Answer
Confidence interval: $4.71\lt µ_1-µ_2\lt6.29$
We are 95% confident that the difference in scores between group 1 and group 2 is between 4.71 and 6.29.
Work Step by Step
$x ̅_1,n_1~and~s_1$ refer to group 1 and $x ̅_2,n_2~and~s_2$ refer to group 2.
$n=200$, so:
$d.f.=n-1=199$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=1.984$
(According to Table VI, for d.f. = 100, the closest value to 199, and area in right tail = 0.025)
$Lower~bound=(x ̅_1-x ̅_2)-t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(23.4-17.9)-1.984\sqrt {\frac{4.1^2}{200}+\frac{3.9^2}{200}}=4.71$
$Upper~bound=(x ̅_1-x ̅_2)+t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(23.4-17.9)+1.984\sqrt {\frac{4.1^2}{200}+\frac{3.9^2}{200}}=6.29$
