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Introduction to Hypothesis Testing:
One Population Value
Chapter 8 Handout
Chapter 8 Summary
Hypothesis Testing for One Population Value:1. Population Mean (
a. (population standard deviation) is given (known): Use z/standard normal/bell shaped distribution
b. (pop std dev) is not given but s (sample std dev) is given Use student’s t distribution
2. Population proportion () Use z/standard normal/bell shaped distribution
3. Population variance ( Use (Chi-Square) distribution
PS: population standard deviation =
A hypothesis is an assumption about the population parameter.
A parameter is a Population mean or proportion
The parameter must be identified before analysis.
I assume the mean GPA of this class is 3.5!
© 1984-1994 T/Maker Co.
What is a Hypothesis?
• States the Assumption (numerical) to be tested
e.g. The average # TV sets in US homes is at least 3 (H0: 3)
• Begin with the assumption that the null hypothesis is TRUE.
(Similar to the notion of innocent until proven guilty)
The Null Hypothesis, H0
•Refers to the Status Quo•Always contains the ‘ = ‘ sign
•The Null Hypothesis may or may not be rejected.
• Is the opposite of the null hypothesise.g. The average # TV sets in US homes
is less than 3 (H1: < 3)
• Challenges the Status Quo
• Never contains the ‘=‘ sign
• The Alternative Hypothesis may or may not be accepted
The Alternative Hypothesis, H1
or HA
Steps:State the Null Hypothesis (H0: 3)State its opposite, the Alternative
Hypothesis (H1: < 3)Hypotheses are mutually exclusive &
exhaustiveSometimes it is easier to form the
alternative hypothesis first.
Identify the Problem
Population
Assume thepopulationmean age is 50.(Null Hypothesis)
REJECT
The SampleMean Is 20
SampleNull Hypothesis
50?20 XIs
Hypothesis Testing Process
No, not likely!
Sample Mean = 50
Sampling DistributionIt is unlikely that we would get a sample mean of this value ...
... if in fact this were the population mean.
... Therefore, we reject the null
hypothesis that = 50.
20H0
Reason for Rejecting H0
• Defines Unlikely Values of Sample Statistic if Null Hypothesis Is True Called Rejection Region of Sampling
Distribution
• Designated a(alpha) Typical values are 0.01, 0.05, 0.10
• Selected by the Researcher at the Start
• Provides the Critical Value(s) of the Test
Level of Significance, a
Level of Significance, aand the Rejection Region
H0: 3
H1: < 30
0
0
H0: 3
H1: > 3
H0: 3
H1: 3
a
a
a/2
Critical Value(s)
Rejection Regions
• Type I Error Reject True Null Hypothesis Has Serious Consequences Probability of Type I Error Is a
Called Level of Significance
• Type II Error Do Not Reject False Null Hypothesis Probability of Type II Error Is b (Beta)
Errors in Making Decisions
H0: Innocent
Jury Trial Hypothesis Test
Actual Situation Actual Situation
Verdict Innocent Guilty Decision H0 True H0 False
Innocent Correct ErrorDo NotReject
H0
1 - a Type IIError (b )
Guilty Error Correct RejectH0
Type IError(a )
Power(1 - b)
Result Possibilities
a
b
Reduce probability of one error and the other one goes up.
a& bHave an Inverse Relationship
• True Value of Population Parameter Increases When Difference Between Hypothesized
Parameter & True Value Decreases
• Significance Level a Increases When aDecreases
• Population Standard Deviation Increases When Increases
• Sample Size n Increases When n Decreases
Factors Affecting Type II Error, b
a
b
b
b
n
3 Methods for Hypotheses Tests
Refer to Figure 8-6 (page 299) for a hypothesis test for means with (pop. std. dev.) is given:
Method 1: Comparing XaX critical) with XMethod 2: Z test, i.e., comparing Za critical) with Z (or Z statistics or Z
calculated)Method 3: Comparing asignificance level) with p-valueYou can modify those three methods for other cases. For example, if is
unknown, you must use student’s t distribution. If you would like to use Method 2, please compare t at critical) with t (or t statistics or t calculated). Refer to Figure 8-8 (page 303).
You always get: • Za critical) from Z distribution• tat critical) from student’s t distribution• .acritical) from distributionYou always get:• Z or Z calculated or Z statistics from sample (page 299 and Figure 8-6)• t or t calculated or t statistics from sample (Figure 8-8, page 299)• .or calculated or statistics from sample (Figure 8-19, page 322)
• Convert Sample Statistic (e.g., ) to Standardized Z Variable
• Compare to Critical Z Value(s) If Z test Statistic falls in Critical Region, Reject
H0; Otherwise Do Not Reject H0
Z-Test Statistics (Known)
Test Statistic
X
n
XXZ
X
X
• Probability of Obtaining a Test Statistic More Extreme or ) than Actual Sample Value Given H0 Is True
• Called Observed Level of Significance Smallest Value of a H0 Can Be Rejected
• Used to Make Rejection Decision If p value a Do Not Reject H0
If p value <a, Reject H0
p Value Test
1. State H0 H0 : 3
2. State H1 H1 :
3. Choosea a = .05
4. Choose n n = 100
5. Choose Method: Z Test (Method 2)
Hypothesis Testing: Steps
Test the Assumption that the true mean # of TV sets in US homes is at least 3.
