Consider testing the hypothesis: H0:ƒÊ=100 vs. Ha:ƒÊ�‚100. If the value of the test statistic is equal to 1.36, then the p-value is:
A. 0.4131
B. 0.1738
C. 0.2066
D. 0.9131
To find the p-value for this hypothesis test, we need to use the test statistic and the appropriate probability distribution. In this case, since the test statistic is given and the sample size is not specified, we will assume that the sample size is large enough to use the normal distribution and the Z-test.
The p-value represents the probability of obtaining a test statistic as extreme or more extreme than the observed test statistic, assuming the null hypothesis is true. In other words, it represents the probability of observing a test statistic as far from the hypothesized value as 1.36, or greater, if the true mean is actually 100.
To determine the p-value, we need to find the area under the normal curve beyond the observed test statistic, in both tails. Since the alternative hypothesis is a two-tailed test (ƒÊ ≠ 100), we will split the significance level equally between the two tails.
Let's calculate the p-value using statistical software, a calculator, or a Z-table:
Using statistical software or a calculator:
P(Z < -1.36) = 0.08687 (left tail)
P(Z > 1.36) = 0.08687 (right tail)
Since this is a two-tailed test, we need to combine the probabilities from both tails:
p-value = 2 * 0.08687 = 0.1738
Therefore, the p-value is 0.1738.
The correct answer is B. 0.1738.
Which test statistic?
If it is a Z score, then .4131 is area between score and mean.