Chapter 10: Q. 10.83 (page 428)

Acute postoperative Days.Refer to Example 10.6 (page 420). The researchers also obtained the following data on the number of acute postoperative days in the hospital using the dynamic and static systems.
At the $5 %$ significance level, do the data provide sufficient evidence to conclude that the mean number of acute post operative days in the hospital is smaller with the dynamic system than with the static system? (Note: role="math" localid="1652859515509" $\bar{x_{1}} = 7.36, s_{1} = 1.22, \bar{x_{2}} = 10.50, s_{2} = 4.59$ ).

Short Answer

Expert verified

The data does not provide sufficient evidence to conclude that the mean number of acute post operative days in the hospital is smaller with the dynamic system than with the static system.

Step by step solution

Step 1. Given information

The significance level is, $α = 0.05$ .

Step 2. State the null and alternative hypothesis:

Null hypothesis:

$H_{0} : μ_{1} = μ_{2}$

That is, there is no evidence that the mean number of acute postoperative days in the hospital is smaller with the dynamic system than with the static system.

$H_{a} : μ_{1} < μ_{2}$

That is, there is evidence that the mean number of acute postoperative days in the hospital is smaller with the dynamic system than with the static system.

Step 3. Compute the value of the test statistic by using Excel add-in (PHStat).

Excel add-in (PHStat) procedure:

In EXCEL, Select Add-Ins > PHStat > Two-Sample tests (Unsummarised Data).
Choose Separate-Variance $t$ test.
In Data enter $0$ under Hypothesized Difference.
Enter $0.05$ under Level of Significance.
Enter $A 1 : A 15$ under Population $1$ Sample cell range.
Enter $B 1 : B 17$ under Population $2$ Sample cell range.
Check First cells in both ranges contain label.
Choose Lower-Tall Test under Test options.
In output options enter Separate-Variance $t$ Test for the difference between two means under title.

Step 4. Excel add-in (PHStat) output:

From the output, the degree of freedom is, $5$ , the value of test statistic is, $- 1.6513$ , the critical value is, $- 2.0150$ and the $P$ -value is, $0.0798$ .

Step 5. Critical value:

From the Excel add-in output, the critical value is, $- 2.0150$ .

$P$ -value:

From the Excel add-in output, the $P$ -value is, $0.0798$ .

Critical value approach:

Here, the value of test statistic does not falls in the rejection region.. That is, $t (= 0 . - 1.6513) > t_{c r i t} (- 2.0150)$ .

Therefore, the null hypothesis is not rejected at $5 %$ level.

Thus, it can be conclude that the test results are not statistically significant at $5 %$ level of significance.

Step 6. P-value approach:

If $P \leq α$ , then reject the null hypothesis. Here, the $P$ -value is, $0.0798$ which is greater than the level of significance. That is, $P (= 0.0798) > α (= 0.05)$ .

Therefore, the null hypothesis is not rejected at $5 %$ level. Thus, it can be conclude that the test results are not statistically significant at $5 %$ level of significance.