lecture 3 - MGCR 271 Business
Sta+s+cs
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Unformatted text preview: MGCR 271 Business
Sta+s+cs
 Make
Inference
 Identify Sampling Distribution Ramnath
Vaidyanathan
 Entity Population (N = 100) Objective of Statistics is to estimate these Parameters!! Property (X) Distribution f(X) !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! !! !! !! !! !! !! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !!! !! !!! !! !!! !!! !!! !!! !!!! !!! !!!! !!! !!!! !!!! !!!! !!!! !!!!!!! !!!! !!!!!!! !!!! !!!!!!! !!!!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! Parameters 0.3 population group ! 0.2 a b c ! ! 0.1 X ∼ N (µ, σ ) !3 !2 !1 0 1 2 3 x Make Inference Population (N = 100) X ∼ N (µ, σ ) Sample (n = 5) Statistic X Inference € x = 2.6 µ =? € What can we say about µ based on the observed value of x? ¯ This is called Statistical Inference 1 Identify Sampling Distribution 2 Make Inference 3 Construct Confidence Intervals Formulate and Test Hypothesis Suppose, the true population mean is μ, what can you say about the distribution of values of the sample mean ¯ X ? 5
 1 Identify Sampling Distribution 2 Make Inference 3 Construct Confidence Intervals Formulate and Test Hypothesis What is a Sampling Distribution 1.5 value Population (N = 100) X ∼ N (µ, σ ) Sample 1.0 (n = 5) Statistic X 1.5 Sampling Distribution variable sample x = 2.3 € 0.5 1.0 x = 2.6 x = 2.4 0.5 ?
 !4 !2 0 population variable sample population € € 0.0 !4 value 0.0 2 4 !All
Possible
Samples!
 x It is the probability distribution of a given € statistic for all possible samples of a given size n. x It depends on (a) the underlying population, (b) the statistic and (c) the sample size. !2 0 2 4 Population (N = 5) Sample (n=1)
 Sample (n = 2) Sample (n = 2) x 1
 2 2 2 3 1
 €€ 2 2 2 3 1
 1
 1
 1
 2
 2
 2
 2
 2
 2
 1
 2
 2
 2
 3
 2
 2
 2
 3 2
 2
 2
 3
 3
 3
 1.5
 1.5
 1.5
 2
 2
 2
 2
 2.5
 2.5
 2.5
 1
 1
 1
 2 1
 1
 1
 2 2 2
 2
 2
 2
 2
 2
 2
 2 2
 2 2 2
 2 2 2 3 3 3 3 3 3 1.5 Properties of Sampling Distribution 1.0 value Ques3ons
 0.5 Sampling Distribution (n=5) variable 1.5 0.0 •  How
would
you
 draw
this?
 •  What
is
the
 shape?
 •  What
is
the
 mean?
 •  What
is
the
 standard
 !4 !2 0 x devia+on?
 sample population Ques3ons
 •  How
does
it
 depend
on
n?
 •  How
does
it
 depend
on
the
 popula+on?
 sample population 1.0 value variable 0.5 2 0.0 !4 !2 4 Sta3s3c
 x 0 2 4 Distribution of all possible values taken by the statistic when all possible samples of a fixed size n are taken from the population. Suppose that the distribution of X across the population is Normal? 10
 1.5 ¯ Distribution of X across all SAMPLES of (µ, σn N size ) σ N (µ, √n ) 1.0 value variable sample population Distribution of X across the POPULATION 0.5 N (µ, σ ) 0.0 σ N (µ, √n ) !4 !2 0 2 4 11
 x Popula+on
Distribu+on:
Normal
 When
a
variable
in
a
popula+on
is
normally
distributed,
the
sampling
 distribu3on
of
the
sample
mean
for
all
possible
samples
of
size
n
is
 also
normally
distributed.

 Sampling
distribu3on
 If
the
popula3on
is
N(µ, σ)
 then
the
sample
means
 distribu3on
is
N(µ, σ/√n).
 Popula3on
 What if the population distribution were not Normal? 13
 Central
Limit
Theorem
 Central
Limit
Theorem:
When
randomly
sampling
from
any
popula+on
 with
mean
µ
and
standard
devia3on
σ,
when
n
is
large
enough,
the
 sampling
distribu3on
of






is
approximately
normal:
~
N
(µ, σ/√n).
 Popula+on
with 
 strongly
skewed 
 distribu+on 
 Sampling
 distribu+on
of


 for
n
=
2
 observa+ons 
 Sampling 
 distribu+on
of 
 for
n
=
10 
 observa+ons 
 Sampling
 distribu+on
of

 



for
n
=
25

 observa+ons
 Law
of
Large
Numbers
 As the number of randomly drawn observations in a sample increases, the mean of the sample gets closer and closer to the population mean m. This is the law of large numbers. It is valid for any population. Note:
We
o)en
intui,vely
expect
predictability
over
a
few
random
observa,ons,
but
it
is
 wrong.
The
law
of
large
numbers
only
applies
to
really
large
numbers.

