Answer:
[tex] SE= \sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}[/tex]
And replacing we got:
[tex] SE= \sqrt{\frac{12^2_1}{105}+\frac{16^2_2}{70}}= 2.24[/tex]
Step-by-step explanation:
We have the following info given:
[tex] n_1= 105, \mu_1 = 88, \sigma_1 =12[/tex]
[tex] n_2= 70, \mu_2 = 78, \sigma_2 =16[/tex]
And for this case we want to find the standard error for the following variable [tex]\bar X_1 -\bar X_2[/tex] and the standard error for this distribution is given by:
[tex] SE= \sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}[/tex]
And replacing we got:
[tex] SE= \sqrt{\frac{12^2_1}{105}+\frac{16^2_2}{70}}= 2.24[/tex]
