Ch 4 - Estimation & Hypothesis One Sample
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Transcript of Ch 4 - Estimation & Hypothesis One Sample
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Chapter a
Inferences Based on a Single Sample:
Estimation with Confidence Intervals
Business Statistics
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Thinking Challenge
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?
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Introduction
to Estimation
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Statistical Methods
Statistical
Methods
EstimationHypothesis
Testing
Inferential
Statistics
Descriptive
Statistics
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Estimation Process
Mean, , isunknown
Population
Sample
Mean
X = 50
Random Sample
I am 95%
confident that
is between 40 &60.
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Unknown Population Parameters Are
Estimated
Estimate Population
Parameter...
with Sample
Statistic
Differences 1 - 2 x1 -x2
Variance 2 s 2
^Proportion p p
Mean x
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Estimation Methods
Estimation
Interval
Estimation
Point
Estimation
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Point Estimation
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Estimation Methods
Estimation
Interval
Estimation
Point
Estimation
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Point Estimation
1. Provides a single value
• Based on observations from one sample
2. Gives no information about how close the value isto the unknown population parameter
3. Example: Sample mean x = 3 is point
estimate of unknown population mean
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Interval Estimation
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Estimation Methods
Estimation
Interval
Estimation
Point
Estimation
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Interval Estimation
1. Provides a range of values
• Based on observations from one sample
2. Gives information about closeness to unknown population
parameter• Stated in terms of probability
– Knowing exact closeness requires knowing unknownpopulation parameter
3. Example: Unknown population mean lies between 50 and 70with 95% confidence
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Key Elements of
Interval Estimation
Sample statistic
(point estimate)Confidence interval
Confidence limit
(lower)
Confidence limit
(upper)
A probability that the population parameter falls
somewhere within the interval.
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Confidence Limits
for Population Mean
x
x
x x
Z X
Z Error
Error X Z
X X Error
Error X
)5(
(4)
(3)
or(2)
)1(
Parameter =
Statistic ± Error
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Many Samples Have Same Interval
x _
X
X = ± Zx
90% Samples
+1.65x-1.65x
95% Samples
+1.96x-1.96x
99% Samples
-2.58x +2.58x
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Confidence Level1. 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%
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Intervals & Confidence Level
x=
1 - /2 /2
X
_
x _
Sampling Distribution of Sample Mean
Large number of intervals
(1 – α)% of
intervals
contain μ
α% do not
Intervals
extend from
X – ZσX to
X + ZσX
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Factors Affecting
Interval Width
1. Data dispersion
• Measured by Intervals extend from
X – ZX toX + ZX
© 1984-1994 T/Maker Co.
3. Level of confidence(1 – )
• Affects Z
2. Sample size
X n
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Confidence Interval Estimates
Confidence
Intervals
Mean Proportion
σ Knownσ
Unknown
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Confidence Interval Estimate
Mean ( Known)
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Confidence Interval Estimates
Confidence
Intervals
Mean Proportion
σ Knownσ
Unknown
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Confidence Interval
Mean ( Known)
1. Assumptions
• Population standard deviation is known
• Population is normally distributed
• If not normal, can be approximated by normaldistribution (n 30)
n Z X
n Z X
2/2/
2. Confidence interval estimate
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Estimation Example
Mean ( Known)
The mean of a random sample of n = 36 is X =50. Set up a 95% confidence interval estimate for if = 10.
27.5373.46
361096.150
361096.150
2/2/
n Z X
n Z X
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Thinking Challenge
You’re a Q/C inspector for
Gallo. The for 2-liter bottles
is .05 liters. A random sample
of 100 bottles showed x = 1.99
liters. What is the 90%
confidence interval estimate of
the true mean amount in 2-
liter bottles? 2 liter
© 1984-1994 T/Maker Co.
