The 98% confidence interval estimate of the population mean is
15.823 < μ < 22.177
In the given situation the wind on a random day in Bryan is normally distributed with the following values;
Standard Deviation = ( δ ) = 5.1 mph
A random day of 16 is taken into account for the consideration of Bryan's average value of 19mph.
n = 16
By taking the confidence level of T - Factor, we get the;
At a 98% confidence level, the t is,
tα /2,df = t₀.₀₄,₂₄ = 2.492 ( df = hours in a day)
Margin of error = E = tα/2,df * (δ /√n)
= 2.492 * (5.1 / √16)
= 3.177
The 98% confidence interval estimate of the population mean is,
x - E < μ < x + E
19 - 3.177 < μ < 19 + 3.177
15.823 < μ < 22.177
The 98% confidence interval estimate of the population mean is
15.823 < μ < 22.177
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