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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.

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  • $\begingroup$ In the first problem, from the wording it sounds as though the (population) standard deviation is known, On the assumption that the distribution is normal, one can use the normal distribution instead of the $t$-distribution, giving a narrower confidence interval. But the normality assumption is not necessarily reasonable, By the way, I doubt the number used, since for the normal one has $2.57$. $\endgroup$ – André Nicolas May 26 '15 at 4:31
  • $\begingroup$ So, in this case it would be "better" to use the t-distribution? $\endgroup$ – Super Rhinocerus May 26 '15 at 4:34
  • $\begingroup$ You have a two-sided test. Thus you have to look at p=0.995 for the corresponding z-value (or x-value). $\endgroup$ – callculus May 26 '15 at 4:37
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    $\begingroup$ It would be better to use the normal. And for the $t$, the number $2.539$ is not right. $\endgroup$ – André Nicolas May 26 '15 at 4:39
  • $\begingroup$ Yes, I looked at the table again - 2.861 should've been my choice? $\endgroup$ – Super Rhinocerus May 26 '15 at 4:41
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You should use the normal distribution, so use the z-score instead of the t-score for your critical value. So 2.575 instead in that first problem. The rest looks great. This is because you are assuming you know $\sigma$, the population standard deviation.

In the second problem, you are supposing you did not know the population standard deviation. Thus since you have a relatively small sample size, you need to use the t-distribution like you did in your first solution you provided. You are spot on about taking the square root of the sample variance to get the sample standard deviation.

Summary: z-scores are when you know $\sigma$ and t-scores are when you don't know $\sigma$.

Edit: after looking at a t-table chart, I think your number for the 99% confidence level on the t-table with 19 as you degrees of freedom is off. I got 2.861. I think you accidentally used the 98% confidence level.

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    $\begingroup$ Ah, I see, thank you (and to the others, too). EDIT: Yes, Andre Nicolas let me know about my error with the t-distribution. $\endgroup$ – Super Rhinocerus May 26 '15 at 4:39
  • $\begingroup$ Thank you for supplying proper details and work showing what you tried and were thinking. It makes it a lot easier to provide helpful responses. $\endgroup$ – CPM May 26 '15 at 4:43
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For the second part, in addition to using the $t$ distribution as you have, since you do not know the population variance, you need to use an unbiased estimate of population variance based on your sample, i.e. $\frac {n}{n-1}\times \text{sample variance}$

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