2 liter
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Confidence Interval
Solution*
998.1982.1
100
05.645.199.1
100
05.645.199.1
2/2/
n Z X
n Z X
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Confidence Interval Estimate
Mean ( Unknown)
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Confidence Interval Estimates
Confidence
Intervals
Mean Proportion
σ Knownσ
Unknown
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Confidence Interval
Mean ( Unknown)
1. Assumptions
• Population standard deviation is unknown
• Population must be normally distributed
2. Use Student’s t– distribution
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Z
t
Student’s t Distribution
0
t (df = 5)
Standard
Normal
t (df = 13)Bell-ShapedSymmetric
‘Fatter’ Tails
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Degrees of Freedom (df)
1. Number of observations that are free to vary aftersample statistic has been calculated
2. Example
Sum of 3 numbers is 6X 1 = 1 (or any number)
X 2 = 2 (or any number)
X 3 = 3 (cannot vary)
Sum = 6
degrees of freedom
= n - 1
= 3 - 1
= 2
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v t .10 t .05 t .025
1 3.078 6.314 12.706
2 1.886 2.920 4.303
3 1.638 2.353 3.182
Student’s t Table
t values
Assume:
n = 3
df = n - 1 = 2 = .10
/2 =.05
t 0
/ 2
/ 2
t 2.920
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Confidence Interval
Mean ( Unknown)
/ 2 / 2
1
S S X t X t
n n
df n
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Estimation Example
Mean ( Unknown)
/ 2 / 2
8 850 2.064 50 2.06425 25
46.69 53.30
S S X t X t
n n
A random sample of n = 25 has x = 50 and s = 8.
Set up a 95% confidence interval estimate for .
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Thinking Challenge
You’re a time study analyst in
manufacturing. You’ve
recorded the following task
times (min.):3.6, 4.2, 4.0, 3.5, 3.8, 3.1.
What is the 90% confidence
interval estimate of thepopulation mean task time?
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Confidence Interval Solution*
x = 3.7
s = 0.38987
• n = 6, df = n - 1 = 6 - 1 = 5
• t.05 = 2.015
• 0.38987 0.38987
•3.7 - 2.015 ------------ ≤ μ ≤ 3.7 + 2.015 -------------
• √ 6 √ 6
• 3.3793 ≤ μ ≤ 4.0207
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Confidence Interval Estimate of
Proportion
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Confidence Interval Estimates
Confidence
Intervals
Mean Proportion
σ Knownσ
Unknown
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Confidence Interval
Proportion
1. Assumptions
• Random sample selected
• Normal approximation can be used if
2 2
ˆ ˆ ˆ ˆ
ˆ ˆ
pq pq p z p p z
n n
2. Confidence interval estimate
ˆ ˆ15 and 15np nq
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Estimation Example
Proportion
A random sample of 400 graduates showed 32 wentto graduate school. Set up a 95% confidence
interval estimate for p.
/ 2 / 2
ˆ ˆ ˆ ˆ
ˆ ˆ
.08 .92 .08 .92.08 1.96 .08 1.96400 400
.053 .107
pq pq p Z p p Z
n n
p
p
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Thinking Challenge
You’re a production manager
for a newspaper. You want to
find the % defective. Of 200
newspapers, 35 had defects.What is the 90% confidence
interval estimate of the
population proportion defective?
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Confidence Interval
Solution*
/ 2 / 2
ˆ ˆ ˆ ˆ
ˆ ˆ
.175 (.825) .175 (.825).175 1.645 .175 1.645
200 200
.1308 .2192
p q p q p z p p z
n n
p
p
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Finding Sample Sizes
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Finding Sample Sizes
for Estimating
2 2
2 22
2
(1)
(2)
( )(3)
( )
x x
x
X SE Z
SE Z Z n
Z n
SE
SE = Sampling Error
I don’t want tosample too much
or too little!
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Sample Size Example
What sample size is needed to be 90% confident the
mean is within 5? A pilot study suggested that thestandard deviation is 45.
