Question 7 of 14, Step 1 of 1 A survey of 84 randomly selected homeowners finds that they spend a mean of $$ \$ 60 $$ per month on home maintenance. Construct a $$ 98 \% $$ confidence interval for the mean amount of money spent per month on home maintenance by all homeowners. Assume that the population standard deviation is \$14 per month. Round to the nearest cent.

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The confidence interval is constructed using the formula

$$
\bar{x} \pm z_{\frac{\alpha}{2}} \left(\frac{\sigma}{\sqrt{n}}\right)
$$

where

- $$\bar{x} = 60$$ (sample mean)
- $$z_{\frac{\alpha}{2}}$$ is the critical value for a $$98\%$$ confidence level. For $$98\%$$, $$\alpha = 0.02$$ and $$\frac{\alpha}{2} = 0.01$$, so $$z_{0.01} \approx 2.3263$$.
- $$\sigma = 14$$ (population standard deviation)
- $$n = 84$$ (sample size)

First, compute the standard error:

$$
\text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{14}{\sqrt{84}}
$$

Calculating $$\sqrt{84} \approx 9.165$$, we have

$$
\frac{14}{9.165} \approx 1.526
$$

Next, calculate the margin of error:

$$
\text{Margin of Error} = 2.3263 \times 1.526 \approx 3.55
$$

Thus, the $$98\%$$ confidence interval is:

$$
60 \pm 3.55
$$

which gives

$$
\text{Lower bound} = 60 - 3.55 = 56.45
$$
$$
\text{Upper bound} = 60 + 3.55 = 63.55
$$

The final $$98\%$$ confidence interval for the mean amount of money spent per month on home maintenance is

$$
(\$56.45, \$63.55)
$$
The $$98\%$$ confidence interval for the mean amount of money spent per month on home maintenance is $$\$56.45$$ to $$\$63.55$$.

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