

normally distributed with = 1𝐽. Find a 95% CI for , the mean impact energy.

• The required quantities are z𝛼/2 = z0.025 = 1.96, 𝑛 = 10, 𝜎 = 1, and 𝑥ҧ = 64.46. The resulting

95% CI is found from Equation 8-1 as follows:

• Interpretation: Based on the sample data, a range of highly plausible values for mean

impact energy for A238 steel at 60°C is 63.84𝐽 ≤ 𝜇 ≤ 65.08𝐽

Sec 8.1.1 Development of the Confidence Interval and Its Basic Properties

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8.1.2 Sample Size for Specified Error on the Mean, Variance Known

x

|x −

If is used as an estimate of , we can be

100(1 − α)% confident that the error

will not

exceed a specified amount E when the sample

size is

z

n =

E

2

Sec 8-1 Confidence Interval on the Mean of a Normal, σ2 Known

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EXAMPLE 8-2 Metallic Material Transition

Consider the CVN test described in Example 8-1.

Determine how many specimens must be tested to ensure

that the 95% CI on for A238 steel cut at 60°C has a length

of at most 1.0J.

The bound on error in estimation E is one-half of the length of

the CI.

Use Equation 8-2 to determine n with E = 0.5, = 1, and

zα/2 = 1.96.

2

z / 2

(1.96)1 = 15.37

=

n =

E

0

.

5

2

Since, n must be an integer, the required sample size is

n = 16.

Sec 8-1 Confidence Interval on the Mean of a Normal, σ2 Known

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12

Large-sample Confidence Interval for 𝝁

Sec 8.1.5 Large-Sample Confidence Interval for 𝜇

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Example 8.4 | Mercury Contamination

• A sample of fish was selected from 53 Florida lakes, and mercury concentration in the

muscle tissue was measured (ppm). The mercury concentration values were

• Find an approximate 95% CI on 𝜇.

Sec 8.1.5 Large-Sample Confidence Interval for 𝜇

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Example 8.4 | Mercury Contamination (ctd.)

• The summary statistics for these data are as follows:

• Because 𝑛 > 40, the assumption of normality is

not necessary to use in Equation 8.11. The

required values are 𝑛 = 53, 𝑥ҧ = 0.5250, 𝑠 = 0.3586,

and 𝑧0.025 = 1.96.

Interpretation: This interval is fairly

wide because there is variability in the

mercury concentration

measurements. A larger sample size

would have produced a shorter

interval.

Sec 8.1.5 Large-Sample Confidence Interval for 𝜇

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Large-Sample Approximate Confidence Interval

Suppose that θ is a parameter of a ̂

probability distribution, and let be an

estimator of θ. Then a large-sample

approximate CI for θ is given by

ˆ − z/2 ˆ ˆ + z/2 ˆ

Sec 8-1 Confidence Interval on the Mean of a Normal, σ2 Known

16

The 𝒕 distribution

Sec 8.2.1 𝑡 distribution

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The Confidence Interval on Mean, Variance

Unknown

https://www.usu.edu/math/cfairbourn/Stat2300/t-table.pdf

Sec 8.2.2 𝑡 Confidence Interval on 𝜇

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Example 8.5 | Alloy Adhesion

• The load at specimen failure is as follows (in megapascals):

• Construct a 95% CI on 𝜇.

• The sample mean is 𝑥ҧ = 13.71 and sample standard deviation is 𝑠 = 3.55. Since 𝑛 = 22,

we have 𝑛 − 1 = 21 degrees of freedom for 𝑡, so 𝑡0.025,21 = 2.080.

Interpretation: The CI is fairly wide

because there is a lot of variability in the

measurements. A larger sample size

would have led to a shorter interval.

Sec 8.2.2 𝑡 Confidence Interval on 𝜇

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Confidence Interval on the Variance and

Standard Deviation of a Normal Distribution

Sec 8.3 Confidence Interval on the Variance and Standard Deviation of a Normal Distribution

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20

Confidence Interval on the Variance and

Standard Deviation of a Normal Distribution

https://people.smp.uq.edu.au/YoniNazarathy/stat_models_

B_course_spring_07/distributions/chisqtab.pdf

Sec 8.3 Confidence Interval on the Variance and Standard Deviation of a Normal Distribution

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Example 8.6 | Detergent Filling

• An automatic filling machine is used to fill bottles with liquid detergent. A random sample of 20 bottles

results in a sample variance of fill volume of 𝑠2 = 0.01532. Assume that the fill volume is approximately

normal. Compute a 95% upper confidence bound.