 What
if
σ
is
Unknown?
 When
σ
is
Unknown?
 The
 sample
 standard
 devia+on
 s
 provides
 an
 es+mate
 of
 the
 popula+on
 standard
devia+on
σ.
 When
 the
 sample
 size
 is
 large,
 the
 sample
 is
 likely
 to
 contain
 elements
 representa+ve
 of
 the
 whole
 popula+on.
 Then
 s
 is
 a
 good
 es+mate
of
σ.

 But
when
the
sample
size
is
small,
 the
 sample
 contains
 only
 a
 few
 individuals.
 Then
 s
 is
 a
 more
 mediocre
es+mate
of
σ.
 Popula+on
 distribu+on
 Large
sample
 Small
sample
 When n is very large, s is a very good estimate of s and the corresponding t distributions are very close to the normal distribution. The t distributions become wider for smaller sample sizes, reflecting the lack of precision in estimating s from s. The
t
Distribu+on
 Suppose
that
a
simple
random
sample
of
size
n
is
drawn
from
 an
N(µ,
σ)
popula+on.

 When
σ
is
known,
the
sampling
distribu+on
is
N(µ, σ/√n).
 When
 σ
 is
 es+mated
 from
 the
 sample
 standard
 devia+on
 s,
 the
 sampling
 distribu+on
 follows
 a
 t
 distribu3on
 with
 degrees
of
freedom
n
−
1.

 
 
 
 
 
















is
the
one‐sample
t
sta3s3c.
 Standardizing
the
Data
 As
with
the
normal
distribu+on,
the
first
step
is
to
standardize
the
 data.
Then
we
can
use
Table
D
to
obtain
the
area
under
the
curve.
 t(µ,s/√n)
 df
=
n
−
1
 s/√n
 1
 t(0,1)
 df
=
n
−
1
 µ 0
 t
 and
the
standard
error
of
the
mean
s/√n
is
its
standard
devia,on
(width).
 Here,
µ
is
the
mean
(center)
of
the
sampling
distribu,on,

 Table entry for p and C is the critical value t * with probability p lying to its right and probability C lying between −t * and t *. Probability p t* TABLE D t distribution critical values ............................................................................................................................... .................................................................................................. Upper tail probability p df 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 .25 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.706 0.703 0.700 0.697 0.695 0.694 0.692 0.691 0.690 0.689 0.688 0.688 0.687 .20 1.376 1.061 0.978 0.941 0.920 0.906 0.896 0.889 0.883 0.879 0.876 0.873 0.870 0.868 0.866 0.865 0.863 0.862 0.861 0.860 .15 1.963 1.386 1.250 1.190 1.156 1.134 1.119 1.108 1.100 1.093 1.088 1.083 1.079 1.076 1.074 1.071 1.069 1.067 1.066 1.064 .10 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 1.372 1.363 1.356 1.350 1.345 1.341 1.337 1.333 1.330 1.328 1.325 .05 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 1.833 1.812 1.796 1.782 1.771 1.761 1.753 1.746 1.740 1.734 1.729 1.725 .025 12.71 4.303 3.182 2.776 2.571 2.447 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 2.120 2.110 2.101 2.093 2.086 .02 15.89 4.849 3.482 2.999 2.757 2.612 2.517 2.449 2.398 2.359 2.328 2.303 2.282 2.264 2.249 2.235 2.224 2.214 2.205 2.197 .01 31.82 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 2.583 2.567 2.552 2.539 2.528 .005 63.66 9.925 5.841 4.604 4.032 3.707 3.499 3.355 3.250 3.169 3.106 3.055 3.012 2.977 2.947 2.921 2.898 2.878 2.861 2.845 .0025 127.3 14.09 7.453 5.598 4.773 4.317 4.029 3.833 3.690 3.581 3.497 3.428 3.372 3.326 3.286 3.252 3.222 3.197 3.174 3.153 .001 318.3 22.33 10.21 7.173 5.893 5.208 4.785 4.501 4.297 4.144 4.025 3.930 3.852 3.787 3.733 3.686 3.646 3.611 3.579 3.552 .0005 636.6 31.60 12.92 8.610 6.869 5.959 5.408 5.041 4.781 4.587 4.437 4.318 4.221 4.140 4.073 4.015 3.965 3.922 3.883 3.850 Identify Sampling Distribution 1.5 value Population (N = 100) X ∼ N (µ, σ ) Sample 1.0 (n = 5) Statistic X 1.5 Sampling Distribution variable sample population 1.0 x = 2.3 € 0.5 x = 2.6 x = 2.4 value variable sample population 0.5 € 0.0 !4 !2 0 2 4 x € Central Limit Theorem:0.0When randomly sampling from any population with mean µ and standard deviation !σ , when2 n is large enough, the 4sampling distri4 ! 0 2 x √ ¯ ¯ bution of the mean X is approximately normal, X ∼ N (µ, σ / n) € Problems 24
 Problem 1 In a large population of adults, the mean IQ is 112 with standard deviation 20. Suppose 200 adults are randomly selected for a market research campaign. The distribution of the sample mean IQ is: A) Exactly normal, mean 112, standard deviation 20 B) Approximately normal, mean 112, standard deviation 20 C) Approximately normal, mean 112 , standard deviation 1.414 D) Approximately normal, mean 112, standard deviation 0.1 Problem 2 



