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]