•

𝑛−1 𝑠 2

𝑛−1 𝑠 2

2

A 95% confidence interval is found from Equation 8.19: 𝜒2

≤ 𝜎 ≤ 𝜒2

𝛼/2,𝑛−1

1−𝛼/2,𝑛−1

2

2

20 − 1 0.0153

20 − 1 0.0153

≤ 𝜎2 ≤

,

i. e.,

1.354 × 10−4 ≤ 𝜎 2 ≤ 4.994 × 10−4

32.852

8.902

• A confidence interval on can be obtained by taking the square root on both sides.

• Practical Interpretation: Therefore, at the 95% level of confidence, the data indicate that the process

standard deviation is between 0.012 and 0.022 fluid ounce. The process engineer or manager now

needs to determine whether a standard deviation this large could lead to an operational problem with

under-or-over-filled bottles

Sec 8.3 Confidence Interval on the Variance and Standard Deviation of a Normal Distribution

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22

Large-Sample Confidence Interval for a

Population Proportion

Sec 8.4 Large-Sample Confidence Interval for a Population Proportion

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23

Example 8.7 | Crankshaft Bearings

• In a random sample of 85 automobile engine crankshaft bearings, 10 have a surface

finish that is rougher than the specifications allow. Construct a 95% two-sided confidence

interval for p.𝑝=10/85=.12

Ƹ

• A 95% two-sided confidence interval for p is computed from Equation 8-23 as

pˆ − z0.025

0.12 − 1.96

pˆ (1 − pˆ )

p pˆ + z0.025

n

pˆ (1 − pˆ )

n

0.12 ( 0.88 )

0.12 ( 0.88 )

p 0.12 + 1.96

85

85

0.0509 p 0.2243

• Practical Interpretation: This is a wide CI. Although the sample size does not appear to be

small (n = 85), the value of 𝑝Ƹ is fairly small, which leads to a large standard error for

contributing to the wide CI.

Sec 8.4 Large-Sample Confidence Interval for a Population Proportion

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24

Applied Statistics and Probability

for Engineers

Seventh Edition

Douglas C. Montgomery

George C. Runger

Chapter 9

Tests of Hypotheses for a Single Sample

Chapter 9 Title Slide

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1

Hypothesis Testing

Two-sided Alternative Hypothesis

Let 𝐻0: 𝜇 = 50 centimeters per second( burning rate of propellant) and 𝐻1: 𝜇 ≠ 50 centimeters per second

• The statement 𝐻0: 𝜇 = 50 is called the null hypothesis.

• The statement 𝐻1: 𝜇 ≠ 50 is called the alternative hypothesis.

One-sided Alternative Hypothesis

• 𝐻0: 𝜇 = 50 centimeters per second

𝐻0: 𝜇 = 50 centimeters per second

or

• 𝐻1: 𝜇 < 50 centimeters per second
𝐻1: 𝜇 > 50 centimeters per second

Sec 9.1.1 Statistical Hypotheses

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2

Hypothesis Testing

Test of a Hypothesis

• 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; if this information is inconsistent with the hypothesis, we will conclude

that the hypothesis is false.

Sec 9.1.2 Tests of Statistical Hypotheses

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3

Tests of Statistical Hypotheses

Sec 9.1.2 Tests of Statistical Hypotheses

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4

Decisions in Hypothesis Testing

Sometimes the type 𝐼 error

probability is called the

significance level, or the error, or the size of the test.

