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: [tex]60,540< \mu<70,060[/tex]Step-by-step explanation:The confidence interval for population mean is given by :-[tex]\overline{x}-z^*\dfrac{\sigma}{\sqrt{n}}< \mu<\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}}[/tex], where [tex]\sigma[/tex] = Population standard deviation.n= sample size[tex]\overline{x}[/tex] = Sample mean z* = Critical z-value .Given :  [tex]\sigma=\$17,000[/tex] n= 49[tex]\overline{x}= \$65,300[/tex] Two-tailed critical value for 95% confidence interval = [tex]z^*=1.960[/tex]Then, the 95% confidence interval would be :-[tex]65,300-(1.96)\dfrac{17000}{\sqrt{49}}< \mu<65,300+(1.96)\dfrac{17000}{\sqrt{49}}[/tex][tex]=65,300-(1.96)\dfrac{17000}{7}< \mu<65,300+(1.96)\dfrac{17000}{7}[/tex][tex]=65,300-4760< \mu<65,300+4760[/tex][tex]=60,540< \mu<70,060[/tex]Hence, the 95​% confidence interval for estimating the population mean [tex](\mu)[/tex] :[tex]60,540< \mu<70,060[/tex]