Question

Chapter 9.1 And 9.2: Problem 15 Previous Problem Problem List Next Problem (1 point) The time needed for college students to complete a certain paper-and-pencil maze follows a normal distribution with a mean of 30 seconds and a standard deviation of 4 seconds. You wish to see if the mean time is changed by vigorous exercise, so you have a group of 25 college students exercise vigorously for 30 minutes and then complete the maze. It takes them an average of 29.2 seconds to complete the maze. Use this information to test the hypotheses H: 4 = 30 H : 30 Conduct a test using a significance level of a 0.01 (a) The test statistic (b) The positive critical value, z = (c) The final conclusion is A. There is not sufficient evidence to reject the null hypothesis that B. We can reject the null hypothesis that = 30 and accept that - 30. 30.

Answer 1

Here Sample Mean x = 29.2. sample size n = 25. population standard deviation os 4 Test statistic is z= Vu (³-M) -√25 (29.2-30) o 4 = -1 or we can take. Z = 1 Because here the di Herence Matters sign doesn't ). o Here 0 =0.01. The test is two tailed test. positive critical value 7 * = Zx . Z -0.005 z* = 2.576 (from 7 table) Cleasly Z= 1 < 7 . we fail to Reject Ho. There is not sufficient evidence the reject the null hypothesis M = 30.