Chapter 8.1, Problem 67E

a.

To determine

To fill: The blanks in thegiven statement with appropriate words.

Answer to Problem 67E

95%, 9.6956 and 15.0084.

Explanation of Solution

.

Given information:

  $\begin{array}{l}\text{95\%}\text{Confidence}\text{interval}\text{=}\left(9.6956,15.0084\right)\\ \text{Sample}\text{mean}\left(\text{<mover accent="true"><mo>x</mo><mo>¯</mo></mover>}\right)=12.352\\ \text{Sample}\text{size}\left(N\right)=23\\ \text{Population}\text{standard}\text{deviation}\left(\sigma \right)=6.5000\\ {\text{SE}}_{Mean}\left(\frac{\sigma }{\sqrt{N}}\right)=1.3553\end{array}$

Confidence interval is the interval estimate of unknown population parameter. It gives the probability that population parameter will fall in this interval.It means that there is 95% surety that population mean weight loss lies between 9.6956 and 15.0084.

Thus, the complete statement is, “ We are $\underset{\_}{95\%}$ confident that the population mean weight loss is between $\underset{\_}{9.6956}$ and $\underset{\_}{15.0084}$ ”.

b.

To determine

To find:The 99% confidence interval.

Answer to Problem 67E

The confidence interval is $\left(8.862,15.842\right)$ .

Explanation of Solution

Given:

Confidence level = 99%

Formula used:

  $\overline{x}-z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}<\mu <\overline{x}+z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}$

Calculation:

The value of $z_{\frac{\alpha }{2}}$ corresponding to $99\%$ confidence level is 2.575, which is obtained from the standard normal table.

The 99% confidence interval using the provided output can be calculated as:

  $\begin{array}{c}\overline{x}-z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}<\mu <\overline{x}+z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}\\ 12.352-2.575\times 1.3553<\mu <12.352+2.575\times 1.3553\\ 12.352-3.490<\mu <12.352+3.490\\ 8.862<\mu <15.842\end{array}$

Hence, the confidence interval is $\left(8.862,15.842\right)$ .

c.

To determine

To find: The sample size to have the margin of error of 1.5 at $95\%$ confidence level.

Answer to Problem 67E

The required sample size is 72.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Margin of error }\left(m\right)=1.5\\ \text{Confidence level = 95\%}\end{array}$

Formula used:

  $n={\left(\frac{z_{\frac{\alpha }{2}}\times \sigma }{m}\right)}^2$

Calculation:

The value of $z_{\frac{\alpha }{2}}$ corresponding to $95\%$ confidence level is 1.96.

The sample size at $95\%$ confidence level can be calculated as:

  $\begin{array}{c}n={\left(\frac{z_{\frac{\alpha }{2}}\times \sigma }{m}\right)}^2\\ ={\left(\frac{1.96\times 6.500}{1.5}\right)}^2\\ =72.137\\ \approx 72\end{array}$

Hence, the sample size is 72.

d.

To determine

To find: The sample size to have the margin of error of 1.5 at $99\%$ confidence level.

Answer to Problem 67E

The required sample size is 125.

Explanation of Solution

Given:

  $\text{Confidence level = 99\%}$

Formula used:

  $n={\left(\frac{z_{\frac{\alpha }{2}}\times \sigma }{m}\right)}^2$

Calculation:

The value of $z_{\frac{\alpha }{2}}$ corresponding to $99\%$ confidence level is 2.575.

The sample size at $99\%$ confidence level can be calculated as:

  $\begin{array}{c}n={\left(\frac{z_{\frac{\alpha }{2}}\times \sigma }{m}\right)}^2\\ ={\left(\frac{2.575\times 6.500}{1.5}\right)}^2\\ =124.508\\ \approx 125\end{array}$

Hence, the sample size is 125.