91Ó°ÊÓ

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.

Null Hypothesis:

$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2$

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.

$H_0:{\mathrm{μ}}_1<{\mathrm{μ}}_2$

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

THe significance level is$\mathrm{Î\pm }=0.005$.

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.

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

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

If $P≤\mathrm{Î\pm }$, 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

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

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.