In testing $H _ { 0 } : p _ { 1 } - p _ { 2 } = 0 \text { versus } H _ { 0} : p _ { 1 } - p _ { 2 } 0 \text {, where } p _ { 1 } \text { and } p _ { 2 }$ denote the two population proportions, the value of the test statistic is found to be z = -1.82. Then, the P-value is
A) .9312
B) .4656
C) .0688
D) .0344
E) .9656

Question

In testing $H _ { 0 } : p _ { 1 } - p _ { 2 } = 0 \text { versus } H _ { 0} : p _ { 1 } - p _ { 2 } > 0 \text {, where } p _ { 1 } \text { and } p _ { 2 }$ denote the two population proportions, the value of the test statistic is found to be z = -1.82. Then, the P-value is
A) .9312
B) .4656
C) .0688
D) .0344
E) .9656

Quizplus · Accepted Answer

The P-value in this context represents the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the assumption that the null hypothesis is true. Given that the test is one-tailed (as indicated by the alternative hypothesis $H_a: p_1 - p_2 > 0$) and the observed z-value is -1.82, which falls in the left tail of the normal distribution, the P-value corresponds to the area to the left of z = -1.82. However, since the interest is in a positive difference (as indicated by the alternative hypothesis), the relevant P-value for decision-making would actually reflect the area to the right of z = -1.82. This is equivalent to 1 minus the cumulative probability up to z = -1.82. The cumulative probability up to z = -1.82 is approximately 0.0344 (reflecting the area to the left), so the P-value for the test is approximately 1 - 0.0344 = 0.9656. However, the correct interpretation of the options given seems to have been misunderstood in the explanation. The correct P-value for a one-tailed test with z = -1.82, aiming to find the area to the right, directly corresponds to the cumulative probability to the left due to the nature of the question and the options provided. Therefore, the correct answer based on the standard normal distribution and the nature of hypothesis testing should be reconsidered in the context of the options given.

In Testing H0:p1−p2=0 versus H0:p1−p2>0, where p1 and p2H _ { 0 } : p _ { 1 } - p _ { 2 } = 0 \text { versus } H _ { 0} : p _ { 1 } - p _ { 2 } > 0 \text {, where } p _ { 1 } \text { and } p _ { 2 }H0​:p1​−p2​=0 versus H0​:p1​−p2​>0, where p1​ and p2​

In Testing $H _ { 0 } : p _ { 1 } - p _ { 2 } = 0 \text { versus } H _ { 0} : p _ { 1 } - p _ { 2 } > 0 \text {, where } p _ { 1 } \text { and } p _ { 2 }$