n= 8
M1 = 20.5
M2= 11.125
S21=
34.29
S22=57.84
Conclusion?
a) Null and alternative hypotheses
Ho: 1 = 2
H1: 1 > 2
b) For a = 0.05 and d.f = n1+ n2 -2
d.f = 8 + 8 -2 = 14
tcritical = ta , d.f = t0.05 , 14
tcritical = 1.761
c) F max = S22/S12
F max = 57.84/34.29
F max = 1.69
d) if t > 1.761 ,we reject the null hypothesis otherwise we fail to reject the null hypothesis.
e). t = M1 - M2/ Sp * sqrt(1/n1+1/n2)
Sp = sqrt((n1-1)s12 + (n2-1)s22)/n1+n2-2))
Sp = 6.79
t = 20.5 - 11.125/6.79*0.5
t = 2.76
Here t = 2.76 > 1.69 , so we reject the null hypothesis.
Conclusion : There is sufficient evidence to support the claim accupuncture treatment decreases cigarette smoking
Independent t-Test: A researcher examines the short-term effect of a single treatment with acupuncture on cigarette...
Question 1: Suppose we are testing the effectiveness of a new hypnosis treatment at reducing cigarette smoking. 20 smokers are recruited. Half are randomly assigned to a control treatment where they are given some information about the dangers of smoking and sent home. The other half receives the new hypnosis treatment. Four weeks later we measure the number of cigarettes each participant smokes in a week. We want to see if people who receive the treatment smoke a different average...