2 22 2
2
22
( ) 1.645 45
219.2 220( ) 5
Z
n SE
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Finding Sample Sizes
for Estimating
ˆ ˆ
ˆ2 2
22
2
ˆ
(1)
(2)
( )(3)
( )
p p
p
p p SE Z
pqSE Z Z
n
Z pqn
SE
SE = Sampling Error
If no estimate of p is
available, use p = q = .5
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Sample Size Example
What sample size is needed to estimate p with 90% confidence
and a width of .03?
(1.645) ² (0.5 0.5)
N = ----------------------- = 3006.69 ≈ 3007 (0.015)²
.03
.0152 2
width
SE
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Thinking Challenge
You work in Human Resources at
Merrill Lynch. You plan to survey
employees to find their average
medical expenses. You want to be95% confident that the sample
mean is within ± $50.
A pilot study showed that was
about $400. What sample size doyou use?
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Sample Size Solution*
2 2
2
2
2 2
2
( )
( )
1.96 400
50
245.86 246
Z n
SE
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Hypothesis Testing Concepts
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Hypothesis Testing
Population
I believe the
population meanage is 50
(hypothesis).
Mean
X = 20
Randomsample
Reject
hypothesis!
Not close.
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What’s a Hypothesis?
A belief about a population
parameter
• Parameter is
population mean,
proportion, variance
• Must be stated
before analysis
I believe the mean GPA of
this class is 3.5!
© 1984-1994 T/Maker Co.
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Null Hypothesis
1. What is tested
2. Has serious outcome if incorrect decision made
3. Always has equality sign: , , or 4. Designated H0 (pronounced H-oh)
5. Specified as H0: some numeric value
• Specified with = sign even if or • Example, H0: 3
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Alternative Hypothesis
1. Opposite of null hypothesis
2. Always has inequality sign: , , or
3. Designated Ha
4. Specified Ha: , , or some value
• Example, Ha: < 3
Id tif i H th
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Identifying Hypotheses
Steps
Example problem: Test that the population mean isnot 3
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)
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What Are the Hypotheses?
State the question statistically:
= 12 State the opposite statistically: 12
Select the alternative hypothesis: Ha: 12
State the null hypothesis: H0: = 12
Is the population average amount of TV
viewing 12 hours?
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What Are the Hypotheses?
State the question statistically:
12 State the opposite statistically: = 12
Select the alternative hypothesis: Ha: 12
State the null hypothesis: H0: = 12
Is the population average amount of TV
viewing different from 12 hours?
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What Are the Hypotheses?
State the question statistically:
20 State the opposite statistically: 20
Select the alternative hypothesis: Ha: 20
State the null hypothesis: H0: 20
Is the average cost per hat less than or equal
to $20?
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What Are the Hypotheses?
State the question statistically:
25 State the opposite statistically: 25
Select the alternative hypothesis: Ha: 25
State the null hypothesis: H0: 25
Is the average amount spent in the bookstore
greater than $25?
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Basic Idea
Sample Means = 50
H0
Sampling Distribution
It is unlikely
that we would
get a samplemean of this
value ...
20
... if in fact this were
the population mean
... therefore, we
reject thehypothesis that
= 50.
