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emmasim [6.3K]
3 years ago
12

A survey of 102 high school students on the amount of money they spend on prom yields an average of $893. Assuming the populatio

n 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.
Mathematics
1 answer:
Mnenie [13.5K]3 years ago
7 0

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]

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