Question

California is 0.5 minutes. To investigate the validity of this claim a random sample of 6 earthquakes were taken and the sample mean and the sample standard deviation were 1.15 and 0.308 and minutes, respectively. Construct a 98% confidence interval and determine if the researcher`s claim can be rejected. g

Answer 1

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

$(x^{-} - t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } , x^{-} + t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } )$

Degrees of freedom ν =n-1 =6-1 =5

Critical value : -

                   $t_{\frac{0.02}{2} } = t_{0.01} = 3.365$

98% confidence interval for the mean is determined by

$(1.15 - 3.365\frac{0.308}{\sqrt{6} } , 1.15+ 3.365\frac{0.308}{\sqrt{6} } )$

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)