Salaries of 43 college graduates who took a statistics course in college have a mean,66,000 , of . Assuming a standard deviation, 18908 , of $, construct a %99 confidence interval for estimating the population mean .

$[58543.42; 73456.58]

Step-by-step explanation:

Hello!

For the variable

X: salary of a college graduate that took a statistics course

Out of n= 43 students, the calculated mean is = $66000

The population standard deviation is δ= $18908

There is no information about the variable distribution, but since the sample size is big enough (n≥30), you can apply the CLT and approximate the distribution of the sample mean to normal ≈N(μ;σ²/n)

Then you can apply the approximation of the standard normal distribution to calculate the 99% CI

± *

[66000±2.586*2883.44]

$[58543.42; 73456.58]

With a 99% confidence level you'd expect that the interval $[58543.42; 73456.58] will include the average salary of college graduates that took a course of statistics.

I hope this helps!

1/4 and 8/3 because:

the reciprocal if 4 is 1/4 because 4 is used as 4/1 and the opposite of that is 1/4. the opposite of 3/8 is 8/3 so that is the answer.

oof cheaters cheat because they’re lazy that’s why i cheat xd

: )

step-by-step explanation: