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:
=28 Companies
=70%=0.70 Success
Sample 2:
=20 Companies
=40%=0.40 Success
Given, assuming =0.01.
Proportion
The standard error (SE)=
=0.1447
Then,
Z-Statistics=
=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:
Level of significance =0.01
Sample (I):
Mean of the sample ()=1.30 million
Standard deviation ()=0.6 million
=28
Sample (II):
Mean of the sample ()=1.90 million
Standard deviation ()=0.55 million
=20
=1.3-1.9=-0.60
Pooled Standard Deviation:
=0.579
Thus,
Standard error=
=0.1698
Degree of freedom (df)=
=28+20-2
=46
Test Statistics (t)=
=-0.3533
T-critical value=t.inv(/2,df) from excell
=2.607
Since |t-stat| > (Critical value)
So, we reject .
Conclusion:
There is enough evidence to reject . So is true.
[i.e, The average finding in incubator B is significantly different from average A]