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Transcript of Statistics for Managers Using Microsoft® Excel 7th Edition Chapter 9 Fundamentals of Hypothesis...
![Page 1: Statistics for Managers Using Microsoft® Excel 7th Edition Chapter 9 Fundamentals of Hypothesis Testing: One Sample Tests Statistics for Managers Using.](https://reader031.fdocuments.in/reader031/viewer/2022032312/56649e0d5503460f94af6ee9/html5/thumbnails/1.jpg)
Statistics for ManagersUsing Microsoft® Excel
7th EditionChapter 9
Fundamentals of Hypothesis Testing:
One Sample Tests
Statistics for Managers Using Microsoft Excel, 7e © 2014 Pearson-Prentice-Hall, Inc. Philip A. Vaccaro , PhD
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Learning Objectives
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The Hypothesis
A hypothesis is a claim (assumption) about a population parameter: population mean ( μ )
pop. proportion ( π )
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The Null Hypothesis, H0
3μ:H0
it is written in terms of the population
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The Null Hypothesis, H0
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The Alternative Hypothesis, H1
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The Alternative Hypothesis, H1
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The Hypothesis Testing Process
Claim: The population mean age is 50. H0: μ = 50, H1: μ ≠ 50
Sample the population and find sample mean.Population
Sample
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The Hypothesis Testing Process
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The Hypothesis Testing Process
Sampling Distribution of X
μ = 50If H0 is true
20X
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The Test Statistic and Critical Values
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The Test Statistic and Critical Values
Critical Values
Distribution of the test statistic
Region of Rejection
Region of Rejection
Do Not Reject Ho
Thecriticalvalues
are statedin either‘z’ or ‘t’values
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Errors in Decision Making
Type I Error Reject a true null hypothesis Considered a serious type of error The probability of a Type I Error is ‘’ Called level of significance of the test Set by researcher in advance:
Type II Error
Failure to reject false null hypothesis The probability of a Type II Error is ‘β’
within your control or dictated by management
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Errors in Decision Making
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Level of Significance, α
H0: μ ≥ 50
H1: μ < 500
H0: μ ≤ 50
H1: μ > 50
a
a
Lower-tail test
0
Upper-tail test
Two-tail test
0
H0: μ = 50
H1: μ ≠ 50
Claim: The population mean age is 50.
/2a
/2a
The probability of rejecting Ho when it is true
typical values
areα =.01α =.02α =.05α =.10
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Hypothesis Testing: σ Known
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Hypothesis Testing: σ Known
Do not reject H0 Reject H0Reject H0
/2
-Z0
H0: μ = 3
H1: μ ≠
3
+Z
/2
Lower critical value
Upper critical value
3
Z
X
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Hypothesis Testing: σ Known
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Hypothesis Testing: σ Known
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Hypothesis Testing: σ Known
Is the test statistic in the rejection region?
Reject H0 Do not reject H0
a = .05/2
-Z= -1.96 0
= .05/2
Reject H0
+Z= +1.96
Here, Z = -2.0 < -1.96, so the test statistic is in the rejection region
a = .05level of
significance
Z 0.06
- 1.9 .0250
Z 0.06
+ 1.9 .9750
.025 .025
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Hypothesis Testing: σ Known
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Hypothesis Testing: σ Known
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Hypothesis Testing: σ Known
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Hypothesis Testing: σ Known ,the ‘p’- Value Approach
* Because we can reject or not reject Ho “at a glance“ .
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Hypothesis Testing: σ Knownp-Value Approach
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Hypothesis Testing: σ Knownp-Value Approach
.0228
/2 = .025
-1.96 0-2.0
Z1.962.0
.0228
/2 = .025
Z 0.00
- 2.00 .0228
Z 0.00
+ 2.00 .9772
Z = 2.84 - 3.00 .8 √100
= - .16 = - 2.00 .08
Z = 3.16 - 3.00 .8 √100
= +.16 = + 2.00 .08
Ho: μ = 3.00
2.84 3.16
(.9772)
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Hypothesis Testing: σ Knownp-Value Approach
Compare the p-value with If p-value < , reject H0
If p-value , do not reject H0
.0228
/2 = .025
-1.96 0
-2.0
Z1.96
2.0
.0228
/2 = .025
The probabilityof seeing a
sample meanof 2.84 or less,
or 3.16 or more , if
the populationmean is really 3.0
is only 4.56%
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Hypothesis Testing: σ KnownConfidence Interval Connections
Simpler,faster
hypothesistest
Canonly be
used witha two-tailhypothesis
test !
Ifa = .05,
constructa 95%
confidenceinterval
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Hypothesis Testing: σ KnownOne Tail Tests
H0: μ ≥ 3
H1: μ < 3
H0: μ ≤ 3
H1: μ > 3
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Hypothesis Testing: σ KnownLower Tail Tests
There is only one critical value, since the rejection area is in
only one tail.
