Question

2. A survey of a new postsurgical treatment was compared with a standard treatment. Seven subjects received the new treatment(cases), while seven others (Controls) received the standard treatment. The recovery times, in days, are given below. Treatment 12 13 15 19 20 21 24 Control 18 23 24 30 32 35 39 Can you conclude that the mean recovery time for those receiving the treatment is less than the mean for those receiving the standard treatment. Use a α=0.05 significance level.

a. Construct a 90% confidence interval for the difference in mean recovery time for those receiving the new treatment and others receiving the standard treatment.

b. Do the hypothesis test.

Answer 1

X :- Represent new treatment

Y :- Represent control treatemt

Mean X̅ = Σ Xi / n
X̅ = 124 / 7 = 17.7143
Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 119.4284 / 7 -1 ) = 4.4615

Mean Y̅ = ΣYi / n
Y̅ = 201 / 7 = 28.7143
Sample Standard deviation SY = √ ( (Yi - Y̅ )2 / n - 1 )
SY = √ ( 327.4284 / 7 -1) = 7.3872

Part a)

H0 :- µ1 = µ2
H1 :- µ1 < µ2

Test Statistic :-
t = (X̅1 - X̅2) / SP √ ( ( 1 / n1) + (1 / n2))



t = ( 17.7143 - 28.7143) / 6.1023 √ ( ( 1 / 7) + (1 / 7 ) )
t = -3.3724

Test Criteria :-
Reject null hypothesis if t < - t(α, n1 + n2 - 2)
Critical value t(α, n1 + n1 - 2) = t( 0.05 , 7 + 7 - 2) = 1.782
t < - t(α, n1 + n2 - 2) = -3.3724 < -1.782
Result :- Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 3.3724 ) = 0.0028
Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.0028 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence to support the claim that the mean recovery time for those receiving the treatment is less than the mean for those receiving the standard treatment.

Part a)

Confidence interval is :-
( X̅1 - X̅2 ) ± t( α/2 , n1+n2-2) SP √( (1/n1) + (1/n2))
t(α/2, n1 + n1 - 2) = t( 0.1/2, 7 + 7 - 2) = 1.782
( 17.7143 - 28.7143 ) ± t(0.1/2 , 7 + 7 -2) 6.1023 √ ( (1/7) + (1/7))
Lower Limit = ( 17.7143 - 28.7143 ) - t(0.1/2 , 7 + 7 -2) 6.1023 √( (1/7) + (1/7))
Lower Limit = -16.8126
Upper Limit = ( 17.7143 - 28.7143 ) + t(0.1/2 , 7 + 7 -2) 6.1023 √( (1/7) + (1/7))
Upper Limit = -5.1874
90% Confidence Interval is ( -16.8126 , -5.1874 )

part b)

Since the values of the confidence interval is negative, we conclude that the claim is true i.e   the mean recovery time for those receiving the treatment is less than the mean for those receiving the standard treatment is true