Question

2. A study on wounds involved subjecting 18 randomly selected anaesthetized crested newts to a razor cut. The healing velocities of the cuts in micrometeres per hour for each newt were 29 27 34 40 22 28 14 35 26 35 12 30 23 18 11 22 23 33

(a) It is believed that the typical wound-healing velocity for crested newts is 30 micrometres/hour. Carry out a test to see if there is any evidence against this claim at level 0.05. Be sure to check all assumptions that you can check.

(b) What type of error (Type 1 or Type 2) could you be making from your conclusion in (a)? Describe that error in the context of the question.

(c) Carry out the test from (a) at level 0.01 using an appropriate condence interval.

Answer 1

Step 1:

Ho: = 30

Ha: 30

Null hypothesis states that the avg wound healing velocity for crested newts is 30 micrometers/hour.

Step 2: Test statistics

n = 18

mean = sum of all terms/ no of terms = 462 / 18   = 25.6667

data	data-mean	(data - mean)2
29	3.3333	11.11088889
27	1.3333	1.77768889
34	8.3333	69.44388889
40	14.3333	205.44348889
22	-3.6667	13.44468889
28	2.3333	5.44428889
14	-11.6667	136.11188889
35	9.3333	87.11048889
26	0.3333	0.11108889
35	9.3333	87.11048889
12	-13.6667	186.77868889
30	4.3333	18.77748889
23	-2.6667	7.11128889
18	-7.6667	58.77828889
11	-14.6667	215.11208889
22	-3.6667	13.44468889
23	-2.6667	7.11128889
33	7.3333	53.77728889

Assuming that the data is normally distributed. Also as the population standard deviation is not given, we will calculate t statistics

Step 3:

Level of significane = 0.05

df = 17

Critical Value of t (Two Tailed): ± 2.11

Since the t stat falls in the rejections area (-2.208 < -2.11), we reject the Null hyothesis.

Hence there is sufficient evidence to believe that avg wound healing velocity for crested is not 30 micrometers/hour.

(b) We might be making Type I error : Reject the Null hypothesis when in reality the Null hypothesis is True.

i.e. The avg wound healing velocity for crested is 30 micrometers/hour but we are rejecting this and saying that it is not 30 micrometers/hours.

(c) t value for 99% of confidence interval at 17 df is TINV(0.01,17) = 2.898

CI = mean +/- E = 19.981 31.353

Since 30 is in the range, we fail to reject the Null hypothesis at 0.01 significane level.