Question

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 3.1 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 ?¯=28.1seconds to complete the maze. Use this information to test the hypotheses
?0:?=30
??:?≠30
Conduct a test using a significance level of ?=0.05.

(a) The test statistic  

(b) The positive critical value, z* =

Answer 1

Solution :

This is the two tailed test,  

The null and alternative hypothesis is ,

H0 :   = 30

Ha :    30

Test statistic = z =

= ( - ) / / n

= (28.1- 30) / 3.1 / 25

Test statistic = z = -3.06

= 0.05

/2 = 0.025

Z/2 = Z0.025  = 1.96

z* = 1.96