Safe flying weight. In response to the increasing weight of airline passengers, in 2003 the Federal Aviation Administration told airlines to assume that passengers average 190 pounds in the summer, including clothing and carry-on baggage. But passengers vary: the FAA gave a mean but not a standard deviation. A reasonable standard deviation is 35 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commute plane carries 19 passengers. What is the approximate probability that the total weight of the passengers exceeds 4000 pounds? Problem 3 
 


 
The academic motivation and study habits of female students as a group are better than those of males. The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures these factors. The distribution of SSHA scores among the women at a college has mean 120 and standard deviation 28, and the distribution of scores among the men has mean 105 and standard deviation 35. A single male student and a single female student are selected at random and given the SSHA test. Assume the scores of the two students are independent. 1.  What are the mean and standard deviation of the difference (female minus male) between their scores? 2.  What’s the probability that the woman chosen scores higher than the man? Assume the distributions are normal. Problem 4 (Mid-Term 2009) The idea of insurance is that we all face risks that are unlikely but carry high cost. Think of a fire destroying your home. So we form a group to share the risk: we all pay a small amount, and the insurance policy pays a large amount to those few of us whose homes burn down. An insurance company looks at the records for millions of homeowners and sees that the mean loss from fire in a year is $250 per house and that the standard deviation is $1000. (The distribution of losses is extremely right-skewed; most people have $0 loss, but a few have large losses.) The company plans to sell fire insurance for $250 plus enough to cover its costs and profit. (a) (6 points) Compute the mean and standard deviation of the average losses from 12 policies. Then explain why selling many thousands of such policies is a safe business. (b) (6 points) If the company sells 10,000 policies, what is the approximate probability that the average loss in a year will be greater than $275? Suppose, the true population mean is μ, what can you say about the range of values the sample mean to take? ¯ X is likely 29
 1 Identify Sampling Distribution 2 Make Inference 3 Construct Confidence Intervals Formulate and Test Hypothesis N (µ, σ ) Construct a Confidence Interval !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! !! ! !! ! !! !! !! ! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !!! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!!!! !!!! !!!! !!!! !!!! ! ! !!!!!!! ! !!!!!!! !! