Question

Consider the following hypothesis test for the mean difference. (Give your answers correct to four decimal places.)

(a) Determine the p-value for H_o: μ_d = 0 and H_a: μ_d > 0, with n = 22 and t = 1.87.


(b) Determine the p-value for H_o: μ_d = 0 and H_a: μ_d ≠ 0, with n = 16 and t = -2.


(c) Determine the p-value for H_o: μ_d = 0 and H_a: μ_d < 0, with n = 32 and t = -2.57.


(d) Determine the p-value for H_o: μ_d = 0.75 and H_a: μ_d > 0.75, with n = 11 and t = 3.6.

Answer 1

Solution:

We have to find p-value for given hypothesis and test statistic values.

We use Excel commands.

Part a) H_o: μ_d = 0 and H_a: μ_d > 0, with n = 22 and t = 1.87.

This is right tailed test.

=T.DIST.RT( x , df )

where

x = t = 1.87 and     df = n - 1= 22 - 1 = 21

Thus

=T.DIST.RT( 1.87 , 21 )

=0.0377

Thus P-value = 0.0377

Part b) H_o: μ_d = 0 and H_a: μ_d ≠ 0, with n = 16 and t = -2.

This is two tailed test and df = n - 1 = 16 - 1 = 15

thus

=T.DIST.2T(2 , 15)

= 0.0639

( note we use absolute t value)

Thus P-value = 0.0639

Part c) H_o: μ_d = 0 and H_a: μ_d < 0, with n = 32 and t = -2.57.

thus is left tailed test. df = n - 1 = 32-1=31

=T.DIST( -2.57 , 31 , TRUE )

= 0.0076

Thus P-value = 0.0076

Part d ) H_o: μ_d = 0.75 and H_a: μ_d > 0.75, with n = 11 and t = 3.6.

This is right tailed test and df = n - 1 = 11 - 1 = 10

Thus

=T.DIST.RT( 3.6 , 10 )

=0.0024

thus P-value = 0.0024