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The missing part of the question is highlighted in bold format
The Wall Street Journal reported that the age at first startup for 90% of entrepreneurs was 29 years of age or less and the age at first startup for 10% of entrepreneurs was 30 years of age or more.
Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is the sample proportion of entrepreneurs whose first startup was at 29 years of age or less. If required, round your answers to four decimal places. np = n(1-p) = E(p) = σ(p) = (b) Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more. If required, round your answers to four decimal places.
Answer:
(a)
np = 180
n(1-p) = 20
E(p) = p = 0.9
σ(p) = 0.0212
(b)
np = 20
n(1 - p) = 180
E(p) = p = 0.1
σ(p) = 0.0212
Step-by-step explanation:
From the given information:
Let consider p to be the sample proportion of entrepreneurs whose first startup was at 29 years of age or less
So;
Given that :
p = 90% i.e p = 0.9
sample size n = 200
Then;
np = 200 × 0.9 = 180
n(1-p) = 200 ( 1 - 0.9)
= 200 (0.1)
= 20
Since np and n(1-p) are > 5 ; let assume that the data follows a normal distribution ;
Then:
The expected value of the sampling distribution of p = E(p) = p = 0.9
Variance
The standard error of σ(p) =
= 0.0212
(b)
Here ;
p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more
p = 10% i.e p = 0.1
sample size n = 200
Then;
np = 200 × 0.1 = 20
n(1 - p) = 200 (1 - 0.1 ) = 180
Since np and n(1-p) are > 5 ; let assume that the data follows a normal distribution ;
Then:
The Expected value of the sampling distribution of p = E(p) = p = 0.1
Variance
The standard error of σ(p) =
= 0.0212