Transcript Exploring Marketing Research William G. Zikmund

Business
Research Methods
William G. Zikmund
Chapter 21:
Univariate Statistics
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UNIVARIATE STATISTICS
• TEST OF STATISTICAL SIGNIFICANCE
• HYPOTHESIS TESTING ONE
VARIABLE AT A TIME
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HYPOTHESIS
• UNPROVEN PROPOSITION
• SUPPOSITION THAT TENATIVELY
EXPLAINS CERTAIN FACTS OR
PHENOMONA
• ASSUMPTION ABOUT NATURE OF
THE WORLD
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HYPOTHESIS
• AN UNPROVEN PROPOSITION OR
SUPPOSITION THAT TENTATIVELY
EXPLAINS CERTAIN FACTS OF
PHENOMENA
• NULL HYPOTHESIS
• ALTERNATIVE HYPOTHESIS
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NULL HYPOTHESIS
• STATEMENT ABOUT THE STATUS QUO
• NO DIFFERENCE
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ALTERNATIVE HYPHOTESIS
• STATEMENT THAT INDICATES THE
OPPOSITE OF THE NULL HYPOTHESIS
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SIGNIFICANCE LEVEL
• CRITICAL PROBABLITY IN CHOOSING
BETWEEN THE NULL HYPOTHESIS
AND THE ALTERNATIVE HYPOTHESIS
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SIGNIFICANCE LEVEL
•
•
•
•
CRITICAL PROBABLITY
CONFIDENCE LEVEL
ALPHA
PROBABLITY LEVEL SELECTED IS
TYPICALLY .05 OR .01
• TOO LOW TO WARRANT SUPPORT
FOR THE NULL HYPOTHESIS
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The null hypothesis that the mean is
equal to 3.0:
H o :   3 .0
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The alternative hypothesis that the
mean does not equal to 3.0:
H 1 :   3 .0
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A SAMPLING DISTRIBUTION
3.0
x
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A SAMPLING DISTRIBUTION
a.025
a.025
3.0
x
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A SAMPLING DISTRIBUTION
LOWER
LIMIT
3.0
UPPER
LIMIT
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Critical values of 
Critical value - upper limit
S
   ZS X or   Z
n
 1 .5 
 3.0  1.96

 225 
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Critical values of 
 3.0  1.960.1
 3.0  .196
 3.196
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Critical values of 
Critical value - lower limit
S
  - ZS X or  - Z
n
 1 .5 
 3.0 - 1.96

 225 
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Critical values of 
 3.0  1.960.1
 3.0  .196
 2.804
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REGION OF REJECTION
LOWER
LIMIT
3.0
UPPER
LIMIT
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HYPOTHESIS TEST  3.0
2.804
3.0
3.196
3.78
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TYPE I AND TYPE II ERRORS
Null is true
Null is false
Accept null
Reject null
Correctno error
Type I
error
Type II
error
Correctno error
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Type I and Type II Errors in Hypothesis Testing
State of Null Hypothesis
in the Population
Decision
Accept Ho
Reject Ho
Ho is true
Ho is false
Correct--no error
Type II error
Type I error
Correct--no error
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CALCULATING ZOBS
x 
zOBS
sx
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Alternate way of testing the
hypothesis
Z obs
X 

SX
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Alternate way of testing the
hypothesis
Z obs
3.78  3.0
3.78  


.1
SX
0.78

.1
 7 .8
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CHOOSING THE APPROPRAITE
STATISTICAL TECHNIQUE
• Type of question to be answered
• Number of variables
– Univariate
– Bivariate
– Multivariate
• Scale of measurement
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PARAMETRIC
STATISTICS
NONPARAMETRIC
STATISTICS
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t-distribution
• Symmetrical, bell-shaped distribution
• Mean of zero and a unit standard deviation
• Shape influenced by degrees of freedom
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DEGREES OF FREEDOM
• Abbreviated d.f.
• Number of observations
• Number of constraints
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Confidence interval estimate
using the t-distribution
  X  t c .l . S X
Upper limit  X  tc.l .
or
Lower limit  X  tc.l .
S
n
S
n
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Confidence interval estimate
using the t-distribution

X
tc.l .
= population mean
= sample mean
= critical value of t at a specified confidence
level
SX
S
n
= standard error of the mean
= sample standard deviation
= sample size
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Confidence Interval using t
  X  t cl s x
X  3 .7
S  2.66
n  17
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upper limit  3 . 7  2 . 12 ( 2 . 66 17 )
 5 . 07
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Lower limit  3 . 7  2 . 12 ( 2 . 66 17 )
 2 . 33
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HYPOTHESIS TEST USING
THE t-DISTRIBUTION
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Univariate hypothesis test
utilizing the t-distribution
Suppose that a production manager believes
the average number of defective assemblies
each day to be 20. The factory records the
number of defective assemblies for each of the
25 days it was opened in a given month. The
mean X was calculated to be 22, and the
standard deviation, S ,to be 5.
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H 0 :   20
H 1 :   20
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SX  S / n
 5 / 25
1
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Univariate hypothesis test
utilizing the t-distribution
The researcher desired a 95 percent
confidence, and the significance level becomes
.05.The researcher must then find the upper
and lower limits of the confidence interval to
determine the region of rejection. Thus, the
value of t is needed. For 24 degrees of
freedom (n-1, 25-1), the t-value is 2.064.
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Lower limit :

  t c.l . S X  20  2.064 5 / 25
 20  2.064 1
 17 .936
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
Upper limit :

  t c.l . S X  20  2.064 5 / 25
 20  2.064 1
 20 .064
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
Univariate hypothesis test
- t-test
tobs
X 
22  20


SX
1
2

1
2
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TESTING A HYPOTHESIS
ABOUT A DISTRIBUTION
• CHI-SQUARE TEST
• TEST FOR SIGNIFANCE IN THE
ANALYSIS OF FREQUENCY
DISTRIBUTIONS
• COMPARE OBSERVED FREQUENCIES
WITH EXPECTED FREQUENCIES
• “GOODNESS OF FIT”
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Chi-Square Test
(Oi  Ei )²
x²  
Ei
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Chi-Square Test
x² = chi-square statistics
Oi = observed frequency in the ith cell
Ei = expected frequency on the ith cell
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Chi-Square Test
- estimation for expected number for each cell
E ij 
R iC
j
n
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Chi-Square Test
- estimation for expected number for each cell
Ri = total observed frequency in the ith row
Cj = total observed frequency in the jth column
n = sample size
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Univariate hypothesis test
- Chi-square Example

O1  E1 

2
X
2
E1

O2  E 2 

2
E2
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Univariate hypothesis test
- Chi-square Example

60  50 

2
X
2
4
50

40  50 

2
50
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HYPOTHESIS TEST OF A
PROPORTION
p is the population proportion
p is the sample proportion
p is estimated with p
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Hypothesis Test of a Proportion
H0 : p  . 5
H1 : p  . 5
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Sp 
0.60.4
100
 .0024
.24

100
 .04899
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.6  .5
p p

Zobs 
.04899
Sp
.1
 2.04

.04899
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Hypothesis Test of a Proportion:
Another Example
n  1,200
p  .20
Sp 
pq
n
Sp 
(.2)(.8)
1200
Sp 
.16
1200
Sp  .000133
Sp  . 0115
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Hypothesis Test of a Proportion:
Another Example
Z
pp
Sp
.20  .15
.0115
.05
Z
.0115
Z  4.348
The Z value exceeds 1.96, so the null hypothesis should be rejected at the .05 level.
Indeed it is significantt beyond the .001
Z
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