Answer:
(6.668, [infinity])
So then we have this condition:
[tex] 6.668 = 6.861 -t_{\alpha/2} \frac{0.440}{\sqrt{16}}[/tex]
And if we solve for the critical value we got:
[tex] t_{\alpha/2}= \frac{\sqrt{16} *(6.861-6.668)}{0.440} = 1.755[/tex]
And we can find the probability accumulated in the left of 1.755 with a distribution with degrees of freedom [tex] df = n-1= 16-1 = 15[/tex] and we got: 0.95, with the following excel code for example:
"=T.DIST(1.755,15,TRUE)"
So then the confidence level would be 95%
Step-by-step explanation:
For this case we have the following info:
[tex]\bar X = 6.861[/tex] represent the sample mean
[tex]s = 0.440[/tex] represent the sample deviation
n =16 represent the sample size
For this case the scientists calculate a lower bound confidence interval given by:
[tex] (\bar X -t_{\alpha/2} \frac{s}{\sqrt{n}} , \infty)[/tex]
And the interval given is:
(6.668, [infinity])
So then we have this condition:
[tex] 6.668 = 6.861 -t_{\alpha/2} \frac{0.440}{\sqrt{16}}[/tex]
And if we solve for the critical value we got:
[tex] t_{\alpha/2}= \frac{\sqrt{16} *(6.861-6.668)}{0.440} = 1.755[/tex]
And we can find the probability accumulated in the left of 1.755 with a distribution with degrees of freedom [tex] df = n-1= 16-1 = 15[/tex] and we got: 0.95, with the following excel code for example:
"=T.DIST(1.755,15,TRUE)"
So then the confidence level would be 95%