Confidence intervals, given mean, deviation, value of 'n'

I am having trouble with confidence intervals for mean. For example:

1) A company claims that their pens will write for over 100 hours. A sample of 50 pens gives a sample mean of 103, with a given stdev of 6. Find a 99% confidence interval for the mean number of hours that pens will last.

I know that 99% is 2.575 for Z(alpha/2), but I don't know about the rest. Another example:

2) Given: n = 50, sample mean: 9.3, stdev: 2.2. Find an 80% confidence interval for this information

Thanks for helping

I have a wonderful Excel program for just about all the stats needs.

I ran your data through it and it told me:

\(\displaystyle \L\\100.81\leq{\mu}\leq{105.19}\)


Running the hypothesis test at alpha = 0.05, we are told to reject H0.

That means there is enough evidence at the 0.005 level to support the claim that the mean writing time is over 100 hours.

So the confidence interval is between those two values? I don't have that program so I don't know how to do it myself.

Also, what about the other example?

Using your values...

Given Standard Deviation = 6
Sample Standard Deviation of the Mean, then = \(\displaystyle \frac{6}{\sqrt{50}}\;=\;0.849\)
Confidence Interval Minimum = 103 - 2.575*0.849 = 100.814
Confidence Interval Minimum = 103 + 2.575*0.849 = 105.186

These appear to be consistent with galactus' views.