The one-sample t-statistic for a test of H_0 : \mu = 42 versus H_1 :\mu < 42 based on n = 15 observations has the

value t = -2.138, where H_0 and H_1 are the null and alternative hypotheses, respectively. The sample size is
denoted by n. Use t-table to determine which two P-values bracket the P-value of the test, Report your answer in
decimal form as listed in the t-table.
() < p-value of the test < (). Fill two blanks?

To determine which two P-values bracket the P-value of the test, you can use the t-table.

First, we need to calculate the degrees of freedom (df) for the test. In this case, since we have a one-sample test with n = 15 observations, the degrees of freedom can be calculated as df = n - 1 = 15 - 1 = 14.

With the t-value given as -2.138, we need to find the corresponding P-value in the t-table.

Looking in the t-table, we find the P-value associated with a t-value of -2.138 and df = 14.

From the t-table, we find that the P-value corresponding to -2.138 and df = 14 is approximately 0.025.

Since this is a one-tailed test (H1: μ < 42), we are interested in the area under the t-distribution curve to the left of -2.138.

Therefore, we can conclude that the P-value of the test is less than 0.025.

Now, let's find the second P-value that brackets the P-value of the test.

Since this is a left-tailed test, the other P-value we need to find is the complement of the P-value we previously found.

Since the area under the t-distribution curve is symmetrical, we know that the P-value we are looking for is 1 - 0.025 = 0.975.

Thus, we can conclude that the two P-values bracketing the P-value of the test are (0.025 < p-value of the test < 0.975).