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Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampl

galben [10]

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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 $\sigma^2=\dfrac{p(1-p)}{n}$

$=\dfrac{0.9(1-0.9)}{200}$

$=\dfrac{0.9(0.1)}{200}$

$\mathbf{=4.5*10^{-4}}$

The standard error of σ(p) = $\sqrt{\sigma ^2}$

$\mathbf{= \sqrt{4.5*10^{-4}}}$

= 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 $\sigma^2=\dfrac{p(1-p)}{n}$

$=\dfrac{0.1(1-0.1)}{200}$

$=\dfrac{0.1(0.9)}{200}$

$\mathbf{=4.5*10^{-4}}$

The standard error of σ(p) = $\sqrt{\sigma ^2}$

$\mathbf{= \sqrt{4.5*10^{-4}}}$

= 0.0212

4 0

3 years ago