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, is changed by vigorous exercise, so you have a group of nine college students exercise vigorously for 30 minutes and then complete the maze. assume that σ remains unchanged at three seconds. the hypotheses you decide to test are H0:μ=30 versus Ha:μ≠30. Suppose it takes the 9 students an average of x̄=32.05 seconds to complete the maze. At 1% significance level, what can you conclude?

(a) H0 should be rejected because the P-value is less than 0.01.

(b) H0 should not be rejected because the P-value is greater than 0.01.

(c) Ha should be rejected because the P-value is less than 0.01.

(d)Ha should not be rejected because the P-value is greater than 0.01.

Answer 1

Solution:

Since the population SD i.e σ is known , we use z test .

H0:μ=30 versus Ha:μ≠30

The test statistic z is given by

z =   

= (32.05 - 30) / (3/9)

= 2.05

Now , observe that ,there is   sign in H1. So , the test is two tailed.

For two tailed test :

p value = 2 * P(Z < -z)

= 2 * P(Z < -2.05)

= 2 * 0.0202

= 0.0404

Given level of significance is 1% i.e. 0.01

p value is greater than 0.01

In this case , we do not reject H0

Answer :

H0 should not be rejected because the P-value is greater than 0.01.

Option (b) is correct.