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:
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/ec89pwyu87c2hxtmvok8buyi0o6xug5o8f.png)
The z-score at 99% confidence level is 2.576
Substituting, we have:
![\begin{gathered} CI\text{ = 39.1 }\pm\text{ }\frac{2.576\text{ }*\text{ 17.4}}{\sqrt[]{12}} \\ =26.161\text{ and 52}.039 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/p0mlojddc0xrcexff3hs5xe01rd54eddu3.png)
Hence, a typical parent would spend between $26.161 and $52.039 or :
