Question

7.56 Heights of Starting Players. In Example $7.5$, we used the definition of the standard deviation of a variable (Definition $3.12$on page $142$) 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 $1,2,3,4,$and $5$. The results are summarized in Table $7.6$on page $298$. Because the sampling is without replacement from a finite population, Equation $(7.1)$can also be used to obtain role="math" localid="1651069065157" ${\sigma }_{\overline{x}}$.
a. Apply Equation $(7.1)$to compute role="math" localid="1651069501306" ${\sigma }_{\overline{x}}$for samples of sizes $1,2,3,4,$and 5. Compare your answers with those in Table $7.6$.
b. Use the simpler formula, Equation $(7.2)$, to compute role="math" localid="1651069072557" ${\sigma }_{\overline{x}}$for samples of sizes $1,2,3,4,$and $5$. Compare your answers with those in Table$7.6.$Why does Equation $(7.2)$ generally yield such poor approximations to the true values?
c. What percentages of the population size are samples of sizes $1,2,3.4,$and $5$?

Short answer

(a) The values of ${\sigma }_{\overline{x}}$ obtained by applying the definition of standard deviation of a variable are almost equal to the values obtained by using Equation $(7.1)$.

(b) The smallest sample size, $n=1,$corresponds to $20\%$of the population size, $N=5$. That is why the equation $7.2$in this problem gives such poor approximations to the true values of ${\sigma }_{\overline{x}}$.

(c) The percentage of the population size are samples of sizes $1,2,3,4$and $5$are $20\%,40\%,60\%,80\%$ and $100\%$.

Step-by-step solution

Part (a) Step 1: Given information

To compute ${\sigma }_{\overline{x}}$for samples of sizes $1,2,3,4,$and $5$and compare the answers with Table $7.6$.

Part (a) Step 2: Explanation

Let, Population standard deviation is $\sigma =3.41$. And the Population size is $N=5$

Sample size	Value of${\sigma }_{\overline{x}}$ obtained by using Equation 7.1. ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\times \sqrt{\frac{N-n}{N-1}}$	Value of${\sigma }_{\overline{x}}$ using definition of standard deviation (As in table 7.6)
1	3.41	3.41
2	2.0881	2.09
3	1.3921	1.39
4	0.8525	0.85
5	0.00	0.00

Part (b) Step 1: Given information

To compute ${\sigma }_{\overline{x}}$for samples of sizes$1,2,3,4,$ and $5$ and compare the answers with Table $7.6$ and explain why the equation $(7.2)$ generally yields such poor approximations to the true values.

Part (b) Step 2: Explanation

Let, the population standard deviation is $\sigma =3.41$.

The appropriate formula for determining the value of ${\sigma }_{\overline{x}}$is equation $7.1$, i.e.,${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\times \sqrt{\frac{N-n}{N-1}}$, when using the basic random sampling without replacement technique to pick the samples from the population.
Equation $7.2$is, ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$
Because the sample size is small in comparison to the population size, there is little difference between sampling without and with replacement. In other words equations $7.1$and $7.2$produce roughly the same value of ${\sigma }_{\overline{X}}$ for small sample sizes.
As a rule of thumb, if the sample size does not exceed $5\%$of the population size (i.e. $n\le 0.05N$), the sample size is small relative to the population size. However, in this case, the smallest sample size, $n=1,$is equal to $20\%$of the population size, $N=5$.

This is why the equation $7.2$in this problem approximates the true values of ${\sigma }_{\overline{x}}$ so poorly.

Part (c) Step 1: Given information

To find the percentages of the population size of samples of sizes $1,2,3,4,$ and $5$.

Part (c) Step 2: Explanation

The population size is $N=5$.

Sample size	Percentage of population size $\frac{n}{N}\times 100\%$
1	20%
2	40%
3	60%
4	80%
5	100%

Hence, the percentages of the population size of samples of sizes $1,2,3,4,$and $5$are obtained.