Transcript Advanced Data Analysis I

Advanced Data Analysis – Mean
Comparisons

Research Issues

Common for marketers to need to compare an empirical
mean to other means

Theoretical – some threshold level that has been established
based on experience


Laczniak Yogurt performs well financially when its customer
satisfaction rating exceeds 5.0 (on a 7-point scale)

Test to see if customer satisfaction > 5.0
Empirical – derived via research in another time, with another
sample, in a different place, etc.

Laczniak Yogurt wants to continuously improve; customer
satisfaction is expected to increase each quarter

Test to see if customer satisfaction this quarter > C. S. last
quarter
Advanced Data Analysis – Mean
Comparisons

Research Issues

Common for marketers to need to compare an
empirical percentage to other percentages

Laczniak Yogurt will only introduce new flavors if they
are deemed to be “acceptable” by 20% of its customers
in a test market


Test to see if Acceptance > 20%
Such issues lend themselves to a specific set of
analysis procedures which involve classical
hypothesis testing procedures
Advanced Data Analysis – Mean
Comparisons

Hypothesis Test Procedures


State Null and alternative hypotheses (Null always
means no relationship – reject it if you can)
Specify the level of statistical significance (a)


This is 1 - Confidence Interval
According to the AMA


Specify the critical region to be used



.05 (in Marketing Research)
Determine the critical value of the key statistic (t)
Collect data and determine test statistic (computed t)
Test hypothesis

Reject or retain the null hypothesis
Advanced Data Analysis – Mean
Comparisons

t-test

Used to test hypotheses that a mean is greater
(less) than or equal to some theoretical value

Laczniak Yogurt is test marketing new flavor
(grape). History tells them that the new flavor will
be successful if they sell, on average, more than
100 cartons in one month in each of the test
markets.
Advanced Data Analysis – Mean
Comparisons
Store
Sales ($000)
1
86
2
97
3
114
4
108
5
123
6
93
7
132
8
116
9
105
10
120
Advanced Data Analysis – Mean
Comparisons

Steps in Hypothesis Testing

State Null and alternative hypotheses (Null always
means no relationship – reject it if you can)



Specify the level of statistical significance (a)


Ho: m ≤ 100
Ha: m >100
.05, obviously
Specify the critical region to be used (Table 4 – P.
828)

Critical t (df = n-1 =10-1=9) = 1.83
Advanced Data Analysis – Mean
Comparisons

Collect data and determine test statistic
 For example




n = 10 (with SRS)
mean = 109.4
s = 14.40
Test hypothesis -- reject or retain the null hypothesis
 Compute t





Where t = computed mean – m/ √ n
109.4 – 100/(14.4/√10)
Computed t = 2.07
Computed t ≥(?) Critical t
2.07 ≥ 1.83 -- Reject the null hypothesis
x
x
Advanced Data Analysis – Mean
Comparisons

What if:



n = 10 (with SRS)
Mean = 103.4
s = 10.0

n = 50 (with SRS)
Mean = 103.4
s = 10.0

Importance of n and s


Advanced Data Analysis – Mean
Comparisons

Hypothesis Tests

t-test

Used to test hypotheses that a mean is greater (less)
than or equal to a different empirically derived value

For example, test the hypothesis is Mean1 > Mean 2
Advanced Data Analysis – Mean
Comparisons

Laczniak Yogurt Company is interested in using a new container
design for its product. Two alternative designs (one red, the
other orange) were chosen to test in 20 stores (ten for each
design) in Iowa. Laczniak management wants to use the design
that lead to the highest sales level in the test markets. Data for
the stores are:
Advanced Data Analysis – Mean
Comparisons
Store
1
2
3
4
5
6
7
8
9
10
Red Container
432
360
397
408
417
380
422
406
400
408
Orange Container
365
405
396
390
404
372
378
410
383
400
Advanced Data Analysis – Mean
Comparisons


Here is a summary of the data:
Red



Orange





Mean = 390.3
s = 15.26
State Null and alternative hypotheses


Mean = 403
s = 20.76
Ho: m1  m2
Ha: m1 > m2
Compute t statistic (t = 1.56); critical t (df = n1 + n2 – 2)
 Critical t = 1.73
Test  If |Computed t| ≥ critical

Is 1.56 ≥ 1.73?

NO -- CANNOT REJECT THE NULL
Advanced Data Analysis – Mean
Comparisons

What if the mean and standard deviation
remained constant and the sample size for
both containers were 25?

What if the means remained the same and s
for both containers was 10.00 (n = 10)?
Advanced Data Analysis – Mean
Comparisons

Research Issues

Common for marketers to need to compare an empirical
percentage to a theoretical percentage (or standard)

Laczniak Yogurt will only introduce new flavors if they are
deemed to be “acceptable” by more than 20% of its customers
in a test market

Assume that Laczniak attempted to introduce a new plum
flavored yogurt

Had a sample of 625 randomly selected customers who
participated in the test market – 140 indicated that they would buy
the new flavor
Advanced Data Analysis – Mean
Comparisons

Steps in Hypothesis Testing

State Null and alternative hypotheses (Null always
means no relationship – reject it if you can)



Specify the level of statistical significance (a)


Ho:  ≤ .20
Ha:  > 20
.05, obviously
Compute test statistic and test hypothesis

z-statistic = p - /sp; where sp = √  (1 - )/ n
Advanced Data Analysis – Mean
Comparisons

Collect data and determine test statistic
 For example




n = 625 (with SRS)
p = 140/625 = .23
sp = √ .2 (.8)/625 = .016
Test hypothesis -- reject or retain the null hypothesis
 Compute z




Where z = p - /sp
.23 – .20/.016
Computed z = 1.88
Computed z ≥ Critical z (1.645 for alpha = .05)


1.88 ≥ 1.645
Reject the null hypothesis
Advanced Data Analysis – Mean
Comparisons

What if the same data were generated, but the
sample size was 225?