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.

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windyyork windyyork · Answer 1 · 2020-04-22T10:57:40+02:00

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

Step-by-step explanation:

Since we have given that

[tex]n_1=13\\\\n_2=17\\\\\bar{x_1}=1.9\\\\\bar{x_2}=5.2[/tex]

and

[tex]s_1=2.1\\\\s_2=3.5[/tex]

So, the standard error for comparing the means :

[tex]SE=\sqrt{\dfrac{s^2_1}{n_1}+\dfrac{s^2_2}{n_2}}\\\\SE=\sqrt{\dfrac{2.1^2}{13}+\dfrac{3.5^2}{17}}\\\\SE=\sqrt{1.0598}\\\\SE=1.029[/tex]

At 95% confidence interval, z = 1.96

So, Confidence interval would be

[tex]\bar{x_1}-\bar{x_2}\pm z\times SE\\\\=(1.9-5.2)\pm 1.96\times 1.0294\\\\=-3.3\pm 2.017624\\\\=(-3.3-2.018,-3.3+2.018)\\\\=(-5.318,-1.282)[/tex]

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