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Transcript
Ch7 Statistics
Dr. Deshi Ye
[email protected]
Outline
State what is estimated
Point estimation
Interval estimation
Compute sample size
Tests of Hypotheses
Null Hypothesis and Tests of Hypotheses
Hypotheses concern one mean
2/139
Why Statistics?
The purpose of most statistical
investigations is to generalize from
information contained in random samples
about the populations from which the
samples were obtained.
How: estimation and tests of Hypothesis
3/139
Thinking
Suppose you’re interested in the
average amount of money that
students in this class (the population)
have on them. How would you find
out?
4/139
Introduction to Estimation
5/139
Statistical Methods
Statistical
Statistical
Methods
Methods
Descriptive
Descriptive
Statistics
Statistics
Inferential
Inferential
Statistics
Statistics
Estimation
Estimation
Hypothesis
Hypothesis
Testing
Testing
6/139
Estimation Process
Population


Mean, , is
unknown

 
Random Sample
Mean 
X = 50
I am 95%
confident that
 is between
40 & 60.
Sample


 
7/139
Estimation Methods
Estimation
Point
Estimation
Interval
Estimation
8/139
Point estimation
We want to know the mean of a population. However, it
is unavailable.
Hence, we choose a sample data, and calculate from the
choosing sample data. Then estimate that the population
is also the same mean.
Point estimation: concerns the choosing of a statistic,
that is, a single number calculated from sample data for
which we have some expectation, or assurance, that is
reasonably close the parameter it is supposed to
estimate.
9/139
Point Estimation
1. Provides Single Value
Based on Observations from 1 Sample
2. Gives No Information about How Close
Value Is to the Unknown Population
Parameter
3. Example: Sample MeanX = 3 Is Point
Estimate of Unknown Population Mean
10/139
Point estimation of a mean
Parameter: Population mean 
Data: A random sample X1 , X 2 , , X n
Estimator:
X
Estimate of standard error:
S
n
11/139
EX
Scientists need to be able to detect small
amounts of contaminants in the
environment. Sample data is listed as
follows:
2.4 2.9 2.7 2.6 2.9 2.0 2.8 2.2 2.4 2.4 2.0
2.5
12/139
EX of point estimation
Compute the point estimator X
and

estimate its standard deviation n (also
called the estimated standard error of X ).
Solution:
29.8
x
 2.483
12
n
s2 
 ( xi  x )
i 1
n 1
n
2

2
2
x

nx
i
i 1
n 1
75.08  (29.8) 2 /12

 0.09788
12  1
s
 0.09788 /12  0.090
Hence the estimated standard deviation is
13/139
n
Bias or Unbias
Question: is the estimation good enough?
EX.
In previous ex, as a check on the current capabilities, the
measurements were made on test specimens spiked
with a known concentration 1.25 ug/l of lead. That is the
readings should average 1.25 if there is no background
lead in the samples.
There appears to be either a bias due to laboratory
procedure or some lead already in the samples before
they were spiked.
14/139
Unbiased estimator
Let
 be the parameter of interest and
ˆ be a statistic.
A statistic ˆ is said to be an unbiased
estimator, or its value an unbiased
estimate, if and only if the mean of the
sampling distribution of the estimator
equals  , whatever the value of  .
15/139
More efficient unbiased estimator
Estimator is not unique: for example: it can be shown
that for a random sample of size n=2, the mean X 2 X
as well as the weighted mean aXa  bX
where a, b are
b
positive constants, are unbiased estimates of the mean
of the population.
1
1
2
2
A statistics ˆ1 is said to be a more efficient unbiased
estimator of the parameter ˆ2 than the statistics if
1. ˆ1 and ˆ2 are both unbiased estimators of
2. the variance of the sampling distribution of the first
estimator is no larger than that of the second and is
smaller for at least one value of  .
16/139
Finding Sample Sizes
I don’t want to
sample too much
or too little!
The error
The error between the estimator and the
quantity it is supposed to estimate is: X  
X 
/ n
is a random variable having approximately
the standard normal distribution
We could assert with probability 1  
inequality
 z / 2
Remind that
that the
X 

 z / 2
/ n
P(Z  z / 2 )  1   / 2
18/139
Maximum error of estimate
The error
| X  |
will be less than
E  z / 2 
with probability
Specially,
1

