Question

If the lifespan of a computer with n=100 sample size, x=6.8 years, s=2.4 years, construct a 99% confidence interval for the population mean. The population is normally distributed

Answer 1

@jim_thompson5910

Answer 2

@dan815

Answer 3

how far did you get?

Answer 4

Let \(L\) denote the lifespan of a computer. You want to find \(a\) and \(b\) such that
\[P(a\le L\le b)=0.99\]
which is the same thing as
\[P(a>L)=0.005~~~\text{and}~~~P(L>b)=0.005\]
Transform the random variable to one with a standard normal distribution, \(Z\):
\[P\left(\frac{a-\overline{x}}{\dfrac{s}{\sqrt{n}}}>\frac{L-\overline{x}}{\dfrac{s}{\sqrt n}}\right)=P\left(\frac{a-6.8}{\dfrac{2.4}{\sqrt{100}}}>Z\right)=0.005\]
A \(z\) table will tell you that, in order to have a probability of 0.5%, you must have
\[\frac{a-6.8}{\dfrac{2.4}{\sqrt{100}}}=-2.58\]
Solve for \(a\). The setup for \(b\) is similar. Alternatively, using the symmetry of the distribution, you can immediately get
\[\frac{a-6.8}{\dfrac{2.4}{\sqrt{100}}}=2.58\]

Table for reference: http://facstaff.unca.edu/dohse/Stat225/Images/Table-z.JPG

Answer 5

Sorry, that last equation should have \(b\), not \(a\).