Question

A survey of 102 high school students on the amount of money they spend on prom yields an average of $893. Assuming the population standard deviation is $450, calculate the 88% confidence interval for the mean amount spent on prom. Round off the margin of error and the limits to the nearest whole number.

Answer 1

Answer:

Margin of error $=E=[69]$

Step-by-step explanation:

Data provided in the question:

$\bar{x}=\ \ 893$

$\sigma=\ \ 450$

$n=102$

At $88 \%$ confidence level the $z$ is.

$\alpha=1-88 \%=1-0.88=0.12$

$\frac{\alpha}{2}=\frac{0.12}{2}=0.06$

$z_{\frac \alpha 2}=20.06=1.555$

$\text { Margin of error }=E=Z_{\frac \alpha 2} \times \frac{\sigma}{\sqrt{n}}$

$E=1.555 \times \frac{450}{\sqrt{102}}$

Margin of error $=E=[69]$