n
.
z / 2  1.96, when   0.05
z / 2  2.575, when   0.01
19/139
EX
An industrial engineer intends to use the mean
of a random sample of size n=150 to estimate
the average mechanical aptitude of assembly
line workers in a large industry. If, on the basis
of experience, the engineer can assume that   6.2
for such data, what can he assert with probability
0.99 about the maximum size of his error.
6.2
E  2.575 
 1.3
150
Thus, the engineer can assert with probability 0.99 that his error is at most
1.3
20/139
Determine of sample size
Suppose that we want to use the mean of
a large random sample to estimate the
mean of a population, and want to be able
to assert with probability 1   that the
error will be at most some prescribed
quantity E. As before, we get
 z / 2   2 
n  (
) 
E


21/139
EX
A research worker want to determine the
average time it takes a mechanic to rotate
the tires of a car, and she wants to be able
to assert with 95% confidence that the
mean of her sample is off by at most 0.5
minute. If she can presume from past
experience that   1.6 minutes, how large a
sample will she have to take?
1.96 1.6 2
n(
)  39.3
0.5
22/139
Continuous
The method discussed so far requires  be
known or it can be approximated with the
sample standard deviation s, thus requiring that
n be large.
Another approach: if it is reasonable to assume
that we are sampling from a normal population,
we get
X 
t
S/ n
is a random variable having the t distribution with
n-1 degree of freedom.
23/139
EX
The first example, n=12,
t0.01  2.718
s
 0.09788 /12  0.090
n
For n=11 degrees of freedom
s
t0.01 
 2.718  0.09  0.2
n
Thus, one can assert with 98% confidence that
the maximum error is within 0.2
X    2.483  1.25  1.233  0.2
24/139
Estimation Methods
Estimation
Point
Estimation
Interval
Estimation
25/139
Interval Estimation
1. Provides Range of Values
Based on Observations from 1 Sample
2. Gives Information about Closeness to
Unknown Population Parameter
Stated in terms of Probability
3. Example: Unknown Population Mean
Lies Between 50 & 70 with 95% Confidence
26/139
Interval Estimation
Sample statistic
(point estimate)
27/139
Interval Estimation
Confidence
interval
Confidence
limit (lower)
Sample statistic
(point estimate)
Confidence
limit (upper)
28/139
Interval Estimation
A probability that the population parameter
falls somewhere within the interval.
Confidence
interval
Confidence
limit (lower)
Sample statistic
(point estimate)
Confidence
limit (upper)
29/139
Confidence Interval Estimates
Confidence
Intervals
Mean
Known
Proportion
Variance
 Unknown
30/139
Confidence Interval Mean ( Known)
Assumptions
Population standard deviation is known
Population is normally distributed
If not normal, can be approximated by normal
distribution (n  30)
Confidence Interval Estimate
X  Z / 2 

n
   X  Z / 2 

n
31/139
Interval Estimation
Interval estimation: with intervals for which
we can assert with a reasonable degree of
certainty that they will contain the
parameter under consideration.
For a large random sample (n > 30) from a
population with the unknown mean and
the known variance. When the observed
value x becomes available, we obtain
x  z / 2 

n
   x  z / 2 

n
32/139
Confidence interval
We can claim with
that the interval
[ x  z / 2 
(1   )100%

n
, x  z / 2 
confidence

n
]
Contains 
It is customary to refer to an interval of this kind as a
confidence interval for  having the degree of (1   )100%
confidence
33/139
Confidence Interval Estimates
Confidence
Intervals
Mean
Known
Proportion
Variance
 Unknown
34/139
Solution for Small Samples
1. Assumptions
Population of X Is Normally Distributed
Use Student’s t Distribution
1. Define variable
X 
T
s/ n
2. T has the Student distribution with n -1 degrees of
freedom (When X is normally distributed)
• There’s a different Student distribution for different degrees
of freedom
• As n gets large, Student distribution approximates a normal
distribution with mean = 0 and sigma = 1
35/139
Small sample (n<30)
Small sample and we assume to get
sampling from a normal distribution
population.
We get the (1   )100% confidence interval
formula
s
s
x  t / 2 
   x  t / 2 
n
n
36/139
EX.
The mean weight loss of n=16 grinding
balls after a certain length of time in mill
slurry is 3.42 grams with a standard
deviation of 0.68 gram. Construct a 99%
confidence interval for the true mean
weight loss of such grinding balls under
the stated condition.
37/139
Confidence Depends on Interval (z)
X =  ± Zx
x_
-2.58x
-1.65x
-1.96x

