p;ease type it 4. In a study on the healing process of bone fractures, researche

Question

p;ease type it

4. In a study on the healing process of bone fractures, researchers observed the time required to completely heal a bone fracture. Ten rats were randomly divided into two groups of five. No treatment was done to one group and a medical treatment was given to the other group. The time spent (in days) to completely heal the fracture for each rat is given below:

Control: 30, 22, 28, 35, 45

Treatment: 33, 40, 24, 25, 24

a) Test to see if the mean duration of healing can be reduced by the medical treatment at = 0.01.

b) Construct a 99% confidence interval for the difference of the means of the two groups.

Explanation / Answer

The null hypothesis,H0: There is no difference in means of the two cases

Alternate hypothesis, Ha: Mean of the treatment group < Mean of control group

Control Group: Mean = 32, Standard deviation = 7.72

Treatment Group: Mean = 29.2, Standard deviation = 6.368

Test statistic = (7.12-8.63) / sqrt((7.72^2/5)+(6.368^2/5)) = -1.15/4.4757 = =0.3376

(a) From a standard t-table, for a 99% confidence interval (alpha =0.01), the one-tailed critical value for 4 degrees of freedom = 3.747

Since the absolute value of the test statistic is less than the critical value, we cannot reject the Null hypothesis. So we can conclude that there is not sufficient evidence to conclude that the mean duration can be reduced by medical treatment.

(b) Difference of the two means = 1.15

We have the critical value for a 2-tailed test =4.604

So the confidence limits = 1.15 +/- 4.604 x sqrt ((7.72^2/5)+(6.368^2/5)) = 1.15 +/- 4.604 x 4.475 = [ -19.45, 21.76]

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.