Due to a lack of general student discussion on the message board my class uses, I've decided to ask this here. I want to know if I proceeded with this question correctly and if my choices were correct. There's also another question I'm unsure about that uses the same general information.
Q: A sample of 20 cigarettes is tested to determine nicotine content and the average value observed was 1.2 mg. Compute a 99% two-sided confidence interval for the mean nicotine content of a cigarette if it is known that the standard deviation of a cigarette's nicotine content is .2 mg.
The degrees of freedom in this question is 19. Couple this with a 99% CI and the value I get from my table is 2.539. I'm using the t-distribution here because the distribution itself is small and because I'm not given the individual values (I think that's the correct reasoning?).
Using the mean, standard deviation, and the result above, I get:
1.2 (+ or -) 2.539 (.2 / sqrt(20) )
Would this be correct?
My second question is: Suppose, from the previous problem, that the sample population variance is not known in advance of the experiment. If the sample variance is .04, compute a 99% two-sided confidence interval for the mean nicotine content.
In this case would I just take the square root of the given variance to obtain the needed standard deviation?
If I'm not carrying out my steps correctly, could I get a quick explanation, please? Thank you.