Question

The difference between the sample means of two populations is 34.6, and the standard deviation of the difference between sample means is 11.9. What is the 95% confidence interval

Answer 1

Answer:

The required interval is $10.8\leq CI \leq 58.4$  

Step-by-step explanation:

Given : The difference between the sample means of two populations is 34.6, and the standard deviation of the difference between sample means is 11.9.

To find : What is the 95% confidence interval lies between ?

Solution :

Let $\mu_1,\mu_2$ were two sample means.

The difference between the sample means of two populations is 34.6

i.e.  $\mu_1-\mu_2=34.6=\mu$

Let $\sigma_1,\sigma_2$ were two standard deviation.

The difference between the standard deviation of two populations is 11.9

i.e.  $\sigma_1-\sigma_2=11.9=\sigma$

The confidence interval at 95% is 2 dimension,

So, The formula to find class interval is

$\mu-2\sigma\leq CI \leq \mu+2\sigma$

$34.6-2(11.9)\leq CI \leq 34.6+2(11.9)$

$34.6-23.8\leq CI \leq 34.6+23.8$      

$10.8\leq CI \leq 58.4$      

Therefore, The required interval is $10.8\leq CI \leq 58.4$