Question

A new online test preparation company compared 3,025 students who had not used its program with 2,150 students who had. Of those students who did not use the online test preparation program, 1,513 increased their scores on the SAT examination compared with 1,100 who did use the program. A significance test was conducted to determine whether there is evidence that the online test preparation company's students were more likely to increase their scores on the SAT exam. What is the p-value for an appropriate hypothesis test?

Answer 1

= 1513/3025 = 0.5002

= 1100/2150 = 0.5116

The pooled sample proportion (P) = ( * n1 + * n2)/(n1 + n2)

                                                       = (0.5002 * 3025 + 0.5116 * 2150)/(3025 + 2150)

                                                       = 0.5049

SE = sqrt(P(1 - P)(1/n1 + 1/n2))

      = sqrt(0.5049 * (1 - 0.5049) * (1/3025 + 1/2150))

      = 0.0141

The test statistic z = ()/SE

                             = (0.5002 - 0.5116)/0.0141

                             = -0.81

P-value = P(Z < -0.81)

             = 0.2090