+1.65x
+2.58x
+1.96x
X
90% Samples
95% Samples
99% Samples
38/139
Confidence Level
1. Probability that the Unknown
Population Parameter Falls Within Interval
2. Denoted (1 - 
 Is Probability That Parameter Is Not Within
Interval
3. Typical Values Are 99%, 95%, 90%
39/139
Intervals & Confidence Level
Sampling
Distribution /2
of Mean
x_
1 -
/2
x = 
_
X
(1 - ) % of
intervals
contain .
Intervals
extend from
X - ZX to
X + ZX
 % do not.
Intervals derived from
many samples
40/139
Confidence interval & level
It is useful to think confidence intervals as a range of
"plausible" values for the parameter.
confidence interval is different from interpreting the
confidence level
suppose we've taken a random sample of 10 ice-cream cones,
and determined that a 95% confidence interval for the mean
caloric contents of a single scoop of ice-cream is (260,310).
Interpret the confidence level: If we repeatedly took samples of
size 10 and then formed confidence intervals, we would expect
95% of them to contain the true (but unknown) mean.
Interpret this particular confidence interval: we are 95% confident
that the true mean caloric content lies between 260 and 310.
41/139
Confidence interval & level
The wider the confidence interval you are willing
to accept, the more certain you can be that the
whole population answers would be within that
range.
For example, if you asked a sample of 1000
people in a city which brand of cola they
preferred, and 60% said Brand A, you can be
very certain that between 40 and 80% of all the
people in the city actually do prefer that brand,
but you cannot be so sure that between 59 and
61% of the people in the city prefer the brand.
42/139
Factors Affecting Interval Width
1. Data Dispersion
Measured by 
2. Sample Size
Intervals Extend from
X - ZX toX + ZX
x  / n
3. Level of Confidence
(1 - )
Affects Z
© 1984-1994 T/Maker
Co.
43/139
Statistical Methods
Statistical
Methods
Descriptive
Statistics
Inferential
Statistics
Estimation
Hypothesis
Testing
44/139
Hypothesis Testing
Population


 


I believe the population
mean age is 50
(hypothesis).
Reject
hypothesis!
Not close.

Random
sample
Mean 
X = 20
45/139
What’s a Hypothesis?
A belief about a
population parameter
I believe the mean GPA
of this class is 3.8!
Parameter is
Population mean,
proportion, variance
Must be stated
before analysis
© 1984-1994 T/Maker Co.
46/139
Tests of Hypotheses
Suppose that a consumer protection agency
wants to test a paint manufacturer’s claim that
the average drying time of his new “fast-drying”
paint is 20 minutes.
It instructs a member of its research staff to paint
each of 36 boards using a different 1-gallon can
of the paint, with the intention of rejecting the
claim if the mean of the drying time exceeds
20.75 minutes. Otherwise, it will accept the claim.
Question: Is it a infallible criterion for accepting
or rejecting the claim?
47/139
EX cont.
Assuming that it is known from past experience that the
standard deviation   2.4
The probability of erroneously rejecting the hypothesis that   20
Let us investigate the possibility that the sample may exceed 20.75
minutes even though the true mean is 20 minutes
z
20.75  20
 1.875
2.4 / 36
Accept the claim
that   20
Reject the claim
that   20
0.0304
x Minutes
  20
20.75
48/139
Another possibility
The procedure fails to detect that
  20
Suppose that the true mean of drying time is   21
Reject the claim
that   20
Accept the claim
that   20
z
20.75  21
 0.625
2.4 / 36
0.2660
20.75
  21
49/139
Type of errors
H: the hypothesis. Ex.
Accept H
  20
Reject H
H is true
Correct decision Type I error
H is false
Type II error
Correct decision
50/139
Type I error: If hypotheses is true but
rejected. Denoted by the letter 
EX.
  0.0304
Type II error: If hypotheses is false but not
rejected. Denoted by the letter 
EX.   0.2660
51/139
7.4 Null Hypotheses
Question: Can we formulate   20 minutes, where
can take on more than one possible value?