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! σ N (µ, √n ) 0.3 ¯ Pr(LC ≤ X ≤ UC ) = C group group a b b c c ! ! ! population 0.2 ¯ Pr(LC ≤ X ≤ UC U= C )C UC 0 0 0.1 C =1−α !1 0 1 −zα 2 0.0 !3 −zα !2 x Lα zC 3 σ µ − zα √n = LC zα µ σ UC = µ + zα √n 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 50 60 80 100 1000 z∗ 0.765 0.741 0.727 0.718 0.711 0.706 0.703 0.700 0.697 0.695 0.694 0.692 0.691 0.690 0.689 0.688 0.688 0.687 0.686 0.686 0.685 0.685 0.684 0.684 0.684 0.683 0.683 0.683 0.681 0.679 0.679 0.678 0.677 0.675 0.674 50% 0.978 0.941 0.920 0.906 0.896 0.889 2.5 0.883 0.879 0.876 0.873 0.870 2.0 0.868 0.866 0.865 0.863 1.5 0.862 0.861 0.860 0.859 0.858 1.0 0.858 0.857 0.856 0.856 0.855 0.5 0.855 0.854 0.854 0.851 0.849 0.0 0.848 0.846 0.0 0.845 0.842 0.841 60% 1.250 1.190 1.156 1.134 1.119 1.108 1.100 1.093 1.088 1.083 1.079 1.076 1.074 1.071 1.069 1.067 1.066 1.064 1.063 1.061 1.060 1.059 1.058 1.058 1.057 1.056 1.055 1.055 1.050 1.047 1.045 1.043 1.042 1.037 1.036 70% 1.638 1.533 1.476 1.440 1.415 1.397 1.383 1.372 1.363 1.356 1.350 1.345 1.341 1.337 1.333 1.330 1.328 1.325 1.323 1.321 1.319 1.318 1.316 1.315 1.314 1.313 1.311 1.310 1.303 1.299 1.296 1.292 0.2 1.290 1.282 1.282 80% 2.353 2.132 2.015 1.943 1.895 1.860 1.833 1.812 1.796 1.782 1.771 1.761 1.753 1.746 1.740 1.734 1.729 1.725 1.721 1.717 1.714 1.711 1.708 1.706 1.703 1.701 1.699 1.697 1.684 1.676 1.671 1.664 0.4 1.660 1.646 1.645 90% 3.182 2.776 2.571 2.447 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 2.120 2.110 2.101 2.093 2.086 2.080 2.074 2.069 2.064 2.060 2.056 2.052 2.048 2.045 2.042 2.021 2.009 2.000 1.990 1.984 C 1.962 1.960 95% 3.482 2.999 2.757 2.612 2.517 2.449 2.398 2.359 2.328 2.303 2.282 2.264 2.249 2.235 2.224 2.214 2.205 2.197 2.189 2.183 2.177 2.172 2.167 2.162 2.158 2.154 2.150 2.147 2.123 2.109 2.099 2.088 0.6 2.081 2.056 2.054 96% 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 2.583 2.567 2.552 2.539 2.528 2.518 2.508 2.500 2.492 2.485 2.479 2.473 2.467 2.462 2.457 2.423 2.403 2.390 2.374 0.8 2.364 2.330 2.326 98% 5.841 4.604 4.032 3.707 3.499 3.355 3.250 3.169 3.106 3.055 3.012 2.977 2.947 2.921 2.898 2.878 2.861 2.845 2.831 2.819 2.807 2.797 2.787 2.779 2.771 2.763 2.756 2.750 2.704 2.678 2.660 2.639 2.626 2.581 2.576 99% 7.453 5.598 4.773 4.317 4.029 3.833 3.690 3.581 3.497 3.428 3.372 3.326 3.286 3.252 3.222 3.197 3.174 3.153 3.135 3.119 3.104 3.091 3.078 3.067 3.057 3.047 3.038 3.030 2.971 2.937 2.915 2.887 1.0 2.871 2.813 2.807 99.5% 10.21 7.173 5.893 5.208 4.785 4.501 4.297 4.144 4.025 3.930 3.852 3.787 3.733 3.686 3.646 3.611 3.579 3.552 3.527 3.505 3.485 3.467 3.450 3.435 3.421 3.408 3.396 3.385 3.307 3.261 3.232 3.195 3.174 3.098 3.091 99.8% 12.92 8.610 6.869 5.959 5.408 5.041 4.781 4.587 4.437 4.318 4.221 4.140 4.073 4.015 3.965 3.922 3.883 3.850 3.819 3.792 3.768 3.745 3.725 3.707 3.690 3.674 3.659 3.646 3.551 3.496 3.460 3.416 3.390 3.300 3.291 99.9% z 32
 ............................................................................................................................... .................................................................................................. T-11 Confidence level C Confidence Interval ¯ C=1-α Pr(LC ≤ X ≤ UC ) = C UC 0 −zα zα ¯ Pr(LC ≤ X ≤ UC ) = C UC α/2 −zα zα 0 zα α/2 33
 What can you now say about the population mean, μ, based on the ¯ sample mean X ? 34
 Discussion Questions •  What
is
the
interpreta+on
of
a
Confidence
Interval?
 •  What
is
the
length
of
a
Confidence
Interval?
 •  How
does
sample
size
affect
a
Confidence
Interval?
 •  How
to
Solve
Problems
involving
Confidence
Intervals?
 35
 Now suppose I claim that the population mean is µ = 5. How ,
 would you go about testing my claim? 36
