Statistics for Business and Economics
Chapter 6
Inferences Based on a Single Sample: Tests of Hypothesis
Learning Objectives
1. Distinguish Types of Hypotheses
2. Describe Hypothesis Testing Process
3. Explain p-Value Concept
4. Solve Hypothesis Testing Problems Based on a Single Sample
5. Explain Power of a Test
Statistical Methods
StatisticalMethods
EstimationHypothesis
Testing
InferentialStatistics
DescriptiveStatistics
Hypothesis Testing Concepts
Hypothesis Testing
Population
I believe the population mean age is 50 (hypothesis).
Mean X = 20
Random sample
Reject hypothesis! Not close.
Reject hypothesis! Not close.
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.
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
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
Identifying HypothesesSteps
Example problem: Test that the population mean is not 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)
• 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?
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?
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?
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?
What Are the Hypotheses?
Basic Idea
Sample Means = 50H0
Sampling Distribution
It is unlikely that we would get a sample mean of this value ...
20
... if in fact this were the population mean
... therefore, we reject the hypothesis that = 50.
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)• Typical values are .01, .05, .10
4. Selected by researcher at start
Rejection Region (One-Tail Test)
Ho
ValueCriticalValue
Sample Statistic
RejectionRegion
NonrejectionRegion
Sampling Distribution
1 –
Level of Confidence
Observed sample statistic
Rejection Region (One-Tail Test)
Sampling DistributionLevel of Confidence
Ho
ValueCriticalValue
Sample Statistic
RejectionRegion
NonrejectionRegion
Sampling Distribution
1 –
Level of Confidence
Observed sample statistic
Rejection Regions (Two-Tailed Test)
Ho
Value CriticalValue
CriticalValue
1/2 1/2
Sample Statistic
RejectionRegion
RejectionRegion
NonrejectionRegion
Sampling Distribution
1 –
Level of Confidence
Observed sample statistic
Ho
Value CriticalValue
CriticalValue
1/2 1/2
Sample Statistic
RejectionRegion
RejectionRegion
NonrejectionRegion
Sampling Distribution
1 –
Level of Confidence
Rejection Regions (Two-Tailed Test)
Sampling DistributionLevel of Confidence
Observed sample statistic
Ho
Value CriticalValue
CriticalValue
1/2 1/2
Sample Statistic
RejectionRegion
RejectionRegion
NonrejectionRegion
Sampling Distribution
1 –
Level of Confidence
Rejection Regions (Two-Tailed Test)
Sampling DistributionLevel of Confidence
Observed sample statistic
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)
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
AcceptH0
1 – Type IIError
()
RejectH0
Type IError ()
Power(1 – )
& Have an Inverse Relationship
You can’t reduce both errors simultaneously!
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
Hypothesis Testing Steps
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
One Population Tests
OnePopulation
Z Test(1 & 2tail)
t Test(1 & 2tail)
Z Test(1 & 2tail)
Mean Proportion Variance
2 Test(1 & 2tail)
Two-Tailed Z Test of Mean ( Known)
One Population Tests
OnePopulation
Z Test(1 & 2tail)
t Test(1 & 2tail)
Z Test(1 & 2tail)
Mean Proportion Variance
2 Test(1 & 2tail)
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 XZ
n
3. Z-Test Statistic
Two-Tailed Z Test for Mean Hypotheses
H0:=0 Ha: ≠ 0
Z0
Reject H0
Reject H
.500 - .025
.475
Z0
= 1
Two-Tailed Z Test Finding Critical Z
What is Z given = .05?
= .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
Two-Tailed Z Test Example
Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified to be 25 grams. Test at the .05 level of significance.
368 gm.368 gm.
Two-Tailed Z Test Solution
• H0:
• Ha:
• • n
• Critical Value(s):
Test Statistic:
Decision:
Conclusion:
= 368
368
.0525
Z0 1.96-1.96
.025
Reject H 0 Reject H0
.025
372.5 3681.50
15
25
XZ
n
Do not reject at = .05
No evidence average is not 368
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 a sample 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 breaking strength?
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 H 0 Reject H0
.025
69.7 70.51
3.5
36
XZ
n
Do not reject at = .05
No evidence average is not 70
One-Tailed Z Test of Mean ( Known)
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 XZ
n
One-Tailed Z Test 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!
.500 - .025
.475
Z0
= 1
One-Tailed Z Test 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 Example
Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified to be 25 grams. Test at the .05 level of significance.
368 gm.368 gm.
One-Tailed Z Test Solution
• H0: • Ha: • = • n = • Critical Value(s):
Test Statistic:
Decision:
Conclusion:
= 368
> 368
.05
25
Z0 1.645
.05
Reject
372.5 3681.50
15
25
XZ
n
Do not reject at = .05
No evidence average is more than 368
One-Tailed Z Test 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. At the .01 level of significance, is there evidence that the miles per gallon is at least 32?
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
XZ
n
Reject at = .01
There is evidence average is less than 32
Observed Significance Levels: p-Values
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 rejected
3. Used to make rejection decision• If p-value , do not reject H0
• If p-value < , reject H0
Two-Tailed Z Test p-Value Example
Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified to be 25 grams. Find the p-Value.
