Question

use the pooled variance method for this problem

[No , A study was undertaken to compare the respiratory notized and nonhypnotized subjects to responses of certain mstructions a control group. ns. The 16 male volunteers were allocated( o an experimental group to be hypnotized or to randomn, Baseline measurements were taken at the eriment. In analyzing the data, the research- start edthat the baseline breathing patterns of the two of the noticed were different; this was surprising, since all the sub- treated the same up to that time. One expla- oposed for this unexpected difference was that evperimental group were more excited in anticipation the experience of being hypnotized. The accompanying resents a summary of the baseline measurements of ventilation (liters of air per minute per square meter dy area). Parallel dotplots of the data are given in the nation pr the following graph.s INote: Formula (6.7.1) yields 14 df.] Experimental Control pose 5.32 5.60 5.74 6.06 6.32 6.34 6.79 7.18 4.50 4.78 4.79 4.86 5.41 5.70 6.08 6.21 tate ose ate 6.169 0.621 5.291 0.652 Section 7.5 One-Tailed t Tests 265 7.0 6.0 E 5.5- 5.0 4.5 Experimental Control (a) Use a t test to test the hypothesis of no difference against a nondirectional alternative. Let a 0.05 (b) Use a t test to test the hypothesis of no difference against the alternative that the experimental conditions produce a larger mean than the control conditions. Let α 0.05.

Answer 1

a)
The Null hypothesis, there is no significant difference in the average baseline measurement between the experimental and control. u1=u2

An Alternative hypothesis, there is a significant difference in the average baseline measurement between the experimental and control. u1 =/= u2

Sp^2 = {(n1-1)*s1^2 + (n2-1)*s2^2} / {n1+n2-2}
Sp^2 = ((8-1)*0.621^2+(8-1)*0.652^2)/(8+8-2)\
Sp^2 = 0.4054

Test statistic, t = (Xbar1 - Xbar2)/sqrt(Sp^2*(1/n1+1/n2))
t = (6.169-5.291)/SQRT(0.40537*(1/8+1/8))
t = 2.7580

df = n1+n2-2
df = 8+8-2
df = 14

critical value: t(a/2, df) = t(0.01/2,14)
t(a/2, df) = t.INV.2T(0.01,14)
t(a/2, df) = 2.9768

p-value = 2*(1-P(T<|t|)
p-value = 2*(1-P(T<abs(2.758))
p-value = T.DIST.2T(abs(2.758),14)
p-value = 0.0154

With t=2.7, p<5%, i reject ho and conclude that there is a significant difference in the average baseline measurement between the experimental and control.

b)
The Null hypothesis, There is no significant difference in the average baseline measurement between the experimental and control. u1=u2
Alternative hypothesis, the average baseline measurement for experimental condition is Greater in comparison to the control condition. u1<u2
Sp^2 = {(n1-1)*s1^2 + (n2-1)*s2^2} / {n1+n2-2}
Sp^2 = ((8-1)*0.621^2+(8-1)*0.652^2)/(8+8-2)
Sp^2 = 0.40537

Test statistic, t = (Xbar1 - Xbar2)/sqrt(Sp^2*(1/n1+1/n2))
t = (6.169-5.291)/SQRT(0.40537*(1/8+1/8))
t = 2.7580

df = n1+n2-2
df = 8+8-2
df = 14.000

critical value: t(a, df) = t(0.01,14)
t(a, df) = ABS(T.INV(0.01,14))
t(a, df) = 2.624494068

p-value = P(T<t)
p-value = P(T<2.758)
p-value = T.DIST(2.758,14,TRUE)
p-value = 0.992299449

With t=2.758, p>5%, i fail to reject Ho. There is no sufficient evidence to support the claim the average baseline measurement for experimental condition is Greater in comparison to the control condition