Hours spent studying in a week:
Sample size: 66
Sample mean: 4.06
Sample standard deviation: 3.67
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?
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'?