368 gm.368 gm.
Two-Tailed Z Test p-Value Solution
Z0 1.50
Z value of sample statistic (observed)
372.5 3681.50
15
25
XZ
n
1/2 p-Value1/2 p-Value
Two-Tailed Z Test p-Value Solution
Z value of sample statistic (observed)
p-value is P(Z -1.50 or Z 1.50)
Z0 1.50-1.50
From Z table: lookup 1.50
.4332
.5000- .4332
.0668
Two-Tailed Z Test p-Value Solution
1/2 p-Value.0668
1/2 p-Value.0668
p-value is P(Z -1.50 or Z 1.50) = .1336
Z value of sample statistic
From Z table: lookup 1.50
.5000- .4332
.0668
Z0 1.50-1.50
Two-Tailed Z Test p-Value Solution
0 1.50-1.50 Z
Reject H0Reject H0
1/2 p-Value = .06681/2 p-Value = .0668
1/2 = .0251/2 = .025
(p-Value = .1336) ( = .05). Do not reject H0.
Test statistic is in ‘Do not reject’ region
One-Tailed Z Test p-Value Example
Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified to be 25 grams. Find the p-Value.
368 gm.368 gm.
One-Tailed Z Test p-Value Solution
Z0 1.50
Z value of sample statistic
372.5 3681.50
15
25
XZ
n
One-Tailed Z Test p-Value Solution
Use alternative hypothesis to find direction
p-Value is P(Z 1.50)
Z value of sample statistic
p-Value
Z0 1.50From Z table: lookup 1.50
.4332
.5000- .4332
.0668
One-Tailed Z Test p-Value Solution
p-Value.0668
Z value of sample statistic
From Z table: lookup 1.50
Use alternative hypothesis to find direction
.5000- .4332
.0668
p-Value is P(Z 1.50) = .0668
Z0 1.50
.4332
= .05
One-Tailed Z Test p-Value Solution
0 1.50 Z
Reject H0
p-Value = .0668
(p-Value = .0668) ( = .05). Do not reject H0.
Test statistic is in ‘Do not reject’ region
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)?
Use alternative hypothesis to find direction
p-Value Solution*
Z0-2.65Z 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.
Two-Tailed t Test of Mean ( Unknown)
One Population Tests
OnePopulation
Z Test(1 & 2tail)
t Test(1 & 2tail)
Z Test(1 & 2tail)
Mean Proportion Variance
2 Test(1 & 2tail)
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
3. t test statisticX
tS
n
t0
Two-Tailed t Test 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 Example
Does an average box of cereal contain 368 grams of cereal? A random sample of 36 boxes had a mean of 372.5 and a standard deviation of 12 grams. Test at the .05 level of significance.
368 gm.368 gm.
Two-Tailed t Test Solution
• H0: • Ha: • = • df = • Critical Value(s):
Test Statistic:
Decision:
Conclusion:
= 368 368
.05
36 - 1 = 35
t0 2.030-2.030
.025
Reject H0 Reject H0
.025
372.5 3682.25
12
36
Xt
S
n
Reject at = .05
There is evidence population average is not 368
Two-Tailed t TestThinking 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 of 64 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.3.25 lb.
Two-Tailed t Test Solution*
• H0: • Ha: • • df • Critical Value(s):
Test Statistic:
Decision:
Conclusion:
= 3.25
3.25
.01
64 - 1 = 63
t0 2.656-2.656
.005
Reject H0 Reject H0
.005
3.238 3.25.82
.117
64
Xt
S
n
Do not reject at = .01
There is no evidence average is not 3.25
One-Tailed t Test of Mean ( Unknown)
One-Tailed t TestExample
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 Solution
• H0: • Ha: • =• df =• Critical Value(s):
Test Statistic:
Decision:
Conclusion:
= 140
< 140
.0520 - 1 = 19
t0-1.729
.05
Reject H0
138.47 1402.57
2.66
20
Xt
S
n
Reject at = .05
There is evidence population average is less than 140
One-Tailed t Test 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 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
Xt
S
n
Do not reject at = .05
There is no evidence average is more than 5
Z Test of Proportion
Data Types
Data
Quantitative Qualitative
ContinuousDiscrete
Qualitative Data
1. Qualitative random variables yield responses that classify
• 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?
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
ˆsample size
xp
n
^
Sampling Distribution of Proportion
1. Approximated by Normal Distribution
– Excludes 0 or n
2. Mean
3. Standard Error
P̂p
Sampling Distribution
where p0 = Population Proportion
.0.0
.1.1
.2.2
.3.3
.0.0 .2.2 .4.4 .6.6 .8.8 1.01.0
PP^̂
P(PP(P^̂ )) ˆ1ˆ3ˆ ppnpn
n
ppp
)1( 00ˆ
Z = 0
zz= 1
Z
Standardizing Sampling Distribution of Proportion
Sampling Distribution
Standardized Normal Distribution
P^P
P
^
^
Zp p p
p p
n
^p
p
^
^
( )1
^0
0 0
One Population Tests
OnePopulation
Z Test(1 & 2tail)
t Test(1 & 2tail)
Z Test(1 & 2tail)
Mean Proportion Variance
2 Test(1 & 2tail)
One-Sample Z Test 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 proportion0
0 0
p̂ pZ
p qn
Hypothesized
population proportion
One-Proportion Z Test Example
The present packaging system produces 10% defective cereal boxes. Using a new system, a random sample of 200 boxes had11 defects. Does the new system produce fewer defects? Test at the .05 level of significance.
