Answer:
The 95% confidence interval of the mean is
(17.8658, 20.5342)
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the mean of the Population = 15.9
Given that the size of the sample 'n' =12
Mean of the sample x⁻ = 19.2
Given that the variance of the sample = 4.41
The standard deviation of the sample (S) = 2.1
<u><em>Step(ii):-</em></u>
<em>Degrees of freedom = n-1</em>
<em>ν = n-1 = 12-1=11</em>
<em>t₀.₀₅ = 2.2010</em>
The 95% confidence interval is determined by
![(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } , x^{-} + t_{0.05} \frac{S}{\sqrt{n} } )](https://tex.z-dn.net/?f=%28x%5E%7B-%7D%20-%20t_%7B0.05%7D%20%5Cfrac%7BS%7D%7B%5Csqrt%7Bn%7D%20%7D%20%2C%20x%5E%7B-%7D%20%2B%20t_%7B0.05%7D%20%5Cfrac%7BS%7D%7B%5Csqrt%7Bn%7D%20%7D%20%29)
![(19.2 - 2.2010 \frac{2.1}{\sqrt{12} } , 19.2 + 2.2010 \frac{2.1}{\sqrt{12} } )](https://tex.z-dn.net/?f=%2819.2%20-%202.2010%20%5Cfrac%7B2.1%7D%7B%5Csqrt%7B12%7D%20%7D%20%2C%2019.2%20%2B%202.2010%20%5Cfrac%7B2.1%7D%7B%5Csqrt%7B12%7D%20%7D%20%29)
(19.2 - 1.3342 , 19.2+1.3342)
(17.8658 , 20.5342)
<u><em>Final answer:-</em></u>
The 95% confidence interval of mean is
(17.8658 , 20.5342)