 1 Identify Sampling Distribution 2 Make Inference 3 Construct Confidence Intervals Formulate and Test Hypothesis What is a Hypothesis? A hypothesis is an assumption or a theory about the characteristics of one or more variables in one or more populations. Is the average weight of candies in the jar equal to 5 g ? Is the population mean µ for the distribution of weights of candies equal to 5g?
 Formulating Hypotheses The statement being tested in a test of significance is called the null hypothesis H0. It is usually a statement of ‘no effect’ or ‘no difference’. The alternative hypothesis is the statement we suspect is true instead of the null hypothesis. It is labeled Ha or H1.
 A hypothesis test or test of significance is designed to assess the strength of the evidence against the null hypothesis. 1 Confidence Interval Method 2 Test Hypothesis 3 Test Statistic Method p-Value Method 1 Confidence Interval Method Fail to Reject H0 0.3 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! !! ! !! ! !! !! !! ! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !!! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!!!! !!!! !!!! !!!! !!!! ! ! !!!!!!! ! !!!!!!! !! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! Fail to RejectFail to Reject H0 H0 0.2 group group a b b c c ! ! ! population H0 : µ = µ0 0.1 α 2 Ha : µ = µ0 0.0 z ∗ C =1−α !1 0 1 α 2 !3 −z ∗ !2 x LC µ0 L∗ zC −z ∗ UC x ¯ 2 3 1 Confidence Interval Method 2 Test Hypothesis 3 Test Statistic Method p-Value Method 2 Test Statistic Method Fail to Reject H0 0.3 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! !! ! !! ! !! !! !! ! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !!! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!!!! !!!! !!!! !!!! !!!! ! ! !!!!!!! ! !!!!!!! !! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! Fail to RejectFail to Reject H0 H0 0.2 group group a b b c c ! ! ! population H0 : µ = µ0 0.1 α 2 Ha : µ = µ0 0.0 z ∗ C =1−α !1 0 1 α 2 !3 −z ∗ !2 x µ0 z∗ z −z ∗ 2 3 x ¯ 1 Confidence Interval Method 2 Test Hypothesis 3 Test Statistic Method p-Value Method 3 p-Value Method Fail to Reject H0 0.3 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! !! ! !! ! !! !! !! ! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !!! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!!!! !!!! !!!! !!!! !!!! ! ! !!!!!!! ! !!!!!!! !! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! Fail to RejectFail to Reject H0 H0 population H0 : µ1 − µ2 = d0 0.2 Ha : µ1 − µ2 = d0 x1 − x2 ¯ ¯ p 2 H0 : µ1 − µ2 = d0 Ha : µ1 − µ2 = d0 ! ! group group a b b c c ! H0 : µ = µ0 0.1 α 2 Ha : µ = µ0 0.0 z ∗ C =1−α !1 0 1 α 2 x1 − x2 ¯ ¯ p 2 !3 −z ∗ !2 x µ0 z∗ z −z ∗ 2 3 x ¯ Summary of Hypothesis Testing Fail to Reject H0 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! Fail to RejectFail to Reject H0 H0 7
 0.3 Reach a H Conclusion0 population 0.2 : µ1 − µ2 = d0 1
 Ha : Formulate µ1 − µ2 =ypothesis H d0 0.1 x1 − x2 ¯ ¯ p 2 H0 : µ = µ0 2
 Choosegroup group a b Significance bc H c: Level a ! ! ! H0 4
 µ1 − µ2 = d0 : µ1 − µ2 = d0 Compute p-Value Ha : µ = µ0 Identify Test 0 Statistic x α 2 α 2 x1 6
 x2 ¯ −¯ p 2 5
 Identify !3 Critical Region 0.0 z∗ −z ∗ !2 !1 1 µ0 z∗ z −z ∗ 2 3 3
 x ¯ Compute Test Score 1
 How to Identifying the Null ? •  Larry’s
