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

Margin of error
`=E=[69]`

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`

`(\alpha)/(2)=(0.12)/(2)=0.06`

`z_(\frac \alpha 2)=20.06=1.555`

`\text { Margin of error }=E=Z_(\frac \alpha 2) * (\sigma)/(√(n))`

`E=1.555 * (450)/(√(102))`

Margin of error
`=E=[69]`