the wind on any random day in bryan is normally distributed with a standard deviation of 5.1 mph. a sample of 16 random days in

Question

2 years ago

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bryan had an average of 19mph. find a 98% confidence interval to capture the true average wind speed in bryan.

1 answer:

Karolina [17] · Answer 1 · 2024-02-14T20:17:45+03:00

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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