Question

Given a sample mean difference of 23, a standard error of 4.2, and a critical t value of 2.262, what are the values for the confidence interval?

Answer 1

Answer: (13.4996, 32.5004)

Step-by-step explanation:

The confidence interval for mean difference is given by :-

$(\overline{x}_1-\overline{x}_2)\pm t^*\times SE$ ,

where $(\overline{x}_1-\overline{x}_2)$ = sample mean difference .

$t^*$= critical t value (two-tailed)

SE= Standard error.

Given : $(\overline{x}_1-\overline{x}_2)=23$

$t^*=2.262$

SE= 4.2

Now, the confidence interval for mean difference will be :-

$23\pm (2.262)\times (4.2)\\\\= 23\pm9.5004\\\\=(23-9.5004,\ 23+9.5004)\\\\=(13.4996,\ 32.5004)$

Hence, the required confidence interval : (13.4996, 32.5004)