Reject H0 Do not reject H0
α
-Z
μ
Z
XCritical value
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Hypothesis Testing: σ KnownUpper Tail Tests
There is only one critical value, since the rejection area is in
only one tail.
Reject H0
Do not reject H0
α
Zμ
Critical value
Z
X
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Hypothesis Testing: σ KnownUpper Tail Test Example
Form the hypothesis test:
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Hypothesis Testing: σ KnownUpper Tail Test Example Suppose that = .10 is chosen for this test Find the rejection region:
Reject H0Do not reject H0
= .10
Z0
Reject H0
1- = .90
Z 0.08
+ 1.2 .8997
Theentirea = .10goes to theupper
tail
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Hypothesis Testing: σ KnownUpper Tail Test Example
What is Z given a = 0.10?
Z .07 .09
1.1 .8790 .8810 .8830
1.2 .8980 .9015
1.3 .9147 .9162 .9177z 0 1.28
.08a = .10
Critical Value = 1.28
.90
.8997
.10
.90
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Hypothesis Testing: σ KnownUpper Tail Test Example
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Hypothesis Testing: σ KnownUpper Tail Test Example
Reach a decision and interpret the result:
= .10
1.280
Reject H0
1- = .90
Z = .88
Do Not Reject Ho
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Hypothesis Testing: σ KnownUpper Tail Test Example
Calculate the p-value and compare to ‘’
Reject H0
= .10
Do not reject H0
1.28
0
Reject H0
Z = .88 .1894
.810610.88)P(Z
6410/
52.053.1ZP
53.1)XP(
p-value = .1894Z 0.08
+ .8 .8106
Z = 53.1 - 52.0 = 1.1 10 / √ 64 1.25
= .88
1 - .8106 = .1894
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“Talk”is notcheap,
especiallywhen
you arebeing
cheated !
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Hypothesis Testing: σ Unknown
df =1df =2df =3df =4df =5df =6
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Hypothesis Testing: σ Unknown
Recall that the ‘t’ test statistic with n-1 degrees of freedom is:
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Hypothesis Testing: σ Unknown Example
_
H0: μ =
168 H1: μ ¹ 168
_
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Hypothesis Testing: σ Unknown Example
H0: μ = 168
H1: μ ≠ 168
Reject H0Reject H0
α/2=.025
-t n-1,α/2
Do not reject H0
0
α/2=.025
-2.0639 2.0639
t n-1,α/2
Determine the regions of rejection
t .025
24 df 2.0639
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Hypothesis Testing: σ Unknown Example
a/2=.025
-t n-1,α/2 0
a/2=.025
-2.0639 2.0639
t n-1,α/21.46Since t df = 24, a=.025 = 1.46 < 2.0639,
Do Not Reject HoRejectHo
RejectHo
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Hypothesis Testing: Connection to Confidence Intervals
For X = 172.5, S = 15.40 and n = 25, the 95% confidence interval is:
166.14 ≤ μ ≤ 178.86
Since this interval contains the hypothesized mean (168), you do not reject the null hypothesis at = .05 _
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Hypothesis Testing: σ Unknown
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Hypothesis Testing:Proportions
Involves categorical variables ( yes/no , male/female )
Two possible outcomes “Success” (possesses a certain characteristic)
“Failure” (does not possesses that characteristic)
Fraction or proportion of the population in the “success” category is denoted by π
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Hypothesis Testing:Proportions
sizesample
sampleinsuccessesofnumber
n
Xp
pμn
)(1σ
p
Sample proportion in the success category is denoted by p
When both nπ and n(1-π) are at least 5, p can be approximated
by a normal distribution with mean and standard deviation
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Hypothesis Testing:Proportions
The sampling distribution of p is approximately normal, so the test statistic is a ‘Z’ value:
n
pZ
)1(
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Hypothesis Testing:Proportions Example
A marketing company claims that it receives 8% responses from its mailings. To test this claim, a random sample of 500 were surveyed with 30 responses. Test at the = .05 significance level.
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Hypothesis Testing:Proportions ExampleH0: π = .08 H1: π ≠ .08
α = .05
n = 500, p = .06
Critical Values: ± 1.96
z0
Reject Reject
.025.025
1.96-1.96
Determine region of rejection
Do Not RejectHo
30 / 500 = .06 = p (sample proportion)
Z 0.06
- 1.9 .0250
Z 0.06
+ 1.9 .9750
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Hypothesis Testing:Proportions Example
Test Statistic: Decision:
Conclusion:
z0
.025.025
1.96-1.96-1.646
Do Not RejectHo
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Potential Pitfalls and Ethical Considerations
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Chapter Summary
In this chapter, we have