Question

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)?

Answer 1

Answer:

(a) $\hat p\sim N(0.90,\ 0.0212^{2}})$

(b) $\hat q\sim N(0.10,\ 0.0212^{2}})$

(c) Not different.

Step-by-step explanation:

The information provided is:

• The age at first startup for 90% of entrepreneurs was 29 years of age or less.
• The age at first startup for 10% of entrepreneurs was 30 years of age or more.

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 $\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

 $\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

(a)

Let p represent the proportion of entrepreneurs whose first startup was at 29 years of age or less.

A sample of n = 200 entrepreneurs is selected.

As n = 200 > 30, according to the Central limit theorem the sampling distribution of sample proportion can be approximated by the normal distribution.

Compute the mean and standard deviation as follows:

$\mu_{\hat p}=p=0.90\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.90(1-0.90)}{200}}=0.0212$

So, $\hat p\sim N(0.90,\ 0.0212^{2}})$.

(b)

Let q represent the proportion of entrepreneurs whose first startup was at 30 years of age or more.

A sample of n = 200 entrepreneurs is selected.

As n = 200 > 30, according to the Central limit theorem the sampling distribution of sample proportion can be approximated by the normal distribution.

Compute the mean and standard deviation as follows:

$\mu_{\hat q}=q=0.10\\\\\sigma_{\hat q}=\sqrt{\frac{q(1-q)}{n}}=\sqrt{\frac{0.10(1-0.10)}{200}}=0.0212$

So, $\hat q\sim N(0.10,\ 0.0212^{2}})$.

(c)

The standard deviation of sample proportions is also known as the standard error.

The standard deviation of p is, 0.0212.

The standard deviation of q is, 0.0212.

Thus, the standard errors of the sampling distributions in parts (a) and (b) are same.