A driver's education course compared 1,500 students who had not taken the course with 1,850 students who had. Of those students who did not take the driver's education course, 1,150 passed the written driver's exam the first time compared with 1,440 of the students who did take the course. A significance test was conducted to determine whether there is evidence that the students who took the driver's education course were more likely to pass the written driver's exam the first time. What is the p-value for an appropriate hypothesis test?

Null Hypothesis, [tex]H_0[/tex] : [tex]p_1 = p_2[/tex] {means students who took the driver's education course and those who didn't took have same chances to pass the written driver's exam the first time}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]p_1 > p_2[/tex] {means students who took the driver's education course were more likely to pass the written driver's exam the first time}

The test statistics used here will be two sample Binomial statistics i.e.;

[tex]\frac{(\hat p_1 - \hat p_2)-(p_1 - p_2)}{\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} } }[/tex] ~ N(0,1)

Here, [tex]\hat p_1[/tex] = No. of students passed the exam ÷ No.of Students that had taken the course

[tex]\hat p_1[/tex] = [tex]\frac{1150}{1850}[/tex] Similarly, [tex]\hat p_2[/tex] = [tex]\frac{1440}{1500}[/tex] [tex]n_1[/tex] = 1,850 [tex]n_2[/tex] = 1,500

Test Statistics = [tex]\frac{(\frac{1150}{1850} -\frac{1440}{1500})-0}{\sqrt{\frac{\frac{1150}{1850}(1- \frac{1150}{1850})}{1850} + \frac{\frac{1440}{1500}(1- \frac{1440}{1500})}{1500} } }[/tex] = -27.38

P-value is given by, P(Z > -27.38) = 1 - P(Z > 27.38)

Now, in z table the highest critical value given is 4.4172 which corresponds to the probability value of 0.0005%. Since our test statistics is way higher than this so we can only say that the p-value for an appropriate hypothesis test is less than 0.999995 .