Question

A study is done to find the 95% confidence interval for the difference in means between the ACT scores between two schools in a given city. One school had a random sample of 39 students had a sample mean of 25.90 with a sample standard deviation of 3.49. The second school had a random sample of 37 students had a sample mean 27.70 with a sample standard deviation of 2.47. What is the lower bound of the 95% confidence interval for the difference in means

Answer 1

95% confidence interval for the difference between the means of ACT scores between two schools, is given by :

`$[(\overline X_1 - \overline X_2)-ME, (\overline X_1 - \overline X_2)+ME]$`

`$\overline X_1 = 25.90 , \ \ \ \overline X_2 = 27.70$`

`$S_1=3.49, \ \ \ S_2 = 2.47$`

`$n_1=39, \ \ n_2=37$`

M.E. , Margin of error,

`$=t_(n_1+n_2-2,0.025) * \left( s \sqrt{(1)/(n_1)+(1)/(n_2)}$`

`$s^2 = ((n_1-1)S_1^2+(n_2-1)S^2_2)/(n_1+n_2-2)$`

`$=(38(3.49)^2+36(2.47)^2)/(39+37-2)$`

= 9.222

s = 3.03

`$t_(74,0.05) = 1.99$`

`$M.E. = 1.99 * 3.03 * \sqrt{(1)/(39)+(1)/(37)}$`

= 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