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Level of Significance
1. Probability
2. Defines unlikely values of sample statistic if null
hypothesis is true
• Called rejection region of samplingdistribution
3. Designated (alpha)
• Typical values are .01, .05, .10
4. Selected by researcher at start
Rejection Region
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Rejection Region
(One-Tail Test)
Ho
Value Critical
Value
Sample Statistic
Rejection
Region
Nonrejection
Region
Sampling Distribution
1 –
Level of Confidence
Observed sample statistic
Rejection Region
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Rejection Region
(One-Tail Test)
Sampling DistributionLevel of Confidence
Ho
Value Critical
Value
Sample Statistic
Rejection
Region
Nonrejection
Region
Sampling Distribution
1 –
Level of Confidence
Observed sample statistic
Rejection Regions
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Rejection Regions
(Two-Tailed Test)
Ho
Value Critical
Value
Critical
Value
1/2 1/2
Sample Statistic
Rejection
Region
Rejection
Region
Nonrejection
Region
Sampling Distribution
1 –
Level of Confidence
Observed sample statistic
Rejection Regions
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Ho
Value Critical
Value
Critical
Value
1/2 1/2
Sample Statistic
Rejection
Region
Rejection
Region
Nonrejection
Region
Sampling Distribution
1 –
Level of Confidence
Rejection Regions
(Two-Tailed Test)
Sampling DistributionLevel of Confidence
Observed sample statistic
Rejection Regions
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Ho
Value Critical
Value
Critical Value
1/2 1/2
Sample Statistic
Rejection
Region
Rejection
Region
Nonrejection
Region
Sampling Distribution
1 –
Level of Confidence
Rejection Regions
(Two-Tailed Test)
Sampling DistributionLevel of Confidence
Observed sample statistic
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Decision Making Risks
Errors in
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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)
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Decision Results
H0: Innocent
Jury Trial
Actual Situation
Verdict Innocent Guilty
Innocent Correct Error
Guilty Error Correct
H0 Test
Actual Situation
Decision H0 True H0
False
Accept
H0
1 – Type II
Error
()
Reject
H0
Type I
Error ()
Power
(1 – )
& Have an
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& Have an
Inverse Relationship
You can’t reduce botherrors simultaneously!
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Factors Affecting 1. True value of population parameter
• Increases when difference with hypothesizedparameter decreases
2. Significance level, • Increases when decreases
3. Population standard deviation,
• Increases when increases
4. Sample size, n
• Increases when n decreases
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Hypothesis Testing Steps
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H0 Testing Steps
State H0
State Ha
Choose
Choose n
Choose test
• Set up critical values
• Collect data
• Compute test statistic
• Make statistical decision
• Express decision
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One Population Tests
One Population
Z Test (1 & 2
tail)
t Test (1 & 2
tail)
Z Test (1 & 2
tail)
Mean Proportion Variance
2 Test (1 & 2
tail)
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Two-Tailed Z Testof Mean ( Known)
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One Population Tests
One Population
Z Test (1 & 2
tail)
t Test (1 & 2
tail)
Z Test (1 & 2
tail)
Mean Proportion Variance
2 Test (1 & 2
tail)
Two-Tailed Z Test
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Two Tailed Z Test
for Mean ( Known)
1. Assumptions
• Population is normally distributed
• If not normal, can be approximated by
normal distribution (n 30)2. Alternative hypothesis has sign
x
x
X X Z
n
3. Z-Test Statistic
Two-Tailed Z Test
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for Mean Hypotheses
H0: = 0 Ha: ≠ 0
Z0
Reject H0
/ 2 / 2
Reject H
Two-Tailed Z Test
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.500
- .025.475
Z 0
= 1
Finding Critical Z
What is Z given = .05?
/ 2 = .025
Z .05 .07
1.6 .4505 .4515 .4525
1.7 .4599 .4608 .4616
1.8 .4678 .4686 .4693
.4744 .4756
.06
1.9 .4750
Standardized Normal
Probability Table (Portion)
1.96 -1.96
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Two-Tailed Z Test Example
Does an average box of cereal
contain 368 grams of cereal? A
random sample of 64 boxes
showed x = 372.5. The companyhas specified to be 15 grams.
Test at the .05 level of
significance.
368 gm.
