SOLUTION: A 90% confidence interval for the difference between the means of two independent populations with unknown population standard deviations is found to be (-0.2, 5.4). Which of the

Algebra ->  Probability-and-statistics -> SOLUTION: A 90% confidence interval for the difference between the means of two independent populations with unknown population standard deviations is found to be (-0.2, 5.4). Which of the       Log On


   



Question 925594: A 90% confidence interval for the difference between the means of two independent populations with unknown population standard deviations is found to be (-0.2, 5.4).
Which of the following statements is/are correct? CHECK ALL THAT APPLY.
A. The standard error of the difference between the two observed sample means is 2.6.
B. A two-sided two-sample t-test testing for a difference between the two population means is rejected at the 10% significance level.
C. A two-sided paired t-test testing for a difference between the two population means is rejected at the 10% significance level.
D. A two-sided two-sample t-test testing for a difference between the two population means is not rejected at the 10% significance level.
E. A two-sided paired t-test testing for a difference between the two population means is not rejected at the 10% significance level.
F. None of the above.
So I am slightly confused with this problem. I know that to conduct a t-test, population means and s.d. are unknown (which they are in this problem?). And since they hinted it at being independent, it cannot be a paired test.
So I rule out C and E.
But when I look at A), I wonder, can I even find SE (standard error?). I am not quite sure how...as well as how I can compute the p-value to test whether the results are statistically significant?

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
rule out C and E and A for sure(we can determine ME not SE)
90% confidence interval is found to be : (-0.2, 5.4).
ME = (5.4-(-.2))/2 = 2.8
%285.4+-+2.8%29+=+2.6
90%CI: is 2.6 ± 2.8
..........
90% confidence interval is found to be: 2.6 ± 2.8
...
unknown population standard deviations...
the critical value used to determine CI would have been as a t-score rather than a z score
two-sided: t-value with a cumulative probability equal to 0.95
......
90% confidence interval is found to be: 2.6 ± 2.8 (rules out B)