In testing $H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } = 0 \text { versus } H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } \neq 0$ the computed value of the test statistic is z = 1.98. The P-value for this two-tailed test is then
A) .4761
B) .0478
C) .0239
D) .2381
E) .2619

Question

Quizplus · Accepted Answer

The P-value for a two-tailed test is found by doubling the area in the tail beyond the observed z-value. Since the z-value is 1.98, we look up the area in the standard normal distribution table for z = 1.98, which gives us the area to the left of z. To find the area in the right tail, we subtract this value from 1 and then double it for a two-tailed test. The correct P-value is closest to .0478, indicating the probability of observing a test statistic as extreme as or more extreme than 1.98 under the null hypothesis.

In Testing H0:μ1−μ2=0 versus H0:μ1−μ2≠0H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } = 0 \text { versus } H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } \neq 0H0​:μ1​−μ2​=0 versus H0​:μ1​−μ2​=0

In Testing $H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } = 0 \text { versus } H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } \neq 0$