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Two-Tailed Z Test Solution
H0:
Ha:
n
Critical Value(s):
Test Statistic:
X - μ 372.5 – 368
Z = --------- = --------------- = 2.4
σ/√ n 15/ √ 64Decision:
Conclusion:
= 368
368
.05
64
Z0 1.96-1.96
.025
Reject H0
Reject H0
.025
Reject at = .05
No evidence average
is 368
Two-Tailed Z Test Thinking
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Two Tailed Z Test Thinking
Challenge
You’re a Q/C inspector. You want to find out
if a new machine is making electrical cords to
customer specification: average breaking
strength of 70 lb. with = 3.5 lb. You take asample of 36 cords & compute a sample
mean of 69.7 lb. At the .05 level of
significance, is there evidence that the
machine is not meeting the average breakingstrength?
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Two-Tailed Z Test Solution*
H0:
Ha:
=
n = Critical Value(s):
Test Statistic:
Decision:
Conclusion:
= 70
70
.05
36
Z0 1.96-1.96
.025
Reject H0
Reject H0
.025
69.7 70.51
3.5
36
X Z
n
Do not reject at = .05
No evidence average
is not 70
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One-Tailed Z Testof Mean ( Known)
One-Tailed Z Test
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One Tailed Z Test
for Mean ( Known)
1. Assumptions• Population is normally distributed
• If not normal, can be approximated by
normal distribution (n 30)2. Alternative hypothesis has < or > sign
3. Z-test Statistic
x
x
X X Z
n
One-Tailed Z Test
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for Mean Hypotheses
H0: = 0 Ha: < 0
Z0
Reject H0
Must be significantly
below
Z0
Reject H0
H0: = 0 Ha: > 0
Small values satisfy H0 .
Don’t reject!
One-Tailed Z Test
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.500
- .025.475
Z 0
= 1
Finding Critical Z
What Is Z given = .025?
= .025
1.96
Z .05 .07
1.6 .4505 .4515 .4525
1.7 .4599 .4608 .4616
1.8 .4678 .4686 .4693
.4744 .4756
.06
1.9 .4750
Standardized Normal
Probability Table (Portion)
One-Tailed Z Test
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Example
Does an average box of cereal
contain more than 368 grams of
cereal? A random sample of 64
boxes showed x = 372.5. Thecompany has specified to be 15
grams. Test at the .05 level of
significance.
368 gm.
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One-Tailed Z Test Solution
H0:
Ha:
=
n = Critical Value(s):
Test Statistic:
X - μ 372.5 – 368
Z = --------- = ---------------
σ/√ n 15/ √ 64= 2.4
Decision:
Conclusion:
= 368
> 368
.05
64
Z0 1.645
.05
Reject
Reject at = .05
No evidence average is 368
One-Tailed Z Test Thinking
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g
Challenge
You’re an analyst for Ford. You want tofind 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. At
the .01 level of significance, is there
evidence that the miles per gallon is atleast 32?
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One-Tailed Z Test Solution*
H0:
Ha:
=
n =
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
= 32
< 32
.01
60
Z0-2.33
.01
Reject
30.7 322.65
3.8
60
X Z
n
Reject at = .01
There is evidence average
is less than 32
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Two-Tailed t Testof Mean ( Unknown)
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One Population Tests
One Population
Z Test (1 & 2
tail)
t Test (1 & 2
tail)
Z Test (1 & 2
tail)
Mean Proportion Variance
2 Test (1 & 2
tail)
t Test for Mean
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( Unknown)
1. Assumptions• Population is normally distributed
• If not normal, only slightly skewed & large
sample (n 30) taken
2. Parametric test procedure
3. t test statistic
X t
S
n
Two-Tailed t Test
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t 0
Finding Critical t Values
Given: n = 3; = .10
/2 = .05
/2 = .05
df = n - 1 = 2
v t .10 t .05 t .025
1 3.078 6.314 12.706
2 1.886 2.920 4.303
3 1.638 2.353 3.182
Critical Values of t Table
(Portion)
2.920 -2.920
Two-Tailed t Test
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Example
Does an average box of
cereal contain 368 grams of
cereal? A random sample
of 25 boxes had a mean of
372.5 and a standard
deviation of 12 grams. Test
at the .05 level of
significance.368 gm.