Null Hypotheses H 0 : (Pronounced H-nought) Usually
require that we hypothesize the opposite of what we
hope to prove.
EX. If we want to show that one method of teaching computer
programming is more efficient than another, we hypothesize that
the two methods are equally effective.
The null hypothesis proposes something initially
presumed true. It is rejected only when it becomes
evidently false.
52/139
Null Hypothesis
1. What is tested
2. has serious outcome if incorrect
decision made
3. Designated H0
4. Specified as H0:   Some Numeric
Value
Specified with = Sign Even if , or 
Example, H0:   3
53/139
Alternative Hypothesis
1. Opposite of Null Hypothesis
2. Always Has Inequality Sign: ,, or 
3. Designated Ha
4. Specified Ha:  < Some Value
Example, Ha:  < 3
 will lead to two-sided tests
<, > will lead to one-sided tests
54/139
Alternative
One-sided alternative: In the drying time
example, the null hypothesis is   20
minutes and the alternative hypothesis is
  20
Two-sided alternative:   0 where 0 is
the value assumed under the null
hypothesis.
55/139
Selecting the null hypothesis
Guideline for selecting the null hypothesis:
When the goal of an experiment is to
establish an assertion, the negation of the
assertion should be taken as the null
hypothesis. The assertion becomes the
alternative hypothesis.
56/139
Identifying Hypotheses Steps
1. Example Problem: Test That the
Population Mean Is Not 3
2. Steps
State the Question Statistically (  3)
State the Opposite Statistically ( = 3)
Must Be Mutually Exclusive & Exhaustive
Select the Alternative Hypothesis (  3)
Has the , <, or > Sign
State the Null Hypothesis ( = 3)
57/139
What Are the Hypotheses?
Is the population average amount of TV
viewing 12 hours?
State the question statistically:  = 12
State the opposite statistically:   12
Select the alternative hypothesis: Ha:  
12
State the null hypothesis: H0:  = 12
58/139
What Are the Hypotheses?
Is the population average amount of TV
viewing different from 12 hours?
State the question statistically:   12
State the opposite statistically:  = 12
Select the alternative hypothesis: Ha:  
12
State the null hypothesis: H0:  = 12
59/139
What Are the Hypotheses?
Is the average cost per hat less than or
equal to $20?
State the question statistically:   20
State the opposite statistically:   20
Select the alternative hypothesis: Ha:  
20
State the null hypothesis: H0:   20
60/139
What Are the Hypotheses?
Is the average amount spent in the
bookstore greater than $25?
State the question statistically:   25
State the opposite statistically:   25
Select the alternative hypothesis: Ha:  
25
State the null hypothesis: H0:   25
61/139
Decision Making Risks
Errors in making decision
1. Type I Error
Reject True Null Hypothesis
Has Serious Consequences
Probability of Type I Error Is (Alpha)
Called Level of Significance
2. Type II Error
Do Not Reject False Null Hypothesis
Probability of Type II Error Is (Beta)
63/139
Decision Results
H0: Innocent
Jury Trial
H0 Test
Actual Situation
Actual Situation
Verdict
Innocent Guilty Decision H0 True
Innocent Correct
Guilty
Error
Error
Do Not
Reject
H0
Correct
Reject
H0
1-
H0
False
Type II
Error
()
Type I Power
Error ()
(1 - )
64/139
Decision Results
H0: Innocent
Jury Trial
H0 Test
Actual Situation
Verdict
Innocent Guilty Decision H0 True
Innocent Correct
Guilty
Actual Situation
Error
Error
Accept
H0
Correct
Reject
H0
1-
H0
False
Type II
Error
()
Type I Power
Error () (1 - )
65/139
 &  have an inverse relationship
You can’t reduce both
errors simultaneously!


66/139
Factors Affecting 
1. True Value of Population Parameter
 Increases When Difference With Hypothesized
Parameter Decreases
2. Significance Level, 
 Increases When Decreases
3. Population Standard Deviation, 
 Increases When  Increases
4. Sample Size, n
 Increases When n Decreases
67/139
Hypothesis Testing
Population


 


I believe the population
mean age is 50
(hypothesis).
Reject
hypothesis!
Not close.

