Answer:
99% Confidence interval: (0.77,0.95)
Step-by-step explanation:
We are given the following in the question:
Sample mean, [tex]\bar{x}[/tex] = 0.86 sec
Sample size, n = 102
Alpha, α = 0.01
Sample standard deviation, s = 0.35 sec
Degree of freedom =
[tex]= n -1\\=102-1\\=101[/tex]
99% Confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 101 and}~\alpha_{0.01} = \pm 2.6253[/tex]
[tex]0.86 \pm 2.6253(\dfrac{0.35}{\sqrt{102}} )\\\\ = 0.86 \pm 0.0909\\\\ = (0.7691 ,0.9509)\approx (0.77,0.95)[/tex]
(0.77,0.95) is the required 99% confidence interval for the true average echo duration μ.