 car
 averages
 31
 miles
 per
 gallon
 on
 the
 highway.
 He
 now
 switches
 to
 a
 new
 motor
oil
that
is
adver+sed
as
increasing
gas
mileage.
Aaer
driving
2500
highway
miles
 with
the
new
oil,
he
wants
to
determine
if
his
gas
mileage
actually
has
increased.
 •  The
 diameter
 of
 a
 spindle
 in
 a
 small
 motor
 is
 supposed
 to
 be
 4
 millimeters.
 If
 the
 spindle
 is
 either
 too
 small
 or
 too
 big,
 the
 motor
 will
 not
 perform
 properly.
 The
 manufacturer
measures
the
diameter
in
a
sample
of
motors
to
determine
whether
the
 mean
diameters
has
moved
away
from
the
target.
 •  The
mean
area
of
the
several
thousand
apartments
in
a
new
development
is
adver+sed
 to
 be
 1400
 square
 feet.
 A
 tenant
 group
 thinks
 that
 the
 apartments
 are
 smaller
 than
 adver+sed.
 They
 hired
 an
 engineer
 to
 measure
 a
 sample
 of
 apartments
 to
 test
 their
 suspicion.
 2
 How to Identify Test Statistic? 1
 Choice of Test Statistic is based on: (a) Number of Samples (a) The Population Parameter (b)  Knowledge about Population Formulate Hypothesis H0 : µ = µ0 Ha : µ = µ0 Z= ¯ X −µ0 √ σ/ n A Test Statistic is a measure of how far the observed sample is from what is expected based on the NULL µ0 x ¯ 2
 …and its Distribution !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! N (0, 1) 0.3 population group ! 0.2 a b c ! ! 0.1 !3 !2 !1 Z= x ¯ X −µ0 √ σ/ n 0 1 2 3 3
 How to Compute Test Score Statistic (St) Test Statistic (TS) Parameter (Pa) z= x−µ0 ¯√ σ/ n Std. Error (Se) 4
 How to Choose Significance Level !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! 0.3 population 1
 0.2 Formulate Hypothesis H0 : µ = µ0 2
 Choose a Significance b c Level ! ! ! group 4
 0.1 Ha : µ = µ0 Identify Test 0 Statistic x α 2 !3 α 2 !2 !1 1 2 z 3 3
 µ0 x ¯ Compute Test Score Hypothesis Testing in a Courtroom Exonerate Innocent Guilty √ Type
II
error α = P(Type I error) β = P(Type II error) Convict Type
I
error √
 H0: The person is innocent Ha: The person is guilty Hypothesis Testing in Drug Introduction Not Approved Not Effective Effective √ Type II error α
=
P(Type
I
error)
 β
=
P(Type
II
error)
 Approved Type I error √ H0:
The
drug
is
not
effec+ve
 Ha:
The
drug
is
effec+ve
 4
 Choose Significance Level Fail to Reject H0 0.3 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! Fail to RejectFail to Reject H0 H0 4
 population 1
 0.2 Formulate Hypothesis H0 : µ = µ0 2
 Choose Significance Level ! ! ! group group a b b c c 0.1 Ha : µ = µ0 Identify Test 0 Statistic x α 2 !3 α 2 0.0 z∗ −z ∗ !2 !1 1 µ0 z∗ z −z ∗ 2 3 3
 x ¯ Compute Test Score 5
 Identify Critical Region Fail to Reject H0 0.3 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! Fail to RejectFail to Reject H0 H0 4
 population 1
 0.2 Formulate Hypothesis H0 : µ = µ0 2
 Identify Significance Level ! ! ! group group a b b c c 0.1 Ha : µ = µ0 Identify Test 0 Statistic x α 2 α 2 5
 Identify !3 Critical Region 0.0 z∗ −z ∗ !2 !1 1 µ0 z∗ z −z ∗ 2 3 3
 x ¯ Compute Test Score 6
 Compute p-Value One‐sided
(one‐ tailed)
test
 Two‐sided
(two‐ tailed)
test
 To
calculate
the
P‐value
for
a
two‐sided
test,
use
the
symmetry
of
the
normal
 curve.
Find
the
P‐value
for
a
one‐sided
test
and
double
it.
 Power of a Test 57
 Type I and Type II Errors Exonerate Innocent Guilty √ Type
