Question

1. (4 points) by what factor should the sample size n be increased to reduce the standard error of the sample mean \\( \overline{x} \\) (of a random sample from a population) to:

a) one - half its original value?
b) one - fourth its original value?
justify your answers.

1. (4 points) recent data from the us census indicates that the mean age of college students is \\( \mu=25 \\) years, with a standard deviation \\( \sigma = 9.5 \\) years. a random sample of 125 students is drawn. what is the probability that the sample mean age of the students is greater than 26 years?

Explanation:

Step1: Recall the formula for standard error

The standard error of the sample mean is given by $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size.

Step2: Solve for part (a)

Let the original standard error be $\sigma_{\bar{X}_1}=\frac{\sigma}{\sqrt{n_1}}$ and the new standard error be $\sigma_{\bar{X}_2}=\frac{1}{2}\sigma_{\bar{X}_1}$. Then $\frac{\sigma}{\sqrt{n_2}}=\frac{1}{2}\times\frac{\sigma}{\sqrt{n_1}}$. Canceling out $\sigma$ (assuming $\sigma
eq0$), we get $\frac{1}{\sqrt{n_2}}=\frac{1}{2\sqrt{n_1}}$. Cross - multiplying gives $2\sqrt{n_1}=\sqrt{n_2}$. Squaring both sides: $n_2 = 4n_1$.

Step3: Solve for part (b)

Let the original standard error be $\sigma_{\bar{X}_1}=\frac{\sigma}{\sqrt{n_1}}$ and the new standard error be $\sigma_{\bar{X}_2}=\frac{1}{4}\sigma_{\bar{X}_1}$. Then $\frac{\sigma}{\sqrt{n_2}}=\frac{1}{4}\times\frac{\sigma}{\sqrt{n_1}}$. Canceling out $\sigma$ (assuming $\sigma
eq0$), we get $\frac{1}{\sqrt{n_2}}=\frac{1}{4\sqrt{n_1}}$. Cross - multiplying gives $4\sqrt{n_1}=\sqrt{n_2}$. Squaring both sides: $n_2 = 16n_1$.

Step4: Solve for part (5)

First, find the standard error of the sample mean: $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{9.5}{\sqrt{125}}\approx\frac{9.5}{11.18}\approx0.85$.
Then, find the z - score: $z=\frac{\bar{X}-\mu}{\sigma_{\bar{X}}}=\frac{26 - 25}{0.85}\approx1.18$.
Using the standard normal distribution table, $P(Z>1.18)=1 - P(Z\leq1.18)$. From the table, $P(Z\leq1.18) = 0.8810$. So $P(Z>1.18)=1 - 0.8810=0.1190$.

Answer:

a) The sample size should be increased by a factor of $4$.
b) The sample size should be increased by a factor of $16$.

1. The probability that the sample mean age of the students is greater than $26$ years is approximately $0.1190$.