Answer: [tex]1.0848<\mu<1.3152[/tex]
Step-by-step explanation:
Confidence interval for population mean is given by :-
[tex]\overline{x}-z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}< mu< \overline{x}+ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]z_{\alpha/2}[/tex] = two -tailed z-value for [tex]{\alpha[/tex] (significance level)
n= sample size .
[tex]\sigma[/tex] = Population standard deviation.
[tex]\overline{x}[/tex] = Sample mean
By considering the given information , we have
[tex]\sigma=0.2\text{ mg}[/tex]
[tex]\overline{x}=1.2\text{ mg}[/tex]
n= 20
[tex]\alpha=1-0.99=0.01[/tex]
Using z-value table ,
Two-tailed Critical z-value : [tex]z_{\alpha/2}=z_{0.005}=2.576[/tex]
The 99 percent two-sided confidence interval for the mean nicotine content of a cigarette will be :-
[tex]1.2- (2.576)\dfrac{0.2}{\sqrt{20}}<\mu<1.2+ (2.576)\dfrac{0.2}{\sqrt{20}}\\\\=1.2- 0.1152<\mu<1.2+ 0.1152\\\\=1.0848<\mu<1.3152 [/tex]
Hence, the 99 percent two-sided confidence interval for the mean nicotine content of a cigarette: [tex]1.0848<\mu<1.3152[/tex]
