A random sample of analyst earnings estimaes has a mean of $2.84 and a standard deviation of $0.40. What probabilistic statement could we make at the 90% confidence level if the sample size were 60? For the above question, what would be the appropriate level of significance? Should it be one-tailed or two-tailed? Thx.

Confidence interval is always 2 tailed. sample size is 60, so by central limit theorem you can assume that the underlying distribution would be approximately normal, and use the normal z-scores

So give me the probabilistic statement… The “mean of the distribution of all possible analyst’s earnings estimate is…” Huh? Silly question.

I think the question would be interpreted as: What would be the 90% confidence interval for the mean of population of analyst estimates? Yeah, indeeds sounds wierd… But the purpose should be just a practice to calculate the confidence intervals. Since sample size is greater than 30 and the population variance is unknow. And We can not assume the analyst’s sampling distribution is normal. So it’s more conservative to use t-distribution. df = n-1 = 59; if two-tailed, alpha/2 = 10%/2 = 0.05. --> t(59, 0.05) approximated as t(60, 0.05) = 2 What bugs me is the answer key seems using alpha = 0.10 as level of significance. If so, then it’s for one-tailed test. --> t(59, 0.1) is approximated as t(60, 0.1) = 1.671. Why?

Hi, You have to use t-distribution( note that the level of significance of test T corresponds to the two-tailed probabilities). df=n-1=59 u have 90% confidence => alpha= 10%=> alpha/2=0.05. t(59,0.05)=1.671 Confidence intervals: 2.84+/- 1.671*0.84 =[1.43636,7.40364] Am I right?

Nope. Gotta use that Sqrt(n) thing…

Using the t-distribution, (two-tailed) std error = .40/square root of 60 => .40/7.74597 => .05164 df = 60-1 = 59, Alpha = 10/2 =.05, the nearest value in t-table would with 60 df and 0.05 which is 1.671 C.I = 2.84 +/- 1.671*.05164 = { 2.92629, 2.75371} Is this right???

swathiraj Wrote: ------------------------------------------------------- > Using the t-distribution, (two-tailed) > > std error = .40/square root of 60 => .40/7.74597 > => .05164 > > df = 60-1 = 59, Alpha = 10/2 =.05, the nearest > value in t-table would with 60 df and 0.05 which > is 1.671 > > C.I = 2.84 +/- 1.671*.05164 = { 2.92629, 2.75371} > > Is this right??? Yes, this is correct answer. The statistical statement is: We are 90% confident the true mean of the population of the analyst estimates is within the range of $2.75 - $2.93. Now I know where my confusion was. We treated the problem as two-tailed test and got the alpha as 0.05. When you check the t-table, which row do you look at? df with alpha for one-tailed test or df with alpha for two-tailed test? 60 df and 0.05 (under one-tail) is 1.671 and 60 df and 0.05 (under two-tail) is 2 .000 Any advice? Thx.

60 and 0.05 gives you the answer… C.I = 2.84 +/- 1.671*.05164 = { 2.92629, 2.75371}

hyang, you’re looking at the wrong column. For t table, if it’s 10%, just look at the “0.10” column. Don’t divide it by 2 as you do with the z table.

quechuong Wrote: ------------------------------------------------------- > hyang, > > you’re looking at the wrong column. > > For t table, if it’s 10%, just look at the “0.10” > column. Don’t divide it by 2 as you do with the z > table. Thanks, you answered my question. It that’s the case, how come both the t and z formulas for confidence intervals show alpha/2?

Would this be a general statement? For one-tailed test (no matter t, z, chi), level of significance is alpha. For two-tailed test (no matter t, z, chi), level of significance is alpha/2.