Hours spent studying in a week:

Sample size: 66

Sample mean: 4.06

Sample standard deviation: 3.67

  1. Construct a confidence interval of 99% for the true average number of hour spent studying in a week. What are the upper and lower bounds?

  2. What is the assumption you must make regarding this population distribution? A) Normal B) t distribution C) Any distribution



For a 99% interval I take the value of z from the z tables where the probability is 0.005 (as 2 * 0.005 = 1%), which is 2.756.

I then multiply this value by (std deviation)/sqrt(sample size) => so 2.756*3.67/sqrt(66).

And this gives me 1.1637. I add an subtract this to the sample mean to get the bounds for a 99% C.I.

And I get 2.8963 and 5.2237. Is that correct?


I am unsure of whether the assumption I must make is A) Normal distribution or C) Any distribution. I think it is A) Normal distribution as the sampling distribution of the sample mean approaches u for large sample sizes...however n=66 doesnt really seem like a 'large' sample size to me so Im not sure about this.


Actually Ive just noticed that the questions gives the SAMLPLE standard deviation...so does this mean I cant use the z test and should use the t test? Altho looking at my stats book it says I can simply substitute s for $\sigma$ if the sample size is large, but is 66 'large'?

  • $\begingroup$ When you used $z$ and $z$-table, you implicitly assumed a normal distribution. $\endgroup$ – Dilip Sarwate Feb 27 '12 at 21:59
  • 1
    $\begingroup$ I would say none of the above for 2... With a large sample, many, but not all, distributions become close enough to normal by the Central Limit Theorem. For those distributions where the Central Limit Theorem may apply, 66 can be considered large or not large, depending on how far away the distributions are from normal. Normal would certainly be good enough, t distribution would probably be fine too. Any distribution is definitely wrong. $\endgroup$ – GeoffDS Feb 27 '12 at 22:00
  • $\begingroup$ It is not clear to me what the problem is. The sample size is $66$. Is that $66$ days? Is it $66$ weeks? Hard to know, but I would worry about a student who studied a mean of $4.06$ hours a week. The standard deviation is quite large compared to the mean, so since one cannot study a negative number of hours, the distribution is far from normal. Averaging over a medium-sized number like $66$ helps a lot, but may not be good enough, particularly for tail probabilities. $\endgroup$ – André Nicolas Feb 27 '12 at 22:24

For many practical purposes, $66$ can be considered "large". The more substantial issue is that for a purpose other the the central limit theorem, $66$ isn't all that large. There is more uncertainty in the estimate of the standard deviation when the sample size is only 66 than when it is, e.g., $500$. For that reason, Student's t-distribution is used. In this case, using Student's distribution with $66-1=65$ degrees of freedom, instead of $2.5758$ you get $2.6536$, so you get a somewhat longer confidence interval.


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