Random
sample
Mean 
X = 20
68/139
Basic Idea
69/139
Basic Idea
Sampling Distribution
 = 50
H0
Sample Mean
70/139
Basic Idea
Sampling Distribution
It is unlikely
that we would
get a sample
mean of this
value ...
20
 = 50
H0
Sample Mean
71/139
Basic Idea
Sampling Distribution
It is unlikely
that we would
get a sample
mean of this
value ...
... if in fact this were
the population mean
20
 = 50
H0
Sample Mean
72/139
Basic Idea
Sampling Distribution
It is unlikely
that we would
get a sample
mean of this
value ...
... therefore,
we reject the
hypothesis
that  = 50.
... if in fact this were
the population mean
20
 = 50
H0
Sample Mean
73/139
Hypothesis testing
1. We formulate a null hypothesis and an appropriate alternative
hypothesis which we accept when the null hypothesis must be
rejected.
2. We specify the probability of a Type I error. If possible, desired, or
necessary, we may also specify the probabilities of Type II errors for
particular alternatives.
3. Based on the sampling distribution of an appropriate statistic, we
construct a criterion for testing the null hypothesis against the given
alternative.
4. We calculate from the data the value of the statistic on which the
decision is to be based.
5. We decide whether to reject the null hypothesis or whether to fail
to reject it.
74/139
Level of Significance
1. Probability
2. Defines Unlikely Values of Sample
Statistic if Null Hypothesis Is True
Called Rejection Region of Sampling
Distribution
3. Designated (alpha)
4. Selected by Researcher at Start
75/139
Level of significance
The probability of a Type I error is also
called the level of significance. Usually, we
set
.
  0.05 or   0.01
Step 2 can often be performed even when
the null hypothesis specifies a range of
values for the parameter.
Ex. The null hypothesis:   20
Then we can claim that   0.0304
In general, we can only specify the maximum probability of
a Type I error, and by again the worst possibility.
76/139
Criterion
One-sided criterion (one-sided test or onetailed test): ex. One-sided alternative   20
Two-sided criterion (two-sided test or twotailed test): ex. two-sided alternative   4
77/139
Rejection Region (One-Tail Test)
78/139
Rejection Region (One-Tail Test)
Sampling Distribution
Level of Confidence
Rejection
Region

1-
Nonrejection
Region
Critical
value
H0
value
Sample Statistic
79/139
Rejection Region (One-Tail Test)
Sampling Distribution
Level of Confidence
Rejection
Region

1-
Nonrejection
Region
Critical
Value
Ho
Value
Sample Statistic
Observed sample statistic
80/139
Rejection Region (One-Tail Test)
Sampling Distribution
Level of Confidence
Rejection
Region

1-
Nonrejection
Region
Ho
Value
Critical
Value
Observed sample statistic
Sample Statistic
81/139
Rejection Region (Two-Tailed Test)
82/139
Rejection Region (Two-Tailed Test)
Sampling Distribution
Level of Confidence
Rejection
Region
Rejection
Region
1-
1/2 
1/2 
Nonrejection
Region
Critical
Value
Ho
Sample Statistic
Value Critical
Value
83/139
Rejection Region (Two-Tailed Test)
Sampling Distribution
Level of Confidence
Rejection
Region
Rejection
Region
1-
1/2 
1/2 
Nonrejection
Region
Critical
Value
Ho
Sample Statistic
Value Critical
Value
84/139
Observed sample statistic
Rejection Region (Two-Tailed Test)
Sampling Distribution
Level of Confidence
Rejection
Region
1/2 
Rejection
Region
1-
1/2 
Nonrejection
Region
Ho
Sample Statistic
Critical Value Critical
Value
Value
85/139
Observed sample statistic
Rejection Region (Two-Tailed Test)
Sampling Distribution
Level of Confidence
Rejection
Region
1/2 
Rejection
Region
1-
1/2 
Nonrejection
Region
Ho
Sample Statistic
Critical Value Critical
Value
Value
86/139
Observed sample statistic
H0 Testing Steps

