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Consider the given question,

No. of states is50.

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.

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

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

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

${\mathrm{σ}}_{\overline{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$

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.

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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.