Chapter 8.3, Problem 67E

(a)

To determine

To calculate: The population mean for the confidence interval of 99%.

Answer to Problem 67E

Thepopulation mean is between −4.5754% to −2.5986%.

Explanation of Solution

Given information:

Sample size $n=47$

Standard deviation $s_x=2.506$

Population mean $\overline{x}=-3.587$

Formula used:

Degree of freedom $df=n-1$

Margin of error $E=t^{*}\frac{s_x}{\sqrt{n}}$

Calculation:

Find the degree of freedom, use the formula $df=n-1$ .

  $\begin{array}{l}df=n-1\\ =47-1\\ =46\end{array}$

The confidence level is 99%.

Convert 99% into decimal.

  $\frac{99}{100}=0.99$

Find the value of the column.

  $\begin{array}{c}\frac{1-c}{2}=\frac{1-0.99}{2}\\ =0.005\end{array}$

From table B find critical value $t^{*}$ , use the row $df$ and column.

Thus, critical value $t^{*}\text{is}2.704$ .

Find the margin of error, use the formula $E=t^{*}\frac{s_x}{\sqrt{n}}$ .

  $\begin{array}{c}E=t^{*}\frac{s_x}{\sqrt{n}}\\ =2.704\times \frac{2.506}{\sqrt{47}}\\ =0.9884\end{array}$

Now, find the population mean.

  $\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ -3.587-0.9884<\mu <-3.587+0.9884\\ -4.5754<\mu <-2.5986\end{array}$

Hence, the required population mean is between -4.5754% to -2.5986%.

(b)

To determine

Explain the reason whether or not the data gives good evidence that on the average nursing mothers lose bone mineral

Answer to Problem 67E

The data gives good evidence that on the average nursing mothers lose bone mineral

Explanation of Solution

From 67(a)confidence interval contain the population mean from -4.5754% to -2.5986%.Since the confidence interval lies below 0.

Hence, the data gives good evidence that on the average nursing mothers lose bone mineral.