II
error α = P(Type I error) β = P(Type II error) Convict Type
I
error √
 H0: The person is innocent Ha: The person is guilty Fail to Reject Ho Reject Ho 0.3 0.2 Ho is TRUE Type I Error (α) group a c population 1. Find critical z ∗ using α σ 2. Calculate x∗ = µ0 + z ∗ √n ¯∗ 1. Find critical z using α ¯ −µa 3. Calculate z ∗ = x /√n σ 1. Find critical = using α σ 2. Calculate xa z ∗ µ0 + z ∗ √n ¯∗ ∗∗ 4. 2. Find β using z ¯ + x∗ ¯∗ 1. Find critical z ∗ x √ α σ using 3. Calculate za =aµ0−µaz ∗ √n σ/ n ∗ 5. Calculate Powerusing α σ 1. Calculate xa =∗x as 1 − β critical z∗ ¯ √ 2. ¯ + 3. z∗ 4. Find β using z aµ0−µaz ∗ √n σ/ n ∗ 1. Find critical z ∗ using α 0.1 σ 2. Calculate x∗ = µ0 + z ∗ √n ¯ 0.0 !3 !2 !1 0 1 2 3 1. Find critical z ∗ using α 2. Calculate x = µ0 + µ0 µa ¯ ∗ za x∗ −µa = ¯ /√ n σ Calculate z ∗ a ∗ x 3. Calculate 0.3 σ z ∗ √n population 4. Find Ha is TRUE 0.2 Type II Power ∗ β Eusing )za rror (β = ∗∗ 4. Find β usingzza
1
- β α 1. Find critical using group a c 3. = x∗ −µa ¯√ σ/ n 2. x∗ ¯∗ + ∗σ 4. 5. Calculate Power¯ 0−µ1z−√n 3. Find β using =aµas n β za z ∗ x / √ a σ ∗ 3. 5. Calculate za =ax asµ1 − β Powerσ − a 4. Find β using z ∗ ¯ /√n ∗ 4. Find β using za 5. Calculate Power as 1 − β ∗ ∗ 5. Calculate Power as 1 − β ∗ 0.1 0.0 !3 σ 5. Calculate x = µ0as z ∗−n 2. Calculate Power + 1 √ β ¯ ∗ 3. Calculate za = !2 !1 0 1 2 5. Calculate Power as 1 − β x x∗ −µa ¯√ σ/ n 3 ∗ 4. Find β using za Problems 60
 Problem
1:
Formulate
Hypotheses
 •  Larry’s
 car
 averages
 31
 miles
 per
 gallon
 on
 the
 highway.
 He
 now
 switches
 to
 a
 new
 motor
oil
that
is
adver+sed
as
increasing
gas
mileage.
Aaer
driving
2500
highway
miles
 with
the
new
oil,
he
wants
to
determine
if
his
gas
mileage
actually
has
increased.
 •  The
 diameter
 of
 a
 spindle
 in
 a
 small
 motor
 is
 supposed
 to
 be
 4
 millimeters.
 If
 the
 spindle
 is
 either
 too
 small
 or
 too
 big,
 the
 motor
 will
 not
 perform
 properly.
 The
 manufacturer
measures
the
diameter
in
a
sample
of
motors
to
determine
whether
the
 mean
diameters
has
moved
away
from
the
target.
 •  The
mean
area
of
the
several
thousand
apartments
in
a
new
development
is
adver+sed
 to
 be
 1400
 square
 feet.
 A
 tenant
 group
 thinks
 that
 the
 apartments
 are
 smaller
 than
 adver+sed.
 They
 hired
 an
 engineer
 to
 measure
 a
 sample
 of
 apartments
 to
 test
 their
 suspicion.
 Problem 2 A phone industry manager thinks that customer monthly cell phone bills have increased, and now average more than $52 per month. The company wishes to test this claim. A random sample of 64 monthly bills has a mean of $53.1. Past company records indicate that the standard deviation is about $10. Suppose that α = .10 is chosen for this test. 62
 Problem 3 The weight of individual candies in the pack is Normally distributed N(65g, 5g). Think of a carton of 12 candies as a random sample of size 12. What is the sampling distribution of the means? Find the middle 95% of the distribution of the sample means. Assume
 that
 μ
 is
 unknown
 but
 σ
 =5
 is
 known.
 You
 buy
 a
 carton
 of
 12
 bars.
 The
 box
 weighs
 770g.
 The
 average
 bar
 weight
 from
 that
 sample
 is
 calculated
 to
 be
 64.2g.
 What
 can
 you
 infer
 about
 the
 mean
 µ
 of
 this
 chocolate
 bar
 popula+on?

