Answer: [tex](2.065,\ 2.735)[/tex]
Step-by-step explanation:
The confidence interval for population mean is given by :-
[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
Given : Significance level : [tex]\alpha: 1-0.9=0.1[/tex]
Critical value : [tex]z_{\alpha./2}=1.645[/tex]
Sample size : [tex]n=55[/tex]
Sample mean : [tex]\overline{x}=2.4[/tex]
Standard deviation : [tex]\sigma = 1.51[/tex]
The 90% confidence interval for the mean number of kills per week will be :-
[tex]2.4\pm (1.645)\dfrac{1.51}{\sqrt{55}}\\\\\approx2.4\pm0.335\\\\=(2.4-0.335,\ 2.4+0.335)=(2.065,\ 2.735)[/tex]
Hence, the 90% confidence interval for the mean number of kills per week by US household cats that hunt outdoors = [tex](2.065,\ 2.735)[/tex]