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: $60,540< \mu<70,060$

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\overline{x}-z^*\dfrac{\sigma}{\sqrt{n}}< \mu<\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}}$

, 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 :-

$65,300-(1.96)\dfrac{17000}{\sqrt{49}}< \mu<65,300+(1.96)\dfrac{17000}{\sqrt{49}}$

$=65,300-(1.96)\dfrac{17000}{7}< \mu<65,300+(1.96)\dfrac{17000}{7}$

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

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

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

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