6. Set Up Critical Value(s) Z = -1.645
7. Collect Data 100 households surveyed
8. Compute Test Statistic Computed Test Stat.= -2
9. Make Statistical Decision Reject Null Hypothesis
10. Express Decision The true mean # of TV set is less than 3 in the US households.
Hypothesis Testing: Steps
Test the Assumption that the average # of TV sets in US homes is at least 3.
(continued)
• Assumptions Population Is Normally Distributed If Not Normal, use large samples Null Hypothesis Has =, , or Sign Only
• Z Test Statistic:
One-Tail Z Test for Mean (Known)
n
xxz
x
x
Z0
a
Reject H0
Z0
Reject H0
a
H0: H1: < 0
H0: 0 H1: > 0
Must Be Significantly Below = 0
Small values don’t contradict H0
Don’t Reject H0!
Rejection Region
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 15 grams. Test at the a0.05 level.
368 gm.
Example: One Tail Test
H0: 368 H1: > 368
_
Z .04 .06
1.6 .4495 .4505 .4515
1.7 .4591 .4599 .4608
1.8 .4671 .4678 .4686
.4738 .4750
Z0
Z = 1
1.645
.50 -.05
.45
.05
1.9 .4744
Standardized Normal Probability Table (Portion)
What Is Z Givena = 0.05?
a = .05
Finding Critical Values: One Tail
Critical Value = 1.645
a= 0.025
n = 25
Critical Value: 1.645
Test Statistic:
Decision:
Conclusion:
Do Not Reject Ho ata = .05
No Evidence True Mean Is More than 368Z0 1.645
.05
Reject
Example Solution: One Tail
H0: 368 H1: > 368 50.1
n
XZ
Z0 1.50
p Value.0668
Z Value of Sample Statistic
From Z Table: Lookup 1.50
.9332
Use the alternative hypothesis to find the direction of the test.
1.0000 - .9332 .0668
p Value is P(Z 1.50) = 0.0668
p Value Solution
0 1.50 Z
Reject
(p Value = 0.0668) (a = 0.05). Do Not Reject.
p Value = 0.0668
a= 0.05
Test Statistic Is In the Do Not Reject Region
p Value Solution
Does an average box of cereal contains 368 grams of cereal? A random sample of 25 boxes showed X = 372.5. The company has specified to be 15 grams. Test at the a0.05 level.
368 gm.
Example: Two Tail Test
H0: 368
H1: 368
a= 0.05
n = 25
Critical Value: ±1.96
Test Statistic:
Decision:
Conclusion:
Do Not Reject Ho at a = .05
No Evidence that True Mean Is Not 368Z0 1.96
.025
Reject
Example Solution: Two Tail
-1.96
.025
H0: 386
H1: 38650.1
2515
3685.372
n
XZ
Two tail hypotheses tests = Confidence Intervals
For X = 372.5oz, = 15 and n = 25,
The 95% Confidence Interval is:
372.5 - (1.96) 15/ 25 to 372.5 + (1.96) 15/ 25
or
366.62 378.38
If this interval contains the Hypothesized mean (368), we do not reject the null hypothesis. It does. Do not reject Ho.
_
Assumptions Population is normally distributed If not normal, only slightly skewed & a large
sample taken
Parametric test procedure
t test statistic
t-Test: Unknown
nSX
t
Example: One Tail t-Test
Does an average box of cereal contain more than 368 grams of cereal? A random sample of 36 boxes showed X = 372.5, ands 15. Test at the a0.01 level.
368 gm.
H0: 368 H1: 368
is not given,
a= 0.01
n = 36, df = 35
Critical Value: 2.4377
Test Statistic:
Decision:
Conclusion:
Do Not Reject Ho at a = .01
No Evidence that True Mean Is More than 368Z0 2.4377
.01
Reject
Example Solution: One Tail
H0: 368 H1: 368 80.1
3615
3685.372
nSX
t
• Involves categorical variables
• Fraction or % of population in a category
• If two categorical outcomes, binomial
distribution Either possesses or doesn’t possess the characteristic
• Sample proportion (p)
Proportions
sizesamplesuccessesofnumber
nXp
Example:Z Test for Proportion
•Problem: A marketing company claims that it receives = 4% responses from its Mailing.
•Approach: To test this claim, a random sample of n = 500 were surveyed with x = 25 responses.
•Solution: Test at the a = .05 significance level.
a = .05n = 500, x = 25 p = x/n = 25/500 = 0.05
Do not reject Ho at Do not reject Ho at a = .05
Z Test for Proportion: Solution
H0: .04
H1: .04
Critical Values: 1.96
Test Statistic:
Decision:
Conclusion:We do not have sufficient
evidence to reject the company’s claim of 4% response rate.
Z p-
(1 - )n
=.05-.04
.04 (1 - .04)500
= 1.14
Z0
Reject Reject
.025.025