Sec 9.1.2 Tests of Statistical Hypotheses

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5

One-Sided and Two-Sided Hypotheses

One-Sided Tests

Two-Sided Test

H0: = 0

or

H1: > 0

H0: = 0

H1: ≠ 0

H0: = 0

H1: < 0
Sec 9.1.3 One-Sided and Two-Sided Hypotheses
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6
𝑷-value
P-value is the observed significance level.
Sec 9.1.4 P-Values in Hypothesis Tests
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7
𝑷-values in Hypothesis Tests
• Consider the two-sided hypothesis test 𝐻0: = 50 against 𝐻1: 50 with 𝑛 = 16 and = 2.5.
Suppose that the observed sample mean is 𝑥ҧ = 51.3 centimeters per second.
• The P-value of the test is the probability above 51.3 plus the probability below 48.7.
Sec 9.1.4 P-Values in Hypothesis Tests
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8
General Procedure for Hypothesis Tests
Sec 9.1.6 General Procedure for Hypothesis Tests
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9
Hypothesis Tests on the Mean
Sec 9.2.1 Hypothesis Tests on the Mean
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10
Example 9.2a | Propellant Burning Rate
• Air crew 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 centimeters per second and the standard deviation is = 2
centimeters per second. The significance level of = 0.05 and a random sample of n =
25 has a sample average burning rate of 51.3 centimeters per second. Draw
conclusions.
• The seven-step procedure is
1.
2.
3.
Parameter of interest: The parameter of interest is , the mean
burning rate.
Null hypothesis: H0: = 50 centimeters per second
Alternative hypothesis: H1: 50 centimeters per second
Sec 9.2.1 Hypothesis Tests on the Mean
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11
Example 9.2b | Propellant Burning Rate
4. Test statistic: The test statistic is z = x − 0
0
/ n
5. Reject H0 if: Reject H0 if the P-value is less than 0.05. The boundaries of the critical
region would be 𝑧0.025 = 1.96 𝑎𝑛𝑑 − 𝑧0.025 = −1.96.
6. Computations: Since x = 51.3 and = 2,
z0 =
51.3 − 50
= 3.25
2/ 25
7. Conclusion: Since z0 = 3.25 and the p-value is = 2[1 − (3.25)] = 0.0012, we reject
𝐻0: = 50 at the 0.05 level of significance.
Practical Interpretation: The mean burning rate differs from 50 centimeters per second,
based on a sample of 25 measurements.
Sec 9.2.1 Hypothesis Tests on the Mean
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12
Large-Sample Test
• A test procedure for the null hypothesis H0: = 0 assuming that the population is
normally distributed and that 2 known is developed. In most practical situations, 2 will be
unknown. Even, we may not be certain that the population is normally distributed.
• In such cases, if n is large (say, n 40) the sample standard deviation s can be
substituted for in the test procedures. Thus, while we have given a test for the mean of
a normal distribution with known 2, it can be easily converted into a large-sample test
procedure for unknown 2 regardless of the form of the distribution of the population.
• Exact treatment of the case where the population is normal, 2 is unknown, and n is small
involves use of the t distribution.
Sec 9.2.3 Large-Sample Test
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13
Hypothesis Tests on the Mean
9.3.1 Hypothesis Tests on the Mean
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14
Example 9.6a | Golf Club Design
• An experiment was performed in which 15 drivers produced by a particular club maker were selected at random
and their coefficients of restitution measured. It is of interest to determine if there is evidence (with = 0.05) to
support a claim that the mean coefficient of restitution exceeds 0.82.
• The observations are:
0.8411 0.8191 0.8182 0.8125 0.8750
0.8580 0.8532 0.8483 0.8276 0.7983
0.8042 0.8730 0.8282 0.8359 0.8660
• The sample mean and sample standard deviation are x = 0.83725 and s = 0.02456. The objective of the
experimenter is to demonstrate that the mean coefficient of restitution exceeds 0.82, hence a one-sided
alternative hypothesis is appropriate.
The seven-step procedure for hypothesis testing is as follows:
1. Parameter of interest: The parameter of interest is the mean coefficient of restitution, .
2. Null hypothesis: H0: = 0.82
3. Alternative hypothesis: H1: 0.82
9.3.1 Hypothesis Tests on the Mean
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15
Example 9.6b | Golf Club Design
4. Test Statistic: The test statistic is
t0 =
x − 0
s/ n
5. Reject H0 if: Reject H0 if the P-value is less than 0.05.
