Q 8.19. - Elementary Statistics

Question

Wedding Costs. Refer to Exercise $8.17.$ Assume that recent wedding costs in the United States are normally distributed with a standard deviation of $\$8100$

a. Determine a $95\%$ confidence interval for the mean cost. $\mu$, of all recent U.S. weddings.

b. Interpret your result in part (a).

c. Does the mean cost of all recent U.S. weddings lie in the confidence interval you obtained in part (a)? Explain your answer.

Step-by-Step Solution

Answer

Part (a) The $95\%$ confidence interval for the mean cost, $\mu$ of all recent $U . S$ weddings is from $22,704.4$ to $29,949.4$

Part (b) The mean cost $\left(\mu \right)$ of all recent U.S. weddings lies within the $95\%$ confidence interval obtained.

Part (c) The mean cost of all recent U.S. weddings lies between $22,704.4$ and $29,949.4$

Part (a) Step 1: Given information

In the United States, wedding costs are generally distributed with a standard deviation of $\$8100$

Part (a) Step 2: Concept

The formula used: $\overline{x}-\frac{2\sigma }{\sqrt{n}}\text{ to }\overline{x}+\frac{2\sigma }{\sqrt{n}}$

Part (a) Step 3: Calculation

Calculate a $95\%$ confidence interval for the average cost of all recent $U . S$ weddings, $\mu$

From the given information, the standard deviation is$\$8,100$

From Exercise 17E, $\overline{x}=26,326.9,n=20$

Empirical rule:

Property 1: Around $68\%$ of the data set is located between $(\overline{x}-s,\overline{ x}+s)$

Property 2: Approximately $95\%$ of the data set is located between $(\overline{x}-2s,\overline{ x}+2s)$

Property 3: Approximately $99.7\%$ of the data set is located between $(\overline{x}-3s,\overline{ x}+3s)$

Using Property $2$, $95\%$ of all observations fall within two standard deviations of the mean on either side.

The $95\%$ confidence interval for the population mean is,

$$\begin{gathered} \overline{x}-\frac{2\sigma }{\sqrt{n}}\text{ to }\overline{x}+\frac{2\sigma }{\sqrt{n}}=26,326.9-\frac{2(8,100)}{\sqrt{20}}\text{ to }26,326.9+\frac{2(8,100)}{\sqrt{20}} \\ =26,326.9-\frac{16,200}{4.4721}\text{ to }26,326.9+\frac{16,200}{4.4721} \\ =26,326.9-36,22.5\text{ to }26,326.9+3,622.5 \\ =22,704.4\text{ to }29,949.4 \end{gathered}$$

Thus, the $95\%$ confidence interval for the mean cost, $\mu$, of all recent $U . S$ weddings is from $22,704.4$ to $29,949.4$

Part (b) Step 1: Explanation

The $95\%$ confidence interval for the mean cost of all recent marriages in the United States is $22,704.4$ to $29,949.4$ The mean cost $\left(\mu \right)$ of all recent weddings in the United States is within the $95\%$ confidence interval.

Part (c) Step 1: Explanation

Because there is only $95\%$ confidence that the mean cost of all recent U.S. weddings is between $22,704.4$ and $29,949.4$ the mean cost of all recent U.S. weddings may or may not be in the confidence interval produced in component (a).