document.write( "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).
\n" );
document.write( "Which of the following statements is/are correct? CHECK ALL THAT APPLY. \r
\n" );
document.write( "\n" );
document.write( " A. The standard error of the difference between the two observed sample means is 2.6.
\n" );
document.write( " B. A two-sided two-sample t-test testing for a difference between the two population means is rejected at the 10% significance level.
\n" );
document.write( " C. A two-sided paired t-test testing for a difference between the two population means is rejected at the 10% significance level.
\n" );
document.write( " 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.
\n" );
document.write( " E. A two-sided paired t-test testing for a difference between the two population means is not rejected at the 10% significance level.
\n" );
document.write( " F. None of the above. \r
\n" );
document.write( "\n" );
document.write( "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. \r
\n" );
document.write( "\n" );
document.write( "So I rule out C and E. \r
\n" );
document.write( "\n" );
document.write( "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? \n" );
document.write( "
Algebra.Com's Answer #561661 by ewatrrr(24785)![]() ![]() You can put this solution on YOUR website! rule out C and E and A for sure(we can determine ME not SE) \n" ); document.write( "90% confidence interval is found to be : (-0.2, 5.4). \n" ); document.write( "ME = (5.4-(-.2))/2 = 2.8 \n" ); document.write( " \n" ); document.write( "90%CI: is 2.6 ± 2.8 \n" ); document.write( ".......... \n" ); document.write( "90% confidence interval is found to be: 2.6 ± 2.8 \n" ); document.write( "... \n" ); document.write( "unknown population standard deviations... \n" ); document.write( "the critical value used to determine CI would have been as a t-score rather than a z score \n" ); document.write( "two-sided: t-value with a cumulative probability equal to 0.95 \n" ); document.write( "...... \n" ); document.write( "90% confidence interval is found to be: 2.6 ± 2.8 (rules out B)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |