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A 50% confidence level has a confidence interval of 0.6745. A 90% confidence level has a confidence interval of 1.96. A 99% c
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Answer #1

The standard error of the mean value is given as:

SE = Z\frac{ \sigma }{ \sqrt[]{n}}

where

Z = confidence interval,

\sigma = standard deviation,

n = number of samples.

According to the given question,

Z (for 90% confidence level) = 1.96

\sigma = 0.46 feet

n = 17

Hence,

SE = \frac{(1.96)*(0.46)}{\sqrt{}17}

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