Solution:
a)
P-value = P(Z > 0.84)
= 1 - P(Z < 0.84 )
= 1 - 0.7995
= 0.2005
b)
P-value = P(Z < z )
= P(Z < -2.38 )
= 0.0087
c)
P-value = 2 * P(Z > z )
= 2 * ( 1 - P(Z < 2.25 ) )
= 2 * ( 1 - 0.9878 )
= 2 * 0.0122
= 0.0244
blu Onfi- mes ately SKILL BUILDER 4 In Exercises 5.56 and 5.57, find the p-value based...
SKILL BUILDER 4 ni In Exercises 5.56 and 5.57, find the p-value based blu on a standard normal distribution for each of the following standardized test statistics. 5.56 (a) z = 0.84 for an upper tail test for a differ- d glad o ence in two proportions onfi mes tely (b) z =-2.38 for a lower tail test for a difference in 0.000714 tion the two means (c) z 2.25 for a two-tailed test for a correlation
Find the p-value based on a standard normal distribution for each of the following standardized test statistics. (a) z = 0.79 for a right tail test for a difference in two proportions Round your answer to two decimal places. p-value = i (b) z = -2.38 for a left tail test for a difference in two means Round your answer to three decimal places. p-value = (c) z = 2.31 for a two-tailed test for a proportion Round your answer...
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...