Question

A research studied the effect that topical application of procyanidin B-2 (PB-2) isolated from apple juice might have on new hair growth. For 6 months they treated one group of 19 balding men twice a day with 1.8 ml of agent containing 1% PB-2, corresponding to 30 mg of PB-2 daily. A second group of 10 balding men served as a control group. They were treated in exactly the same way, except that the agent contained no PB-2. No other hair care products except shampoos and rinses were permitted during the study. Before and after the six-month period, hairs at a predetermined site were clipped from each participating subject and the diameters of the collected hairs were measured. The change in total hairs per .25 cm2 for each of the participants was recorded and the summary result is presented in table below. Group n Mean Variance Treatment 19 6.68 11.1 Control 10 0.08 20.8 (a).Test for any difference on the average weight gains between two groups using α = 0.01. State the null and alternate hypotheses, and report the value of the test statistic, and the critical value used to conduct the test. Report your decision regarding the null hypothesis and your conclusion in the context of the problem. Assume population variances are equal. (b). Find a 95% confidence interval for µ1 − µ2 and interpret this

Answer 1

(a)

H0: Null Hypothesis: (There is no difference on the average weight gains between two groups)

μι = μ2

HA: Alternative Hypothesis: (There is difference on the average weight gains between two groups)

ul + u2

sp = V (nl - 1)s12 + (n2 - 1) s22 nl + n2 - 2

(19 - 1) x 11.1+(10 - 1) x 20.8 19 + 10-2 = 3.786

SE = SpV1/nl + 1/n2

= 3.786 X 1/19 + 1/10 = 1.4791

Test Statistic is given by:

t = (6.68-0.08)/1.4791

= 4.4621

$\alpha$ = 0.01

ndf = n1 + n2 - 2 = 19 + 10 - 2 = 27

From Table, critical values of t = $\pm$ 2.7707

Since calculated value of t = 4.4621 is greater than critical value of t = 2.7707, the difference is significant. Reject null hypothesis.

The data support the claim that there is difference on the average weight gains between two groups.

(b)

$\alpha$ = 0.05

ndf = 27

From Table, critical values of t = $\pm$ 2.0518

Confidence Interval:

(6.68 - 0.08) $\pm$ (2.0518 X 1.4791)

= 6.6 $\pm$ 3.0348

= ( 3.5652 ,9.6348)

Confidence Interval:

3.5652 <
μl – μ2
< 9.6348

Interpretation:

95% Confidence Interval (3.5652 ,9.6348) is a range of values that we can be 95% certain contains the true mean difference of the population.