I used Z value to calculate confidence interval. How do I know to use t distribution instead of z value? -------------------------------------------- What is the approximate 99 percent confidence interval for the population mean based on a sample of 60 returns with a mean of 7 percent and a sample standard deviation of 25 percent? A) 1.584% to 14.584%. B) -1.584% to 15.584%. C) 0.546% to 13.454%. D) 1.546% to 13.454%. Your answer: B was correct! The standard error for the mean = s/(n)0.5 = 25%/(60)0.5 = 3.227%. The critical value from the t-table should be based on 60 – 1 = 59 df. Since the standard tables do not provide the critical value for 59 df the closest available value is for 60 df. This leaves us with an approximate confidence interval. Based on 99 percent confidence and df = 60, the critical t-value is 2.660. Therefore the 99 percent confidence interval is approximately: 7% ± 2.660(3.227) or 7% ± 8.584% or -1.584% to 15.584%.

Always use a T-value with sample standard deviation, although for n large t -> z. Note Z value here would be 2.58. Not much different.

Use T-table … since sample size is high (n > 30) and population variance in unknown S.E = sample mean/ SQRT(sample size) S.E. = 25/SQRT(60) = 3.2274% t-statistic(59,0.01) = 2.6618 C.I = (point statistic) ± (rel factor)*(S.E.) C.I = 7 ± (2.6618 * 3.2274) = 7 ± 8.59069332 -1.59069332% to 15.59069332% Answer is B. - Dinesh S

Thanks for replies. Now I understood that whenever I see sample variance or sample standard deviation, I will use value from t table. On the other hand if I see population variance or population standard deviation, I will use Z value. I also did a mistake to not to squarte root the variance. I need to be careful next time.