Question

A sample of 14 amusement park visitors spent a mean of $23.96 on food and drinks. The standard deviation of this sample was $3.61. Find the 95% confidence interval of the mean amount spent on food and drinks by all visitors to the amusement park. Round your answers to the nearest cent.

Answer 1

For a confidence level of 95%, you can use the next formula:

`\begin{gathered} (\bar{X}-Z_(\alpha/2)\cdot\frac{\sigma}{\sqrt[]{n}},\bar{X}+Z_(\alpha/2)\cdot\frac{\sigma}{\sqrt[]{n}}) \\ \text{Where, }\bar{X}\text{ is the mean, }\sigma\text{ is the standard deviation, n is the number of the sample} \\ \text{And for a confidence level of 95\% the interval is 1-}\alpha=0.95 \\ \text{Then }\alpha=1-0.95=0.05 \\ \text{And }\alpha/2=0.05/2=0.025 \end{gathered}`

If you look at the standard normal probability table, the Z value for 0.025 is -1.96 and 0.95+0.025=0.975, the Z-score for this value is 1.96 then:

`\begin{gathered} (23.96-1.96_{}\cdot\frac{3.61}{\sqrt[]{14}},23.96+1.96_{}\cdot\frac{3.61}{\sqrt[]{14}}) \\ (23.96-1.89,23.96+1.89) \end{gathered}`

The interval confidence is then (22.07, 25.85)