Question

Salaries of 49 college graduates who took a statistics course in college have a​ mean, x overbar​, of $ 65 comma 300. Assuming a standard​ deviation, sigma​, of ​$17 comma 805​, construct a 95​% confidence interval for estimating the population mean mu.

Answer 1

Answer:
`60,540< \mu<70,060`

Explanation:

The confidence interval for population mean is given by :-

`\overline{x}-z^*(\sigma)/(√(n))< \mu<\overline{x}+z^*(\sigma)/(√(n))`

, where
`\sigma` = Population standard deviation.

n= sample size

`\overline{x}` = Sample mean

z* = Critical z-value .

Given :
`\sigma=\$17,000`

n= 49

`\overline{x}= \$65,300`

Two-tailed critical value for 95% confidence interval =
`z^*=1.960`

Then, the 95% confidence interval would be :-

`65,300-(1.96)(17000)/(√(49))< \mu<65,300+(1.96)(17000)/(√(49))`

`=65,300-(1.96)(17000)/(7)< \mu<65,300+(1.96)(17000)/(7)`

`=65,300-4760< \mu<65,300+4760`

`=60,540< \mu<70,060`

Hence, the 95​% confidence interval for estimating the population mean
`(\mu)` :

`60,540< \mu<70,060`