Answer:
The 95% of confidence interval are
(15.8663 ,16.2337)
Explanation:
Step(i):-
Size of the Population = 16
Size of the sample (n) = 4
The mean of the sample = 16.05 oz
The standard deviation of the sample (S) = 0.1 oz
Degrees of freedom =n-1 =4-1 =3
t₀.₀₅,₃ = 3.1824
Step(ii):-
The 95% of confidence interval is determined by
![(x^(-) - Z_(0.05) (S)/(√(n) ) , x^(-) -+Z_(0.05) (S)/(√(n) ))](https://img.qammunity.org/2022/formulas/mathematics/college/34dn7b02lu6p4vp220ua9axq9xrk1ryeg5.png)
![(16.05 - 3.1824(0.1)/(√(3) ) , 16.05 -+3.1824 (0.1)/(√(3) ))](https://img.qammunity.org/2022/formulas/mathematics/college/fap6vj3cxan9xstgji2k8en1fd80mu75d5.png)
(16.05 - 0.1837 ,16.05 +0.1837 )
(15.8663 ,16.2337)
Final answer:-
The 95% of confidence interval are
(15.8663 ,16.2337)