Given Information:
Number of lithium batteries = n = 16
Mean life of lithium batteries = μ = 645 hours
Standard deviation of lithium batteries = σ = 31 hours
Confidence level = 95%
Required Information:
Confidence Interval = ?
Answer:
![CI = 628.5 \: to \: 661.5 \: hours](https://tex.z-dn.net/?f=CI%20%3D%20628.5%20%5C%3A%20to%20%5C%3A%20661.5%20%5C%3A%20hours)
Step-by-step explanation:
The confidence interval is given by
![CI = \mu \pm t_{\alpha/2} (\frac{\sigma}{\sqrt{n}})](https://tex.z-dn.net/?f=CI%20%3D%20%5Cmu%20%5Cpm%20t_%7B%5Calpha%2F2%7D%20%28%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%29)
Where μ is the mean life of lithium batteries, σ is the standard deviation, n is number of lithium batteries selected, and t is the critical value from the t-table with significance level of
tα/2 = (1 - 0.95) = 0.05/2 = 0.025
and the degree of freedom is
DoF = n - 1 = 16 - 1 = 15
The critical value (tα/2) at 15 DoF is equal to 2.131 (from the t-table)
![CI = 645 \pm 2.131(\frac{31}{\sqrt{16}})](https://tex.z-dn.net/?f=CI%20%3D%20645%20%5Cpm%202.131%28%5Cfrac%7B31%7D%7B%5Csqrt%7B16%7D%7D%29)
![CI = 645 \pm 2.131(7.75})](https://tex.z-dn.net/?f=CI%20%3D%20645%20%5Cpm%202.131%287.75%7D%29)
![CI = 645 \pm 16.515](https://tex.z-dn.net/?f=CI%20%3D%20645%20%5Cpm%2016.515)
![CI = 645 - 16.515 \: and \: 645 + 16.515](https://tex.z-dn.net/?f=CI%20%3D%20645%20-%2016.515%20%5C%3A%20and%20%5C%3A%20645%20%2B%2016.515)
![CI = 628.5 \: to \: 661.5 \: hours](https://tex.z-dn.net/?f=CI%20%3D%20628.5%20%5C%3A%20to%20%5C%3A%20661.5%20%5C%3A%20hours)
Therefore, the 95% confidence interval is 628.5 to 661.5 hours
What does it mean?
It means that we are 95% confident that the mean life of 16 lithium batteries is within the interval of (628.5 to 661.5 hours)
5/6 is 1/6 more than 4/6. All you had to do is subtract 4/6 from 5/6 to get <u><em>1/6</em></u>.
Let's solve your equation step-by-step.
-2/3p+ 1/6= 7/10
Answer:
p= -4/5
Answer:
5cm in the diagram
Step-by-step explanation:
30/6=5