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Taking u 2
¼ U p , we got the following approach for the solution of the inequality:
u 2
¼ t 2
a 2 z 2
2 abz þ b 2
α
u 2
¼
2
2 z 2
þ β
a 2 z 2
2 abz þ b 2
u 2
2 z 2
u 2
2
α
β
0 ¼
2
α
2 z 2
þ β
0 ¼ z 2 a 2
u 2
2
Þ z 2 aðÞþb 2
u 2
2
ð
α
β
b 2
u 2
2
2 ab
β
0 ¼ z 2
z
2 þ
2 :
a 2
u 2
α
a 2
u 2
α
The solution formula for quadratic equations
r
v 2
4 w
v
2
z u , o ¼
ð 11
5 Þ
:
with
b 2
u 2
2
2 ab
β
v ¼
2 , w ¼
2 ,
ð 11
6 Þ
:
a 2
u 2
a 2
u 2
α
α
leads to the desired confidence interval:
z u d þ 1 z o
z u 1 d z o 1
ð 11
7 Þ
:
:
We now turn to the implementation. First, we need to determine the values
a , b ,
. Here EX A is “value per visit” of the control group and EX B the similar
value for the recommendation group. For the calculation of the confidence interval
for the average order revenue, similarly “avg. order value” has to be used.
A confidence interval “CRO” can also be determined.
The variance D 2 X A of the control group and D 2 X B of the recommendation group
can be calculated via the quadratic sums of the order revenues:
α
,
β
D 2 X A ¼ EX A EX A
2
ð
Þ
:
2
D 2 X B ¼ EX B EX B
ð
Þ
The values n A and n B depend on the target quantity “visits” or “orders.”
The quantile of the normal distribution U p , and hence u , depends on the desired
confidence level and is a constant:
• 90 % confidence interval: U p ¼ U 0,95 ¼ 1.6449
• 95 % confidence interval: U p ¼ U 0,975 ¼ 1.9600
• 99 % confidence interval: U p ¼ U 0,995 ¼ 2.5758
Therewith, using ( 11.6 ), we can compute p , q , and by ( 11.5 ) and ( 11.7 ), we get
the desired interval.
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