Question

Neurosurgery Operative Times. In Example 10.83, we conducted a nonpooled t -test, at the5% significance level, to decide whether the mean operative time is less with the dynamic system than with the static system.
a. Using a pooled t-test, repeat that hypothesis test.
b. Compare your results from the pooled and nonpooled tt -tests.

c. Which test do you think is more appropriate, the pooled or nonpooled tt -test? Explain your answer.

Short answer

(a) The data provides the sufficient evience to concludethat the mean number of acute postoperativedays spent in the hospital is smaller with the dynamic systemthan with the static system.

(b) The null hypothesis is rejected by using pooled t-test and it is not rejected by using nonpooled t-test.

(c)

The non-pooled t-test is appropriate, because the standard deviations for the two variables are not equal. And the two smple sizes are also different.

Step-by-step solution

Part (a) Step 1. GIven Information.

We conducted a nonpooled t -test, at the5% significance level, to decide whether the mean operative time is less with the dynamic system than with the static system.

Part (a) Step 2. State the null and alternative hypothesis.

Null Hypothesis:

$H_0:{\mu }_1={\mu }_2$

The data does not provide the sufficient evience to concludethat the mean number of acute postoperativedays spent in the hospital is smaller with the dynamic systemthan with the static system.

$H_0:{\mu }_1<{\mu }_2$

The data provides the sufficient evience to concludethat the mean number of acute postoperativedays spent in the hospital is smaller with the dynamic systemthan with the static system

Part (a) Step 3. Decide the significance test.

THe significance level is$\alpha =0.005$.

Part (a) Step 4. MINITAB procedure.

Compute the value of the test static and P-value by using MINITAB procedure:

1. Choose the Stat > Base Statistics > 2-sample r.

2. Choose samples in different columns.

3. In First,enter the column Dynamic and in second, enter the column Static.

4. Select Assume equalvariances.

5. Choose option.

6. In confidence level, enter 95.

7. In Alternative, select less than.

8. Click OK in all the dialog boxes.

Part (a) Step 5. MINITAB output.

From the MINITAB output, the value of test statistic is-2.45.

Part (a) Step 6. P-level approach.

From the MINITAB output, the value of P is 0.012.

If $P\le \alpha$, reject the hypothesis.

Here the P-value is less than the level of significance.

Thus, the null hypothesis is rejected at 5% level.

Interpretation:

The data provides the sufficient evience to concludethat the mean number of acute postoperativedays spent in the hospital is smaller with the dynamic systemthan with the static system

Part (b) Step 1. Comparison.

The null hypothesis is rejected by using pooled t-test and it is not rejected by using nonpooled t-test.

Part (c) Step 1. Explanation.

The non-pooled t-test is appropriate, because the standard deviations for the two variables are not equal.

And the two smple sizes are also different.