A medical team was asked to study the difference in systolic blood pressure (BP) between men (µ1) and women (µ2) with heart trouble. The BP measurements are normally distributed. The variance of the men’s group is equal to thatof the women’s group. For a random sample of 4 men, the sample average is 175 and the sample standard deviation is 16. For a random sample of 5 women, the sample average is 145 and the sample standard deviation is 12.
What is the 95% confidence interval estimate for the difference between the two means?
A 
−4.20 to 33.20 

B 
−0.65 to 29.65 

c 
5.05 to 34.05 

d 
8.02 to 51.98 
Solution:
Given in the question
95% confidence interval for differences of the mean can be
calculated as
(X1bar  X2bar)+/talpha/2* sqrt((Sp^2((1/n1)+(1/n2)))
Here we will use the pooled variance tests as both men and women
variances are equal.
Number of men = 4
X1bar = 175
S1 = 16
Number of womens = 5
X1bar = 145
S1 = 12
Sp^2 can be calculated as
Sp^2 = ((n11)S1^2) + ((n21)*S2^2))/((n11)+(n21)) =
(((41)*16*16) + ((51)*12*12))/((41)+(5)) = (768+576)/7 =
192
alpha = 0.05, alpha/2 = 0.025 and df = n1 + n2 2 = 4+52 = 7
So talpha/2 = 2.365
95% confidence interval for differences of the mean can be
calculated as
(X1bar  X2bar)+/talpha/2* sqrt((Sp^2((1/n1)+(1/n2)))
(175145)+/ 2.365*sqrt(192*((1/4)+(1/5))
30+/ 2.365*sqrt(86.4)
30+/21.98
So 95% confidence interval is 8.02 to 51.98
So its correct answer is D. i.e. 8.02 to 51.98
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