Two-Tailed t Test
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Solution
H0:
Ha:
=
df = Critical Value(s):
Test Statistic:
X – μ 372.5 – 368
t = ------- = -------------- = 1.875
s/√n 12/√ 25 Decision:
Conclusion:
= 368
368
.05
25 - 1 = 24
t0 2.064-2.064
.025
Reject H0
Reject H0
.025
Do not reject at = .05
There is no evidence
population average is not
368
Two-Tailed t Test
Thi ki Ch ll
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Thinking Challenge
You work for the FTC. A
manufacturer of detergent claims
that the mean weight of detergent is
3.25 lb. You take a random sample of16 containers. You calculate the
sample average to be 3.238 lb. with a
standard deviation of .117 lb. At the
.01 level of significance, is the
manufacturer correct?3.25 lb.
Two-Tailed t Test
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Solution*
H0:
Ha:
df Critical Value(s):
Test Statistic:
X – μ 3.238 – 3.25
t = -------- = ----------------
s/√n 0.117/√16
= -0.410
Decision:
Conclusion:
= 3.25
3.25
.01
16 - 1 = 15
t0 2.947-2.947
.005
Reject H0
Reject H0
.005 Do not reject at = .01
There is no evidence
average is not 3.25
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One-Tailed t Testof Mean ( Unknown)
One-Tailed t Test
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Example
Is the average capacity of batteries at least 140 ampere-
hours? A random sample of 20
batteries had a mean of 138.47 and a standard deviation of
2.66. Assume a normal
distribution. Test at the .05 level of significance.
One-Tailed t Test
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Solution
H0:
Ha:
=
df = Critical Value(s):
Test Statistic:
Decision:
Conclusion:
= 140
< 140
.05
20 - 1 = 19
t 0-1.729
.05
Reject H0
138.47 1402.57
2.66
20
X t
S
n
Reject at = .05
There is evidence population
average is less than 140
One-Tailed t Test
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Thinking Challenge
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 of significance, is
there evidence that the average bear
sales per store is more than 5 ($00)?
One-Tailed t Test
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Solution*
H0:
Ha:
=
df = Critical Value(s):
Test Statistic:
Decision:
Conclusion:
= 5
> 5
.05
10 - 1 = 9
t0 1.833
.05
Reject H0
6.4 51.31
3.373
10
X t
S
n
Do not reject at = .05
There is no evidence
average is more than 5
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Observed Significance Levels: p-Values
V l
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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
for which H0 can be rejected3. Used to make rejection decision
• If p-value , do not reject H0
• If p-value < , reject H0
Two-Tailed Z Test
p Val e E ample
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p-Value Example
Does an average box of cereal
contain 368 grams of cereal? A
random sample of 36 boxes
showed x = 372.5. The
company has specified to be
15 grams. Find the p-Value.
368 gm.
Two-Tailed Z Test
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p-Value Solution
Z 0 1.8 Z value of sample
statistic (observed)
X - μ 372.5 - 368
Z = ---------------- = --------------------- = 1.8
σ/√ n 15/ √ 36
Two-Tailed Z Test
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1/2 p-Value 1/2 p-Value
p-Value Solution
Z value of sample
statistic (observed)
p-value is P(Z -1.80 or Z 1.80)
Z 0 1.80 -1.80
From Z table:
lookup 1.50
.4641
.5000
- .4641
.0359
Two-Tailed Z Test
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p-Value Solution
1/2 p-Value
.0359
1/2 p-Value
.0359
p-value is P(Z -1.80 or Z 1.80) = .0718
Z value of sample
statistic
From Z table:
lookup 1.50
.5000
- .4641
.0359
Z 0 1.80 -1.80
One-Tailed Z Test
p Value Example
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p-Value Example
Does an average box of cereal
contain more than 368
grams of cereal? A random
sample of 36 boxes showed x = 372.5. The company has
specified to be 15 grams.
