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 seconds. You wish to see if the mean time μ μ is changed by vigorous exercise, so you have a group of 15 college students exercise vigorously for 30 minutes and then complete the maze. It takes them an average of x¯=27.2 x ¯ = 27.2 seconds to complete the maze. Use this information to test the hypotheses H0:μ=30 H 0 : μ = 30 Ha:μ≠30 H a : μ ≠ 30 Conduct a test using a significance level of α=0.01 α = 0.01 . (a) The test statistic (b) The positive critical value, z* =??

Answer 1

Solution,

The null and alternative hypothesis is ,

H0 :   = 30

Ha :    30

= 27.2

= 3

n = 15

a) Test statistic = z =

= ( - ) / / n

= (27.2 - 30 ) / 3 / 15

Test statistic = z = -3.61

b) This is the two tailed test .

= 0.01

/2 = 0.005

Z/2 = Z0.005  = 2.576

z* = 2.576