In: Statistics and Probability
Incubator A has a 70% success ratio getting companies to survive at least 4 years from inception. The average venture funding of the 28 companies reaching that 4 year mark, is 1.3 million dollars with a standard deviation of 0.6 million.
Incubator B has a 40% success ratio getting companies to survive at least 4 years from inception. The average venture funding of the 20 companies reaching that 4 year mark, is 1.9 million dollars with a standard deviation of 0.55 million
Answer:
Given that,
You and a group of friends wish to start a company. You have an
idea, and you are comparing startup incubators to apply to.
(Startup incubators hold classes and help startups to contact
venture capitalists and network with one another) Assume funding is
normally distributed.
Incubator A has a 70% success ratio getting companies to survive at
least 4 years from inception. The average venture funding of the 28
companies reaching that 4-year mark, is 1.3 million dollars with a
standard deviation of 0.6 million.
Incubator B has a 40% success ratio getting companies to survive at
least 4 years from inception. The average venture funding of the 20
companies reaching that 4-year mark, is 1.9 million dollars with a
standard deviation of 0.55 million
(a).
Are the success ratios significantly different? (note the count is given but not N, how do you find N?):
Test of significance for difference of properties.
Given that,
Sample 1:
Sample 2:
=0.1447
Then,
=2.073
P-value:
P-value=P(Z > 2.073)+P(Z < 2.073)
=0.038452
Since p-value 0.038 is greater than 0.01 so we do not reject
.
Conclusion:
The success ratio are not significantly different.
(b).
Is the average funding in incubator B significantly
different from the average funding in a? (use
=0.01). Assume a normal distribution:
Sample (I):
Mean of the sample ()=1.30
million
Standard deviation ()=0.6
million
Sample (II):
Mean of the sample ()=1.90
million
Standard deviation ()=0.55
million
Pooled Standard Deviation:
=0.579
Thus,
=0.1698
=28+20-2
=46
=-0.3533
T-critical value=t.inv(
/2,df)
from excell
=2.607
Since |t-stat| > (Critical value)
Conclusion:
There is enough evidence to reject
. So
is true.
[i.e, The average finding in incubator B is significantly different from average A]