A simple random sample of 40 salaries of NCAA football coaches has a mean of $415,953. Assume that σ = $463,364. Construct a 95% confidence interval estimate of the mean salary of an NCAA football coach. $346,352 < μ < $485,554

$295,433 < μ < $536,473

$272,355 < μ < $559,551

$415,952 < μ < $415,954

ci = [271,953 , 559,953]

Step-by-step explanation:

x = mean: $415953

z = 95% : z value of 95% is 1.960

s = standard deviation = $463364

n = number of observations: 40

X +- Z * S/sqrt N

415953 + - 1.96 * 463364/sqrt40

= 415953 + - 144000

[271,953 , 559,953]

or

step-by-step explanation:

we need to solve for 'x' the following equation:  

we have that:

⇒

⇒ ⇒

factorizing, we get that:

  ⇒

then, the solutions are: or

nothing extraneous has been found!