Find the p-Value.368 gm.
One-Tailed Z Test
V l S l i
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p-Value Solution
Z 0 1.80
Z value of sample
statistic
X - μ 372.5 - 368Z = ---------------- = --------------------- = 1.8
σ/√ n 15/ √ 36
One-Tailed Z Test
V l S l ti
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p-Value Solution
p-Value
.0359
Z value of sample
statistic
From Z table:
lookup 1.80
Usealternative
hypothesis
to find
direction
.5000
- .4641
.0359
p-Value is P(Z 1.80) = .0359
Z 0 1.80
.4641
Two-Tailed Z Test
V l S l ti
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p-Value Solution
0 1.80 -1.80 Z
Reject H0 Reject H0
1/2 p-Value = .03591/2 p-Value = .0359
1/2 = .0251/2 = .025
(p-Value = .0718) ( = .05).
Do not reject H0.
Test statistic is in ‘Do not reject’ region
One-Tailed Z Test
V l S l ti
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p-Value Solution
Usealternative
hypothesis
to find
direction
p-Value is P(Z 1.80)
Z value of sample
statistic
p-Value
Z 0 1.80
From Z table:
lookup 1.80
.4641
.5000
- .4641
.0359
One-Tailed Z Test
V l S l ti
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= .05
p-Value Solution
0 1.50 Z
p-Value = .0359
(p-Value = .0359) < ( = .05).Do not Reject H0.
Test statistic is in ‘Do not reject’ region
p-Value
Thi ki Ch ll
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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 standarddeviation 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 ofsignificance (p-Value)?
p-Value
S l ti *
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Usealternative
hypothesis
to find
direction
Solution*
Z 0 -2.65 Z value of sample
statistic
From Z table:
lookup 2.65
.4960
p-Value
.004 .5000- .4960
.0040
p-Value is P(Z -2.65) = .004.
p-Value < ( = .01). Reject H0.
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Z Test of Proportion
Data Types
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Data Types
Data
Quantitative Qualitative
ContinuousDiscrete
Qualitative Data
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Qualitative Data
1. Qualitative random variables yield responses thatclassify
• e.g., Gender (male, female)
2. Measurement reflects number in category
3. Nominal or ordinal scale
4. Examples
• Do you own savings bonds?
• Do you live on-campus or off-campus?
Proportions
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Proportions1. Involve qualitative variables
2. Fraction or percentage of population in a
category
3. If two qualitative outcomes, binomial
distribution
• Possess or don’t possess characteristic
4. Sample Proportion ( p)number of successes
ˆ
sample size
x p
n
^
Sampling Distribution
of Proportion
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of Proportion
1. Approximated by Normal
Distribution
Excludes 0 or n
2. Mean
3. Standard Error
n
p p p
)1( 00ˆ
ˆ P p
Sampling Distribution
where p0 = Population Proportion
.0
.1
.2
.3
.0 .2 .4 .6 .8 1.0
P^
P(P^
) ˆ1ˆ3ˆ p pn pn
Standardizing Sampling
Distribution of Proportion
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Z= 0
z= 1
Z
Distribution of Proportion
Sampling
Distribution
Standardized Normal
Distribution
P^ P
P
^
^
Z p p p
p p
n
^ p
p
^
^
( )1
^0
0 0
One Population Tests
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One Population Tests
One Population
Z Test (1 & 2
tail)
t Test (1 & 2
tail)
Z Test (1 & 2
tail)
Mean Proportion Variance
2 Test (1 & 2
tail)
One-Sample Z Test
for Proportion
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for Proportion
1. Assumptions
• Random sample selected from a binomial population
• Normal approximation can be used if
0 0ˆ ˆ15 and 15np nq
2. Z-test statistic for proportion
0
0 0
ˆ p p Z
p q
n
Hypothesized population
proportion
One-Proportion Z Test
Example
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Example
The present packaging systemproduces 10% defective cereal
boxes. Using a new system, a
random sample of 200 boxeshad 11 defects. Does the new
system produce fewer defects?
