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

[kennedy0]

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

 

[galactus]

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.

 

[kennedy0]

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?

 

[tkhunny]

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.