Question

The director of research and development is testing a new medicine. She wants to know if there is evidence at the 0.01 level that the medicine relieves pain in less than 348 seconds. For a sample of 14 patients, the average time in which the medicine relieved pain was 343 seconds with a variance of 484. Assume the population distribution is approximately normal.

Step 1 of 5: State the null and alternative hypotheses

Step 2 of 5: Find the value of the test statistic. Round your answer to three decimal places.

Step 3 of 5: Specify if the test is one-tailed or two-tailed.

Step 4 of 5: Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Step 5 of 5: Make the decision to reject or fail to reject the null hypothesis.

Answer 1

Step 1 of 5: State the null and alternative hypotheses

Null Hypothesis : H0 : the medicine doesn't relieves pain in less than 348 seconds. 348 seconds

Alternative Hypothesis : Ha : the medicine doesn't relieves pain in less than 348 seconds. 348 seconds

Step 2 of 5: Find the value of the test statistic. Round your answer to three decimal places.

Here variance = s2 = 484

standard error = sqrt(s2/n) = sqrt(484/14) = 5.88

= 343 secs

Test statistic

t = (343 - 348)/5.88 = -0.850

Step 3 of 5: Specify if the test is one-tailed or two-tailed.

The test is a one tailed test.

Step 4 of 5: Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Here degree of freedom = dF = n -1 = 14 - 1 = 13

critical test statistic for 0.01 level of significance.

tcritical = TINV(0.02, 13) = 2.650

so here decision rule is that if l t l >  tcritical , we will reject the null hypothesis.

Step 5 of 5: Make the decision to reject or fail to reject the null hypothesis.

Here as we know that l t l < tcritical , we would fail toreject the null hypothesis and conclude that the medicine doesn't relieves pain in less than 348 seconds