Question

In Example 7.5, we used the definition of the standard deviation of a variable to obtain the standard deviation of the heights of the five starting players on a men's basketball team and also the standard deviation of $\overline{x}$ for samples of sizes 1,2,3,4,5.The results are summarized in Table 7.6on page 298. Because the sampling is without replacement from a finite population, Equation (7.1) can also be used to obtain ${\sigma }_{\overline{x}}$.

Part (a): Apply Equation (7.1) to compute ${\sigma }_{\overline{x}}$ for sample sizes of 1,2,3,4,5. Compare your answers with those in Table 7.6.

Part (b): Use the simpler formula, Equation (7.2) to compute ${\sigma }_{\overline{x}}$ for samples of sizes 1,2,3,4,5.Compare your answers with those in Table 7.6. Why does Equation (7.2)generally yield such poor approximations to the true values?

Part (c): What percentages of the population size are samples of sizes 1,2,3,4,5.

Short answer

Part (a): The values of ${\sigma }_{\overline{x}}$ obtained by using equation 7.1are approximately equal to the actual values of ${\sigma }_{\overline{x}}$obtained by using the definition of standard deviation of a variable.

Part (b): When the sample size is small relative to the population size, there is little difference between sampling without and with relatpacement, i.e. equation 7.1 and 7.2provide approximately same value of ${\sigma }_{\overline{x}}$for relatively small sample sizes compared to population size.

Here the smallest sample size, i.e., $n=1$ is equal to 20%of the population size $N=5$.

Part (c): The population size are samples of sizes 1,2,3,4,5 is given below,

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

Population standard deviation is $\sigma =3.41$.

Population size$N=5$.

Part (a) Step 1. Make a table.

Construct a table,

We can see that the values of ${\sigma }_{\overline{x}}$ obtained by using equation 7.1are approximately equal to the actual values of data-custom-editor="chemistry" ${\sigma }_{\overline{x}}$obtained by using the definition of standard deviation of a variable.

Part (b) Step 1.

Consider the table,

We have used simple random sampling without replacement procedure to draw the samples from the population and for this procedure, the appropriate formula for obtaining the value of ${\sigma }_{\overline{x}}$ is equation 7.1 i.e.,${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\times \sqrt{\frac{N-n}{N-1}}$.

We know that, when the sample size is small relative to the population size, there is little difference between sampling without and with relatpacement, i.e. equation 7.1 and 7.2provide approximately same value of ${\sigma }_{\overline{x}}$for relatively small sample sizes compared to population size.

We say that the sample size is small relative to the population size if the size of the sample does not exceed 5% of the size of the population $\left(n\le 0.05N\right)$. But, here the smallest sample size, i.e., $n=1$ is equal to 20%of the population size $N=5$.

That is why the equation 7.2 yield such poor approximations to the true values of the ${\sigma }_{\overline{x}}$in this problem.

Part (a) Step 1. Find the percentages of the population size are samples of sizes 1,2,3,4,5.

The population size is $N=5$.