State H0

Set up critical values

State Ha

Collect data

Choose 

Compute test statistic

Choose n

Make statistical decision

Choose test

Express decision
87/139
One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
c2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
88/139
7.5 Hypothesis concerning one
mean
Suppose we want to test on the basis of
n=35 determinations and at the 0.05 level
of significance whether the thermal
conductivity of a certain kind of cement
brick is 0.340, as has been claimed. We
can expect that the variability of such
determinations is given by   0.010
89/139
Solution
1) Null hypothesis:   0.340
alternative hypothesis:   0.340
2) Level of significance:   0.05
3) Criterion:
X 
Z
/ n
z
 z / 2
z / 2
90/139
Criterion Region for testing
(Normal population and known)
Alternative
hypothesis
Reject null
hypothesis if
  0
Z   z
  0
Z  z
  0
  0
Z   z / 2 or Z  z / 2
91/139
Critical values
  0.05
One-sided
alternatives
-1.645
1.645
Two-sided
alternatives
-1.96
1.96
  0.01
-2.33
2.33
-2.575
2.575
92/139
EX. Cont.
3) Criterion: Reject the null hypothesis if
Z  1.96 or Z  1.96
where
4) Calculations:
X 
Z
/ n
0.343  0.34
z
 1.77
0.01/ 35
5) Decision: The null hypothesis
cannot be rejected.
93/139
P-value
P value (or tail probability): the probability
of getting difference between x and 0
greater than or equal to that actually
observed. P(Z  z )
EX. In above example
2(1  0.9616)  0.078
½ P-value
1.77
½ P-value
1.77
94/139
P-Value
1. Probability of Obtaining a Test Statistic
More Extreme (or than Actual Sample
Value Given H0 Is True
2. Called Observed Level of Significance
Smallest Value of  H0 Can Be Rejected
3. Used to Make Rejection Decision
If p-Value  , Do Not Reject H0
If p-Value < , Reject H0
95/139
P value
P value for a given test statistic and null
hypothesis: The P value is the probability
of obtaining a value for the test statistic
that is as extreme or more extreme than
the value actually observed. Probability is
calculated under the null hypothesis.
96/139
EX
A process for producing vinyl floor
covering has been stable for a long period
of time, and the surface hardness
measurement of the flooring produced has
a normal distribution with mean 4.5 and
standard deviation 1.5. A second shift has
been hired and trained and their
production needs to be monitored.
Consider testing the hypothesis
H 0 :   4.5 versus H1 :   4.5
97/139
A random sample of hardness
measurements is made of n=25 vinyl
specimens produced by the second shift.
Calculate the P value when using the test
statistic
X 
Z
If
/ n
X  3.9
98/139
Solution
The observed value of the test statistic is
z
3.9  4.5
 2.0
1.5 / 25
Since the alternative hypothesis is two-sided, we
must consider large negative value for Z as well as
large positive values
P( Z  2.0)  P( Z  2.0)  0.0228
Consequently, the P value is 0.0456
  0.05
The small P value suggests the mean of the second
99/139
shift is not at the target value of 4.5
P value
To understand P values, you have to understand fixed level testing.
With fixed level testing, a null hypothesis is proposed (usually, specifying no
treatment effect) along with a level for the test, usually 0.05. All possible
outcomes of the experiment are listed in order to identify extreme outcomes
that would occur less than 5% of the time in aggregate if the null hypothesis
were true.
This set of values is known as the critical region. They are critical because if
any of them are observed, something extreme has occurred. Data are now
collected and if any one of those extreme outcomes occur the results are said
to be significant at the 0.05 level. The null hypothesis is rejected at the 0.05
level of significance and one star (*) is printed somewhere in a table. Some
investigators note extreme outcomes that would occur less than 1% of the
time and print two stars (**) if any of those are observed.
Many researchers quickly realized the limitations of reporting only whether a
result achieved the 0.05 level of significance. Was a result just barely
significant or wildly so? Would data that were significant at the 0.05 level be
significant at the 0.01 level? At the 0.001 level? Even if the result are wildly
statistically significant, is the effect large enough to be of any practical
importance?
100/139
P value
Observed significance level (or P value)--the smallest
fixed level at which the null hypothesis can be rejected. If
your personal fixed level is greater than or equal to the P
value, you would reject the null hypothesis.
If your personal fixed level is less than to the P value,
you would fail to reject the null hypothesis.
For example, if a P value is 0.027, the results are
significant for all fixed levels greater than 0.027 (such as
0.05) and not significant for all fixed levels less than
0.027 (such as 0.01). A person who uses the 0.05 level
would reject the null hypothesis while a person who uses
the 0.01 level would fail to reject it.
101/139
P-Value Thinking Challenge
You’re an analyst for Ford.
You want to find out if the
average miles per gallon of
Escorts is at least 32 mpg.
Similar models have a standard
deviation of 3.8 mpg. You take
a sample of 60 Escorts &
compute a sample mean of 30.7
mpg. What is the value of the
observed level of significance
(p-Value)?
102/139
p-Value Solution*
p-Value is P(Z  -2.65) = .004.
p-Value < ( = .01). Reject H0.