6. Computations: Since x = 0.83725, s = 0.02456, = 0.82, and n = 15, we have
t0 =
0.83725 − 0.82
= 2.72
0.02456/ 15
7. Conclusions: From Appendix A Table II, for a t distribution with 14 degrees of freedom, t0 = 2.72
falls between two values: 2.624, for which = 0.01, and 2.977, for which = 0.005. Since, this is a
one-tailed test the P-value is between those two values, that is, 0.005 < P < 0.01. Therefore, since
P < 0.05, we reject H0 and conclude that the mean coefficient of restitution exceeds 0.82.
Practical Interpretation: There is strong evidence to conclude that the mean coefficient of restitution
exceeds 0.82.
9.3.1 Hypothesis Tests on the Mean
Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved
16
Hypothesis Tests on the Variance
9.4.1 Hypothesis Tests on the Variance
Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved
17
Example 9.8 | Automated Filling
• An automated filling machine is used to fill bottles with liquid detergent. A random sample of 20 bottles results in
a sample variance of fill volume of s2 = 0.0153 (fluid ounces)2. If the variance of fill volume exceeds 0.01 (fluid
ounces)2, an unacceptable proportion of bottles will be underfilled or overfilled. Is there evidence in the sample
data to suggest that the manufacturer has a problem with underfilled or overfilled bottles? Use = 0.05, and
assume that fill volume has a normal distribution.
• Using the seven-step procedure results in the following:
1
9.4.1 Hypothesis Tests on the Variance
Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved
18
Large-Sample Tests on a Proportion
9.5.1 Large-Sample Tests on a Proportion
Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved
19
Example 9.10 | Automobile Engine Controller
A semiconductor manufacturer produces controllers used in automobile engine applications. The customer
requires that the process fallout or fraction defective at a critical manufacturing step not exceed 0.05 and that the
manufacturer demonstrate process capability at this level of quality using = 0.05. The semiconductor
manufacturer takes a random sample of 200 devices and finds that four of them are defective. Can the
manufacturer demonstrate process capability for the customer?
1. Parameter of Interest: The parameter of interest is the process fraction defective p.
2. Null hypothesis: H0:p = 0.05
3. Alternative hypothesis: H1: p < 0.05
4. Test Statistic:
z0 =
x − np0
np0 (1 − p0 )
where x = 4, n = 200, and p0 = 0.05.
5. Reject H0 if: Reject H0: p = 0.05 if the p-value is less than 0.05.
4 − 200(0.05)
6. Computations: The test statistic is z 0 = 200(0.05)(0.95) = −1.95
7. Conclusions: Since z0 = −1.95, the P-value is (−1.95) = 0.0256, so we reject H0 and
Practical
Interpretation: We
conclude that the
process is capable
conclude that the process fraction defective p is less than 0.05.
9.5.1 Large-Sample Tests on a Proportion Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved
20
Large-Sample Tests on a Proportion
Another form of the test statistic Z0 is
9.5.1 Large-Sample Tests on a Proportion Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved
21
Testing for Goodness of Fit
• Based on chi-square distribution
• Requires a random sample of size n from the 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 df for the critical value is k – p – 1 where p represents the number of parameters of
the hypothesized distribution estimated by sample statistics and k represents the number
of categories.
9.7 Testing for Goodness of Fit
Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved
22
Example 9.12a | Printed Circuit Board
Defects-Poisson Distribution
• The number of defects in printed circuit boards is hypothesized to follow
a Poisson distribution. A random sample of n = 60 printed boards has
been collected, and the following number of defects observed.
• The estimate of the mean number of defects per board is the sample
average, (32·0 + 15·1 + 9·2 + 4·3)/60 = 0.75. From the Poisson
distribution with parameter 0.75, we may compute pi, the theoretical,
hypothesized probability associated with the ith class interval. We may
find the pi as follows:
e −0.75 (0.75)0
p = P ( X = 0) =
= 0.472
1
0!
e −0.75 (0.75)1
p 2 = P ( X = 1) =
= 0.354
1!
e −0.75 (0.75)2
p 3 = P ( X = 2) =
= 0.133
2!
p 4 = P ( X 3) = 1 − ( p1 + p 2 + p3 ) = 0.041
9.7 Testing for Goodness of Fit
Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved
23
Example 9.12b | Printed Circuit Board
Defects-Poisson Distribution
• The expected frequencies are computed by multiplying the sample size n
