Question

As reported by the U.S. Census Bureau in Educational Attainment in the United States, the percentage of adults in each state who have completed a bachelor's degree is provided on the Weiss Stats site. Use the technology of you choice to solve the following problems.

Part (a): Obtain the standard deviation of the variable "percentage of adults who have completed a bachelor's degree" for the population of 50 states.

Part (b): Consider simple random samples without replacement from the population of 50 states. Strictly speaking, which is the correct formula for obtaining the standard deviation of the sample mean- Equation (7.1) or Equation (7.2)? Explain your answer.

Part (c): Referring to part (b), obtain R for simple random samples of size 30 by using both formulas. Why does Equation (7.2) provide such a poor estimate of the true value given by Equation (7.1)?

Part (d): Referring to part (b), obtain R for simple random samples of size 2 by using both formulas. Why does Equation (7.2) provide a somewhat reasonable estimate of the true value given by Equation (7.1)?

Short answer

Part (a): The standard deviation of the variable "percentage of adults who have completed a bachelor's degree" for the population of 50 states is 4.733.

Part (b): The correct formula for obtaining the standard deviation of the sample mean- Equation (7.1) is ${\sigma }_{\overline{x}}=\left(\sqrt{\frac{N-n}{N-1}}\right)\left(\frac{\sigma }{\sqrt{n}}\right)$.

The correct formula for obtaining the standard deviation of the sample mean- Equation (7.2) is ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

Part (c): When the sample size is 5% or less than the population size, it is expected to have both the results to be approximately close. But, here the sample size is 60% of the population and hence the result using equation 7.1 is significantly less than 1. Moreover, there is a large discrepancy between the two values.

Part (d): When the sample size is 5% or less than the population size, it is expected to have both the results to be approximately close. Hence, the two values are approximately close.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

No. of states is50.

Part (a) Step 2. Find the standard deviation of the variable.

The standard deviation of the variable "percentage of adults who have completed a bachelor's degree" for the population of 50 states.

Using Excel, the standard deviation is4.733.

Part (b) Step 1. Determine which is the correct formula for obtaining the standard deviation of the sample mean.

When sampling is done without replacement from a finite population, then the standard deviation for sample mean is obtained using equation 7.1,

${\sigma }_{\overline{x}}=\left(\sqrt{\frac{N-n}{N-1}}\right)\left(\frac{\sigma }{\sqrt{n}}\right)$

When sampling is done with replacement from a finite population, then the standard deviation for sample mean is obtained using equation 7.1,

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

For given situation, the sampling is done without replacement. Hence, it can be concluded that equation 7.1, is correct to obtain the standard deviation for sample mean.

Step (c) Step 1. Determine the standard deviation of the sample using equation 7.1.

The standard deviation $\left({\sigma }_{\overline{x}}\right)$ of the sample mean using Equation 7.1,

Substitute $$\begin{gathered} N=50, \\ n=30, \\ \sigma =4.733 \end{gathered}$$ in equation 7.1,

localid="1652635664606" role="math" $$\begin{gathered} {\sigma }_{\overline{x}}=\left(\sqrt{\frac{50-30}{50-1}}\right)\left(\frac{4.733}{\sqrt{30}}\right) \\ =\left(\sqrt{\frac{20}{49}}\right)\left(\frac{4.733}{\sqrt{5.4772}}\right) \\ =0.552 \end{gathered}$$

The standard deviation $\left({\sigma }_{\overline{x}}\right)$ of the sample mean using Equation 7.2,

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\left(\frac{4.733}{\sqrt{30}}\right) \\ =0.864 \end{gathered}$$

Step (c) Step 2. Find the sample size of 30 is 60% of the population size.

The sample size of 30 is $60\%$of the population size. Then,

$$\begin{gathered} \frac{Samplesize}{Populationsize}\times 100=\frac{30}{50}\times 100 \\ =60\% \end{gathered}$$

When the sample size is $5\%$or less than the population size, it is expected to have both the results to be approximately close. But, here the sample size is $60\%$ of the population and hence the result using equation 7.1 is significantly less than 1. Moreover, there is a large discrepancy between the two values.

Part (d) Step 1. Determine the standard deviation of the sample mean using equation 7.1 for sample size 2.

The standard deviation $\left({\sigma }_{\overline{x}}\right)$ of the sample mean using Equation 7.1for sample size 2,

Substitute $$\begin{gathered} N=50, \\ n=2, \\ \sigma =4.733 \end{gathered}$$ in equation 7.1,

$$\begin{gathered} {\sigma }_{\overline{x}}=\left(\sqrt{\frac{50-2}{50-1}}\right)\left(\frac{4.733}{\sqrt{2}}\right) \\ =\left(\sqrt{\frac{48}{49}}\right)\left(\frac{4.733}{\sqrt{1.4142}}\right) \\ =3.312 \end{gathered}$$

The standard deviation $\left({\sigma }_{\overline{x}}\right)$ of the sample mean using Equation 7.2for sample size 2,

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{2}} \\ =\frac{4.733}{\sqrt{2}} \\ =3.347 \end{gathered}$$

Step (d) Step 2. Find the sample size of 2 is 4% of the population size.

The sample size of 2 is $4\%$ of the population size,

$$\begin{gathered} \frac{Samplesize}{Populationsize}\times 100=\frac{2}{50}\times 100 \\ =4\% \\ <5\% \end{gathered}$$

When the sample size is $5\%$ or less than the population size, it is expected to have both the results to be approximately close. Hence, the two values are approximately close.