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

Solution :

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,

$=t_{n_1+n_2-2,0.025} \times \left( s \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

$s^2 = \frac{(n_1-1)S_1^2+(n_2-1)S^2_2}{n_1+n_2-2}$

  $=\frac{38(3.49)^2+36(2.47)^2}{39+37-2}$

 = 9.222

s = 3.03

$t_{74,0.05} = 1.99$

$M.E. = 1.99 \times 3.03 \times \sqrt{\frac{1}{39}+\frac{1}{37}}$

        = 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