Solution :
95% confidence interval for the difference between the means of ACT scores between two schools, is given by :
[tex]$[(\overline X_1 - \overline X_2)-ME, (\overline X_1 - \overline X_2)+ME]$[/tex]
[tex]$\overline X_1 = 25.90 , \ \ \ \overline X_2 = 27.70$[/tex]
[tex]$S_1=3.49, \ \ \ S_2 = 2.47$[/tex]
[tex]$n_1=39, \ \ n_2=37$[/tex]
M.E. , Margin of error,
[tex]$=t_{n_1+n_2-2,0.025} \times \left( s \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$[/tex]
[tex]$s^2 = \frac{(n_1-1)S_1^2+(n_2-1)S^2_2}{n_1+n_2-2}$[/tex]
[tex]$=\frac{38(3.49)^2+36(2.47)^2}{39+37-2}$[/tex]
= 9.222
s = 3.03
[tex]$t_{74,0.05} = 1.99$[/tex]
[tex]$M.E. = 1.99 \times 3.03 \times \sqrt{\frac{1}{39}+\frac{1}{37}}$[/tex]
= 1.4
Therefore, 95% CI = [(25.90-27.70) - 1.4 , (25.90-27.70) + 1.4]
= [-3.2, -0.4]
Therefore, the lower bound is -3.2
