Given:
number of people (n) = 12
mean = 39.1
standard deviation = 17.4
99% confidence level
Using the confidence level formula, we can find the estimate of how much a typical parent would spend on their child's birthday:
[tex]\begin{gathered} CI\text{ = x }\pm\text{ }\frac{z\varphi}{\sqrt[]{n}} \\ \text{where x is the mean} \\ z\text{ is the z-score at 99\% confidence interval} \\ \varphi\text{ is the standard deviation} \\ n\text{ is the number of people asked} \end{gathered}[/tex]The z-score at 99% confidence level is 2.576
Substituting, we have:
[tex]\begin{gathered} CI\text{ = 39.1 }\pm\text{ }\frac{2.576\text{ }\times\text{ 17.4}}{\sqrt[]{12}} \\ =26.161\text{ and 52}.039 \end{gathered}[/tex]Hence, a typical parent would spend between $26.161 and $52.039 or :
[tex]39.1\text{ }\pm\text{ 12.939}[/tex]