Answer:
(a) ![\hat p\sim N(0.90,\ 0.0212^{2}})](https://tex.z-dn.net/?f=%5Chat%20p%5Csim%20N%280.90%2C%5C%200.0212%5E%7B2%7D%7D%29)
(b) ![\hat q\sim N(0.10,\ 0.0212^{2}})](https://tex.z-dn.net/?f=%5Chat%20q%5Csim%20N%280.10%2C%5C%200.0212%5E%7B2%7D%7D%29)
(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](https://tex.z-dn.net/?f=%5Cmu_%7B%5Chat%20p%7D%3Dp)
The standard deviation of this sampling distribution of sample proportion is:
![\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Chat%20p%7D%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
(a)
Let <em>p</em> represent the proportion of entrepreneurs whose first startup was at 29 years of age or less.
A sample of <em>n</em> = 200 entrepreneurs is selected.
As <em>n</em> = 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](https://tex.z-dn.net/?f=%5Cmu_%7B%5Chat%20p%7D%3Dp%3D0.90%5C%5C%5C%5C%5Csigma_%7B%5Chat%20p%7D%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.90%281-0.90%29%7D%7B200%7D%7D%3D0.0212)
So,
.
(b)
Let <em>q</em> represent the proportion of entrepreneurs whose first startup was at 30 years of age or more.
A sample of <em>n</em> = 200 entrepreneurs is selected.
As <em>n</em> = 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](https://tex.z-dn.net/?f=%5Cmu_%7B%5Chat%20q%7D%3Dq%3D0.10%5C%5C%5C%5C%5Csigma_%7B%5Chat%20q%7D%3D%5Csqrt%7B%5Cfrac%7Bq%281-q%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.10%281-0.10%29%7D%7B200%7D%7D%3D0.0212)
So,
.
(c)
The standard deviation of sample proportions is also known as the standard error.
The standard deviation of <em>p</em> is, 0.0212.
The standard deviation of <em>q</em> is, 0.0212.
Thus, the standard errors of the sampling distributions in parts (a) and (b) are same.