 63
 Backup 64
 A
two‐tail
or
two‐sided
test
of
the
popula+on
mean
has
these
null
and
 alterna+ve
hypotheses:
 H0
:

µ
=
[a
specific
number]


Ha
:

µ
≠
[a
specific
number]
 Tes+ng
a
Hypothesis
 A
one‐tail
or
one‐sided
test
of
a
popula+on
mean
has
these
null
and
 alterna+ve
hypotheses:
 H0
:


µ
=
[a
specific
number]


Ha
:


µ
<
[a
specific
number]





OR
 H0
:


µ
=
[a
specific
number]


Ha
:


µ
>
[a
specific
number]
 The
FDA
tests
whether
a
generic
drug
has
an
absorp+on
extent
similar
to
the
 known
absorp+on
extent
of
the
brand‐name
drug
it
is
copying.
Higher
or
lower
 absorp+on
would
both
be
problema+c,
thus
we
test:
 



H0
:
µgeneric
=
µbrand
 
Ha
:
µgeneric
≠
µbrand
 
two‐sided
 Formulate and Test Hypothesis Fail to Reject H0 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! Fail to RejectFail to Reject H0 H0 7
 0.3 Reach a H Conclusion0 population 0.2 : µ1 − µ2 = d0 1
 Ha : Formulate µ1 − µ2 =ypothesis H d0 0.1 x1 − x2 ¯ ¯ p 2 H0 : µ = µ0 2
 Identifygroup group a b Significance bc H c: Level a ! ! ! H0 4
 µ1 − µ2 = d0 : µ1 − µ2 = d0 Compute p-Value Ha : µ = µ0 Identify Test 0 Statistic x α 2 α 2 x1 6
 x2 ¯ −¯ p 2 5
 Identify !3 Critical Region 0.0 z∗ −z ∗ !2 !1 1 µ0 z∗ z −z ∗ 2 3 3
 x ¯ Compute Test Score Formulate Hypothesis !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! 0.3 population 1
 0.2 Formulate Hypothesis group group a b b c c ! ! ! H0 : µ = µ0 0.1 Ha : µ = µ0 0.0 !3 !2 !1 0 1 2 3 x µ0 Identifying the Null Hypothesis •  Larry’s
 car
 averages
 31
 miles
 per
 gallon
 on
 the
 highway.
 He
 now
 switches
 to
 a
 new
 motor
oil
that
is
adver+sed
as
increasing
gas
mileage.
Aaer
driving
2500
highway
miles
 with
the
new
oil,
he
wants
to
determine
if
his
gas
mileage
actually
has
increased.
 •  The
 diameter
 of
 a
 spindle
 in
 a
 small
 motor
 is
 supposed
 to
 be
 4
 millimeters.
 If
 the
 spindle
 is
 either
 too
 small
 or
 too
 big,
 the
 motor
 will
 not
 perform
 properly.
 The
 manufacturer
measures
the
diameter
in
a
sample
of
motors
to
determine
whether
the
 mean
diameters
has
moved
away
from
the
target.
 •  The
mean
area
of
the
several
thousand
apartments
in
a
new
development
is
adver+sed
 to
 be
 1400
 square
 feet.
 A
 tenant
 group
 thinks
 that
 the
 apartments
 are
 smaller
 than
 adver+sed.
 They
 hired
 an
 engineer
 to
 measure
 a
 sample
 of
 apartments
 to
 test
 their
 suspicion.
 p-value in One-Sided and Two-Sided tests One‐sided
(one‐ tailed)
test
 Two‐sided
(two‐ tailed)
test
 To
calculate
the
P‐value
for
a
two‐sided
test,
use
the
symmetry
of
the
normal
 curve.
Find
the
P‐value
for
a
one‐sided
test
and
double
it.
 Confidence Level C = 1 - α √ µ − zα σ / n ≤ x ≤ √ + zα σ / n ¯µ µ − zα σ / n ≤ x ≤ µ + zα σ / n ¯ √ √ µ − √ σ / n ≤ x ≤√ + zα σ / n zα ¯µ µ + zα σ / n ≤ x + z σ / n µ¯ α √ µ ≤ x + zα σ / √ ¯ n √ √ µ − zα σ / xn ≤α¯ / µ + zµ σ / n ¯−z x≤ n≤ α σ √ x − zα σ / √ ≤ µ ¯ n µ ≤ x + zα σ / n ¯ √ √ x − zα σ / n ≤ µ ≤ x + zα σ / n ¯ ¯ √ √ 70
 Solving Problems on CI 1. Find zα such that Pr(Z ≤ zα ) = 1+C 2 from the Normal Distribution Table σ σ 2. The Confidence Interval (CI) for Level C is given by x − zα √n , x − zα √n ¯ ¯ σ 3. The Length of the CI is 2zα √n 71
 Inference Up to now: Given a sequence of random variables with distribution F, what can we say about the mean of a sample? What we really want: Given the mean of a sample, what can we say about the underlying sequence of randomvariables? 72
 Inference Up to now: Given a sequence of random variables with distribution F, what can we say about the mean of a sample? What we really want: Given the mean of a sample, what can we say about the underlying sequence of random variables? 73
 Marijuana Up to now: If I told you that smoking pot was Binomially distributed with probability p, you could tell me what I would expect to see if I sampled n people at random. What we want: If I sample n people at random and find out m of them smoke pot, what does that tell me about p? 74
 interval from one experiment 12 11 10 9 8 Horizontal line = true value 50 100 150 200 expt 12 11 10 9 8 50 100 150 200 75
 expt value 12 11 10 9 8 50 100 150 200 expt There are 13 red lines and 200 experiments. Is this an ok interval? 76
 ...
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This note was uploaded on 08/31/2010 for the course MANAGEMENT MGCR 271 taught by Professor Vaidyanathan during the Summer '10 term at McGill.

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