Question

A study of bulimia among college women studied the connection between childhood sexual abuse and a measure of family cohesion​ (the higher the​ score, the greater the​ cohesion). The sample mean on the family cohesion scale was 1.9 for 13 sexually abused students ​(sequals2.1​) and 5.2 for 17 nonabused students ​(sequals3.5​). a. Find the standard error for comparing the means. b. Construct a​ 95% confidence interval for the difference between the mean family cohesion for sexually abused students and​ non-abused students. Interpret.

Answer 1

Answer: a) 1.029, b) (-5.318, -1.282).

Explanation:

Since we have given that

`n_1=13\\\\n_2=17\\\\\bar{x_1}=1.9\\\\\bar{x_2}=5.2`

and

`s_1=2.1\\\\s_2=3.5`

So, the standard error for comparing the means :

`SE=\sqrt{(s^2_1)/(n_1)+(s^2_2)/(n_2)}\\\\SE=\sqrt{(2.1^2)/(13)+(3.5^2)/(17)}\\\\SE=√(1.0598)\\\\SE=1.029`

At 95% confidence interval, z = 1.96

So, Confidence interval would be

`\bar{x_1}-\bar{x_2}\pm z* SE\\\\=(1.9-5.2)\pm 1.96* 1.0294\\\\=-3.3\pm 2.017624\\\\=(-3.3-2.018,-3.3+2.018)\\\\=(-5.318,-1.282)`

Hence, a) 1.029, b) (-5.318, -1.282).