The information comes from this problem statement:

“In 1975, it was estimated that 65% of the US consumers regularly used coupons when shopping. A research company investigates whether this has changed. In a sample of 1034 people, 796 reported regularly using discount coupons.”

I calculated the observation value to be .76983 (796/1034), and the standard error was given to me as .0148 . I calculated the z value as (.76983-0.65)/.0148. This gives me a ridiculously large Z-score of 8.10. Did I calculate this wrong? How do I find the p value if the z-score is so big it’s not even on the standard normal probability table? Please give me a detailed, step by step answer. Thanks.

## What Is The Z-score/ Two Tail P-value?

October 11th, 2012 admin

Ans:

Ho: p= 0.65

Ha: p not = 0.65

p-hat = 796/1034 = 0.7698

z = (0.7698 – 0.65)/sqrt(0.65*(1-0.65)/1034)

= 8.0765

P-value = P(z > 8.0765) + P(z < -8.0765) = 0.0 (Ans.) From TutorTeddy