= 60 times the probabilities pi. That is, Ei = npi. The expected frequencies
follow:
• Because the expected frequency in the last cell is less than 3, we combine
the last two cells.
9.7 Testing for Goodness of Fit
Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved
24
Example 9.12c | Printed Circuit Board
Defects-Poisson Distribution
The seven-step hypothesis-testing procedure may now be applied, using = 0.05, as follows:
1. Parameter of interest: The variable of interest is the form of the distribution of defects in printed circuit
boards.
2. Null hypothesis: H0: The form of the distribution of defects is Poisson.
3. Alternative hypothesis: H1: The form of the distribution of defects is not Poisson.
k
(
oi − Ei )2
4. Test statistic: The test statistic is =
Ei
i =1
5. Reject H0 if: Reject H0 if the P-value is less than 0.05.
6. Computations:
2
2
2
32
−
28.32
15
−
21.24
13
−
10.44
χ20 =
+
+
= 2.94
28.32
21.24
10.44
2
2
7. Conclusions: We find from Appendix Table III that = 2.71 and = 3.84
Because = 2.94 lies
between these values, we conclude that the P-value is between 0.05 and 0.10. Therefore, since the P-value
exceeds 0.05 we are unable to reject the null hypothesis that the distribution of defects in printed circuit boards
is Poisson. The exact P-value computed from Minitab is 0.0864.
9.7 Testing for Goodness of Fit
Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved
25
Goodness of Fit Test Example
• Acme Toy Company prints baseball cards.
• The company claims that 30% of the cards are rookies, 60% veterans but
not All-Stars, and 10% are veteran All-Stars.
• Suppose a random sample of 100 cards has 50 rookies, 45 veterans, and
5 All-Stars. Is this consistent with Acme's claim? Use a 0.05 level of
significance.
Copyright @ 2019 John Wiley & Sons, Inc. All Rights Reserved
26
Goodness of Fit Test Example
The seven-step hypothesis-testing procedure may now be applied, using = 0.05, as follows:
1. Parameter of interest: The variable of interest is the form of the distribution of players on the cards.
2. Null hypothesis: H0: The proportion of rookies, veterans, and All-Stars is 30%, 60% and 10%, respectively..
3. Alternative hypothesis: H1: The proportion of rookies, veterans, and All-Stars is not 30%, 60% and 10%.
4. Test statistic: The test statistic is
=
k
(oi − Ei )2
i =1
Ei
5. Reject H0 if: Reject H0 if the P-value is less than 0.05.
6. Computations:
50 − 30 2
45 − 60 2
5 − 10 2
+
+
= 19.58
30
60
10.44
7. Conclusions: We find from Appendix Table III that χ20.005,2 = 10.597. Because χ20 = 19.58 is greater than
10.597, we conclude that the P-value is less than 0.005. Therefore, we reject the null hypothesis that 30% of
the cards are rookies, 60% veterans but not All-Stars, and 10% are veteran All-Stars.
χ20 =
9.7 Testing for Goodness of Fit
Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved
27
Goodness-of-Fit Test:
Normality
… the test investigates
if the observed frequencies in a frequency distribution
match the theoretical normal distribution
to determine the mean and standard deviation of the frequency distribution
- Compute the z-value for the lower class limit and the upper class
limit for each class
- Determine Ei for each category
- Use the chi-square goodness-of-fit test to
Oi coincides with Ei
15 - 28
determine if
Goodness-of-Fit Test:
Normality
•
A sample of 500 donations to the Arthritis Foundation is reported in the
following frequency distribution
Amount Spent
$14
•
•
15 - 29
Oi
20
60
140
120
90
70
Is it reasonable to conclude that the distribution is normally distributed with a mean of $10 and a
standard deviation of $2?
Use the .05 significance level
15 - 30
Amount Spent
Oi
$14
70
Total
500
Area
Ei
(Oi- Ei )2/Ei
To compute Ei for the first class,
first determine the z - value
X −
6 − 10
z =
=
= − 2 . 00
2
find the probability of a z - value less than –2.00
P( z < −2.00) = .0228
15 - 31
15 - 32
Amount Spent
Oi
Area
$14
70
.02
Total
500
Ei
(Oi- Ei )2/Ei
The expected frequency is the probability of a
z-value less than –2.00 times the sample size
E1 = (. 0228 )( 500 ) = 11 . 40
The other expected frequencies
are computed similarly
15 - 33
15 - 34
Amount Spent
Oi
Area
Ei
(Oi- Ei )2/Ei
$14
70
.02
11.40
301.22
Total
500
500
336.33
H0: The observations follow the normal distribution
H0: The observations do NOT follow the normal
distribution
= 0.05
H0 is rejected if 2 = 336.33 is greater
df = 6-2-1=3
than20.05,3 =7.815 from table
H0: is rejected.
The observations do NOT follow the normal distribution
15 - 35

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