Q:

# 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.

Accepted Solution

A:
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 $$(\mu)$$ :$$60,540< \mu<70,060$$