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
11.10ˆ 200 2.12
.10 .90200
p pZ
p qn
Reject at = .05
There is evidence new system < 10% defective
One-Proportion Z Test Thinking Challenge
You’re an accounting manager. A year-end audit showed 4% of transactions had errors. You implement new procedures. A random 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*
• H0: • Ha: • = • n = • Critical Value(s):
Test Statistic:
Decision:
Conclusion:
p = .04
p .04
.05
500
Z0 1.96-1.96
.025
Reject H 0 Reject H0
.025
0
0 0
25.04ˆ 500 1.14
.04 .96500
p pZ
p qn
Do not reject at = .05
There is evidence proportion is not 4%
Calculating Type II Error Probabilities
Power of Test
1. Probability of rejecting false H0
• Correct decision
2. Designated 1 -
3. Used in determining test adequacy
4. Affected by• True value of population parameter
• Significance level • Standard deviation & sample size n
Finding PowerStep 1
X0 = 368
Reject H0
Do NotReject H0
Hypothesis:H0: 0 368Ha: 0 < 368 = .05
Draw15
25
n
Finding PowerSteps 2 & 3
Xa = 360
‘True’ Situation: a = 360 (Ha)
Draw
Specify
1-
X0 = 368
Reject H0
Do NotReject H0
Hypothesis:H0: 0 368Ha: 0 < 368 = .05
Draw15
25
n
Finding PowerStep 4
363.065363.065
065.363
25
1564.13680
n
ZX L
XXaa = 360= 360
‘True’ Situation: a = 360 (Ha)
Draw
Specify1-1-
X0 = 368
Reject H0
Do NotReject H0
Hypothesis:H0: 0 368Ha: 0 < 368 = .05
Draw15
25
n
Finding PowerStep 5
363.065363.065
065.363
25
1564.13680
n
ZX L
065.363
25
1564.13680
n
ZX L
XXaa = 360= 360
‘True’ Situation: a = 360 (Ha)
Draw
Specify
X0 = 368
Reject H0
Do NotReject H0
Hypothesis:H0: 0 368Ha: 0 < 368 = .05
Draw15
25
n
= .154
1- =.846
Z Table
Power Curves
Power Power
Power
Possible True Values for aPossible True Values for a
Possible True Values for a
H0: 0 H0: 0
H0: =0
= 368 in Example
Chi-Square (2) Test of Variance
One Population Tests
OnePopulation
Z Test(1 & 2tail)
t Test(1 & 2tail)
Z Test(1 & 2tail)
Mean Proportion Variance
2 Test(1 & 2tail)
Chi-Square (2) Testfor Variance
1. Tests one population variance or standard deviation
2. Assumes population is approximately normally distributed
3. Null hypothesis is H0: 2 = 02
4. Test statistic
Hypothesized pop. variance
Sample variance
2
2
2
1)
(n S
0
Chi-Square (2) Distribution
Select simple randomsample, size n.Compute s2
Compute 2 =(n-1)s 2 /2
Astronomical numberof 2 values
PopulationSampling Distributionsfor Different SampleSizes
21 2 30
What is the critical 2 value given:Ha: 2 > 0.7
n = 3 =.05?
Finding Critical Value Example
20
Upper Tail AreaDF .995 … .95 … .051 ... … 0.004 … 3.8412 0.010 … 0.103 … 5.991
2 Table (Portion)
df = n - 1 = 25.991
Reject
= .05
Finding Critical Value Example
What is the critical 2 value given:Ha: 2 < 0.7
n = 3 =.05?
What do you do if the rejection region is on the left?
What is the critical 2 value given:Ha: 2 < 0.7
n = 3 =.05?
Finding Critical Value Example
.103 20
Upper Tail AreaDF .995 … .95 … .051 ... … 0.004 … 3.8412 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
Is the variation in boxes of cereal, measured by the variance, equal to 15 grams? A random sample of 25 boxes had a standard deviation of 17.7 grams. Test at the .05 level of significance.
Chi-Square (2) Test Solution
• H0: • Ha: • = • df = • Critical Value(s):
Test Statistic:
Decision:
Conclusion:
2 = 15
2 15
.0525 - 1 = 24
2200
/2 = .025
39.36412.401
= 33.42
2 22
2 20
( 1) (25 1) 17.7
15
n S
Do not reject at = .05
There is no evidence 2 is not 15
Conclusion
1. Distinguished Types of Hypotheses
2. Described Hypothesis Testing Process
3. Explained p-Value Concept
4. Solved Hypothesis Testing Problems Based on a Single Sample
5. Explained Power of a Test