The current rate of success is 70% for 200 students. When the new test is implemented, the rate increases to 85% for the 50 students who pass the test and are allowed into the program. Is this difference significant?

Answer 1

I will get you started.

You want see if .85 is different from .70

The numerator of your fraction will be:

.85 - .70 Always subtract the population proportion from the sample proportion.

The .15 has to be divided by the standard deviation. Do you have that formula?

The value you obtain after the division is a z-score.

Depending on how your instructor teaches this concept, obtaining a large z-score might be sufficient to say that there is a difference. Some instructors ask you to find the p-value from a table or your calculator and then compare that p-value with a given alpha of perhaps .05 (which is common). If you p-value is smaller than .05 then you can say there is a difference.

Answer 2

X1 and X2 are the numbers of successes in samples 1 and 2, respectively.
p = (140 + 43) / (200+50) = 183 /250 =0.73
z = (0.7-0.85) /√ (0.73)(1-0.73)(1/200 +1/50) = - 0.15 / √ (0.005) = -0.15 / 0.07 = -2.14
To have a proportional difference to be considered significant (at a significant level of 0.05 and df (248), the z value must exceed -3.30. Because −2.14 does not exceed this value, it can be concluded that there is a significant improvement in the training program's success after the new test.