Use
alternative
hypothesis
to find
direction

p-Value
.004

.5000
- .4960
.0040
.4960
-2.65 0
Z value of
sample statistic

Z
From Z table:
lookup 2.65
103/139
One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
c2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
104/139
t Test for Mean ( Unknown)
1. Assumptions
Population Is Normally Distributed
If Not Normal, Only Slightly Skewed & Large
Sample (n  30) Taken
2. Parametric Test Procedure
105/139
Statistic for small sample
The test of null hypothesis   0
statistic
on the
X 
T
S/ n
106/139
  0
Criterion Region for testing
(Statistic for small sample )
Alternative
hypothesis
Reject null
hypothesis if
  0
T  t
  0
T  t
  0
T  t / 2 or T  t / 2
107/139
Two-Tailed t Test
Finding Critical t Values
Given: n = 3;  = .10
0
t
108/139
Two-Tailed t Test
Finding Critical t Values
Given: n = 3;  = .10

 /2 = .05
0
t
 /2 = .05
109/139
Two-Tailed t Test
Finding Critical t Values
Given: n = 3;  = .10
df = n - 1 = 2 

 /2 = .05
0
t
 /2 = .05
110/139
Two-Tailed t Test
Finding Critical t Values
Given: n = 3;  = .10
df = n - 1 = 2 


 /2 = .05
Critical Values of t Table
(Portion)
v
t.10
t.05
t.025
1 3.078 6.314 12.706
2 1.886 2.920 4.303
0
t
3 1.638 2.353 3.182
 /2 = .05
111/139
Two-Tailed t Test
Finding Critical t Values
Given: n = 3;  = .10
df = n - 1 = 2 


 /2 = .05
Critical Values of t Table
(Portion)
v
t.10
t.05
t.025
1 3.078 6.314 12.706
2 1.886 2.920 4.303
-2.920 0 2.920 t
 /2 = .05
3 1.638 2.353 3.182

112/139
One-Tailed t Test
You’re a marketing analyst
for Wal-Mart. Wal-Mart had
teddy bears on sale last week.
The weekly sales ($ 00) of
bears sold in 10 stores was: 8
11 0 4 7 8 10 5 8 3.
At the .05 level, is there
evidence that the average bear
sales per store is more than 5
($ 00)?
113/139
One-Tailed t Test
Solution*
H0:
Ha:
=
df =
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
114/139
One-Tailed t Test
Solution*
H0:  = 5
Ha:  > 5
=
df =
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
115/139
One-Tailed t Test
Solution*
H0:  = 5
Ha:  > 5
 = .05
df = 10 - 1 = 9
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
116/139
One-Tailed t Test
Solution*
H0:  = 5
Ha:  > 5
 = .05
df = 10 - 1 = 9
Critical Value(s):
Test Statistic:
Decision:
Reject
.05
0 1.8331
Conclusion:
t
117/139
One-Tailed t Test
Solution*
H0:  = 5
Ha:  > 5
 = .05
df = 10 - 1 = 9
Critical Value(s):
Test Statistic:
X   6.4  5
t

 1.31
S
3.373
n
10
Decision:
Reject
.05
0 1.8331
Conclusion:
t
118/139
One-Tailed t Test
Solution*
H0:  = 5
Ha:  > 5
 = .05
df = 10 - 1 = 9
Critical Value(s):
Reject
.05
0 1.8331
Test Statistic:
X   6.4  5
t

 1.31
S
3.373
n
10
Decision:
Do not reject at  = .05
Conclusion:
t
119/139
One-Tailed t Test
Solution*
H0:  = 5
Ha:  > 5
 = .05
df = 10 - 1 = 9
Critical Value(s):
Reject
.05
0 1.8331
t
Test Statistic:
X   6.4  5
t

 1.31
S
3.373
n
10
Decision:
Do not reject at  = .05
Conclusion:
There is no evidence
average is more than 5
120/139