Question

In: Statistics and Probability

The Wall Street Journal reported that the age at first startup for 90% of entrepreneurs was...

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. (a) 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. np = n(1-p) = E(p) = σ(p) = (c) Are the standard errors of the sampling distributions different in parts (a) and (b)? Justify your answer. The input in the box below will not be graded, but may be reviewed and considered by your instructor.

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Expert Solution

a) p is the sample proportion of entrepreneurs whose first startup was at 29 years of age or less

Here we have to consider p = 0.9 (given)

sample size n = 200.

np = n * p = 200 * 0.9 = 180.0000

n(1-p) = 200*(1-0.9) = 20

here np and n(1-p) both are greater than 5 hence we can assume the data follows normal distribution.

Expected value of the sampling distribution of p = E(p̄) = p = 0.9

Variance = p(1-p)/n = (0.9*0.1)/200 = 4.5*10-4

standard error = 0.0212

b) p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more.

p = 0.1

sample size n = 200

np = 200*0.1 = 20

n(1-p) = 200*(1-0.1) = 180

here np and n(1-p) both are greater than 5 hence we can assume the data follows normal distribution.

Expected value of the sampling distribution of p = E(p̄) = p = 0.1

Variance = p(1-p)/n = (0.9*0.1)/200 = 4.5*10-4

standard error = 0.0212

c) Yes the standard errors in both the parts are same. because for a given sample size n if the p value of one sample is equal to (1-p) value of another sample then the standard error of both the samples will be same.


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