Question

New Mobile Homes. Refer to Examples 8.1 and 8.2. Use the data in Table \(8.1\) on page 315 to obtain a \(99.7 \%\) confidence interval for the mean price of all new mobile homes. (Hint: Proceed as in Example 8.2, but use Property 3 of the empirical rule on page 271 instead of Property 2.)

Short answer

$(\$59678,\$66878)$

Step-by-step solution

Given information

We have a sample of new mobile house pricing in size 36.

Concept

The formula used:$\overline{x}=\frac{1}{n}\times \sum _{i=1}^n{x}_i$and${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Calculation

We have a sample of new mobile house pricing in size $36$

Let $x_i$be the $i-th$observation on price from the sample, $i=1,2,\dots \dots .,36$From the data we have $\sum _{i=1}^{36}{x}_i=2278.0$

$\therefore$Sample mean price $\overline{x}=\frac{1}{n}\times \sum _{i=1}^n{x}_i$

$$\begin{gathered} =\frac{1}{36}\times 2278.0 \\ =\$63.278\text{Thousand} \\ =\$63,278 \end{gathered}$$

Let the population mean be $\mu$

Population standard deviation

$\sigma =\$7.2$Thousand

$=\$7200$

$\therefore$Standard deviation of the sample mean

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{7200}{\sqrt{36}} \\ =\$1200 \end{gathered}$$

This sample can now be considered a large sample because $36$is greater than $30$

The sample mean $\overline{x}$follows a normal distribution with mean $\mu$and standard deviation${\sigma }_{\overline{X}}=\$1200$ according to the central limit theorem.

Calculation

Therefore, $99.74\%$the Confidence interval of the population mean $\mu$is given by $\left(\overline{x}-3{\sigma }_{\overline{x}},\overline{x}+3{\sigma }_{\overline{x}}\right)$

$\therefore 99.74\%$ Confidence interval of $\mu$

$$\begin{gathered} =\left(\overline{x}-3{\sigma }_{\overline{x}},\overline{x}+3{\sigma }_{\overline{x}}\right) \\ =(63278-3\times 1200,63278+3\times 1200) \\ =(\$59678,\$66878) \end{gathered}$$