Question

When a surgeon repairs injuries, sutures (stitched knots) are used to hold together and stabilize the injured area. If these knots elongate and loosen through use, the injury may not heal properly because the tissues would not be optimally positioned. Researchers at a university tied a series of different types of knots with two types of suture material, Maxon and Ticron.

Suppose that 112 tissue specimens were available and that for each specimen the type of knot and suture material were randomly assigned. The investigators tested the knots to see how much the loops elongated. The elongations (in mm) were measured and the resulting data are summarized below. For purposes of this exercise, assume it is reasonable to regard the elongation distributions as approximately normal.

	Maxon
Types of knot	n	x	sd
Square (control)	10	10.7	0.1
Duncan Loop	15	11.6	0.3
Overhand	15	11.0	0.9
Roeder	10	13.5	0.1
Snyder	10	13.5	2.0

	Ticron
Types of knot	n	x	sd
Square (control)	10	2.6	0.06
Duncan Loop	11	11.5	0.40
Overhand	11	8.1	1.00
Roeder	10	5.0	0.04
Snyder	10	8.1	0.06

(a)

Is there a significant difference in mean elongation between the square knot and the Duncan loop for Maxon thread? (Use α = 0.05. Use a statistical computer package to calculate the P-value.Use μ_{Square Knot} − μ_{Duncan Loop}. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.)

t= df = P-value =

(b)

Is there a significant difference in mean elongation between the square knot and the Duncan loop for Ticron thread? (Use α = 0.05. Use a statistical computer package to calculate the P-value.Use μ_{Square Knot} − μ_{Duncan Loop}. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.)

t= df = P-value =

(c)

For the Duncan loop knot, is there a significant difference in mean elongation between the Maxon and Ticron threads? (Use α = 0.05. Use a statistical computer package to calculate the P-value.Use μ_Maxon − μ_Ticron. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.)

t= df = P-value =

Answer 1

(a)

Following is the output of independent sample t test generated by excel:

The test statistics is

t = -10.76

The degree of freedom is

df = 18

The p-value is

p-value = 0.000

(b)

The test statistics is

t = -72.90

The degree of freedom is

df = 10

The p-value is

p-value = 0.000

(c)

The test statistics is

t = 0.70

The degree of freedom is

df = 17

The p-value is

p-value = 0.495