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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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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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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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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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 Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved 6 𝑷-value P-value is the observed significance level. Sec 9.1.4 P-Values in Hypothesis Tests Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved 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 Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved 8 General Procedure for Hypothesis Tests Sec 9.1.6 General Procedure for Hypothesis Tests Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved 9 Hypothesis Tests on the Mean Sec 9.2.1 Hypothesis Tests on the Mean Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved 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 Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved 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 Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved 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 Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved 13 Hypothesis Tests on the Mean 9.3.1 Hypothesis Tests on the Mean Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved 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 Copyright © 2019 John Wiley & Sons, Inc. All Rights Reserved 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 than20.05,3 =7.815 from table H0: is rejected. The observations do NOT follow the normal distribution 15 - 35