Question

# A medical team was asked to study the difference in systolic blood pressure (BP) between men...

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 = ((n1-1)S1^2) + ((n2-1)*S2^2))/((n1-1)+(n2-1)) = (((4-1)*16*16) + ((5-1)*12*12))/((4-1)+(5-)) = (768+576)/7 = 192
alpha = 0.05, alpha/2 = 0.025 and df = n1 + n2 -2 = 4+5-2 = 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)))
(175-145)+/- 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