Test at the .05 level of
significance.
One Proportion Z Test Solution
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One-Proportion Z Test Solution
H0:
Ha:
=
n = Critical Value(s):
Test Statistic:
Decision:
Conclusion:
p = .10
p < .10
.05
200
Z0-1.645
.05
Reject H0
0
0 0
11.10
ˆ 200 2.12.10 .90
200
p p Z
p q
n
Reject at = .05
There is evidence new
system < 10% defective
One-Proportion Z Test Thinking
Challenge
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Challenge
You’re an accounting manager. A
year-end audit showed 4% of
transactions had errors. You
implement new procedures. Arandom sample of 500 transactions
had 25 errors. Has the proportion
of incorrect transactions changed
at the .05 level of significance?
One-Proportion Z Test Solution*
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One-Proportion Z Test Solution
H0: Ha:
=
n = Critical Value(s):
Test Statistic:
Decision:
Conclusion:
p = .04
p .04
.05
500
Z0 1.96-1.96
.025
Reject H0
Reject H0
.025
0
0 0
25.04
ˆ 500 1.14.04 .96
500
p p Z
p q
n
Do not reject at = .05
There is evidence
proportion is not 4%
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Chi-Square (2) Testof Variance
One Population Tests
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One Population Tests
One Population
Z Test (1 & 2
tail)
t Test (1 & 2
tail)
Z Test (1 & 2
tail)
Mean Proportion Variance
2 Test (1 & 2
tail)
Chi-Square (2) Test
for Variance
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for Variance
1. Tests one population variance or standarddeviation
2. Assumes population is approximately normally
distributed3. Null hypothesis is H0: 2 = 0
2
4. Test statistic
Hypothesized pop. variance
Sample variance
2
2
2
1)
(n S
0
Chi-Square (2) Distribution
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C Squa e ( ) st but o
Select simple random
sample, size n.
Compute s 2
Compute 2 = (n-1)s 2 / 2
Astronomical number
of 2 values
Population Sampling Distributions
for Different Sample
Sizes
2 1 2 3 0
Finding Critical Value Example
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What is the critical 2 value given
Ha: 2 > 0.7
n = 3
=.05?
Finding Critical Value Example
2 0
Upper Tail Area
DF .995 … .95 … .05
1 ... … 0.004 … 3.841
2 0.010 … 0.103 … 5.991
2 Table
(Portion)
df = n - 1 = 25.991
Reject
= .05
Finding Critical Value Example
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Finding Critical Value Example
What is the critical 2 value given:
Ha: 2 < 0.7
n = 3
=.05?
What do you do
if the rejectionregion is on the
left?
Finding Critical Value Example
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What is the critical 2 value given:Ha: 2 < 0.7
n = 3
=.05?
Finding Critical Value Example
.103 2 0
Upper Tail Area
DF .995 … .95 … .05
1 ... … 0.004 … 3.841
2 0.010 … 0.103 … 5.991
2 Table
(Portion)
Upper Tail Area
for Lower Critical
Value = 1-.05 = .95 = .05
Reject H0
df = n - 1 = 2
Chi-Square (2) Test Example
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q ( ) p
Is the variation in boxes ofcereal, measured by the
variance, equal to 15 grams?
A random sample of 25 boxeshad a standard deviation of
17.7 grams. Test at the .05
level of significance.
Chi-Square (2) Test
Solution
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Solution
H0: Ha:
=
df = Critical Value(s):
Test Statistic:
Decision:
Conclusion:
2 = 152 15
.05
25 - 1 = 24
/2 = .025
= 33.42
2 22
2 2
0
( 1) (25 1) 17.7
15
n S
Do not reject at = .05