Answer:
99% Confidence interval: (1196,1973)
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 20
Mean, μ = 1584
Standard Deviation, σ = 607
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 19 and}~\alpha_{0.01} = \pm 2.86[/tex]
[tex]1584 \pm 2.86(\dfrac{607}{\sqrt{20}} )\\\\ = 1584 \pm 388.18\\ = (1195.82 ,1972.18)\\ \approx (1196,1973)[/tex]
Thus, there is a 99% chance that it will break in approximately 1196 to 1972