Respuesta :
Answer:
98% confidence interval for the mean is
(0.72689 , 1.57311)
Step-by-step explanation:
Given random sample size 'n' = 6
Given sample mean x⁻ = 1.15
Given sample standard deviation 'S' = 0.308
98% confidence interval for the mean is determined by
[tex](x^{-} - t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } , x^{-} + t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } )[/tex]
Degrees of freedom ν =n-1 =6-1 =5
Critical value : -
[tex]t_{\frac{0.02}{2} } = t_{0.01} = 3.365[/tex]
98% confidence interval for the mean is determined by
[tex](1.15 - 3.365\frac{0.308}{\sqrt{6} } , 1.15+ 3.365\frac{0.308}{\sqrt{6} } )[/tex]
On calculation , we get
( 1.15 - 0.42311 , 1.15+ 0.42311)
(0.72689 , 1.57311)
Final answer:-
98% confidence interval for the mean is
(0.72689 , 1.57311)
