Question

Chapter 5, Section 1, Exercise 013 Find the p-value based on a standard normal distribution for each of the following standardized test statistics (a) Z = 0.89 for a right tail test for a difference in two proportions Round your answer to two decimal places p-value0.20 the absolute tolerance is +/-0.01 (b) Z-_2.39 for a left tail test for a difference in two means Round your answer to three decimal places. p-value- the absolute tolerance is +/-0.001 (c) = 2.18 for a two-tailed test for a proportion Round your answer to three decimal places. p-value-

Answer 1

A).

Consider the given problem here “Φ(z)” measure the probability from “- infinity” to “z”.

=> for “z=0.89”, the “Φ()” value is “Φ(0.89) = 0.8132671”, => for right tail test the “p-value” is given by, “p=1 - Φ(0.89) = 0.1867329 = 0.19”, => p-value=0.19.

B).

Now, if “z = (-2.39)”, => the “Φ()” value is given by, => Φ(-2.39) = 1- Φ(2.39) = 1-0.9915758, => the “p-value” for the lest tailed test for “z=(-2.39)” is given by, => P(-2.39)= Φ(-2.39).

=> P(-2.39)= 1- 0.9915758 = 0.0084242 = 0.008, => p-value=0.008.

C).

Now, if “z = 2.18”, => the “Φ()” value is given by, => Φ(2.18) = 0.9853713, => the “p-value” for the two-tailed test for “z=2.18” is given by, => p(2.39)=2*(1-0.9853712).

=> P(2.18) = 0.0292576 = 0.029, => p-value=0.029.