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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

Answers:

1.

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?

2.

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.

EDIT:

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'?

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    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.2012-02-27

1 Answers 1

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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.