The owner of a local golf course wants to estimate the diffe

The owner of a local golf course wants to estimate the difference between the average ages of males and females that play on the golf course. He randomly samples 29 men and 26 women that play on his course. He finds the average age of the men to be 43.608 with a standard deviation of 9.616. The average age of the women was 44.085 with a standard deviation of 5.243. If a 95% confidence interval is calculated to estimate the difference between the two average ages, what is the margin of error? Assume both population standard deviations are equal.

A. 3.554

B. 2.123

C. 4.161

D. 2.006

E. 4.259

Solution

b) WHEN SD ARE EQUAL

ME = t a/2 * S * Sqrt ( 1 / n1 + 1 / n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
X (Mean)=43.608; Standard Deviation (s.d1)=9.616
Number(n1)=29
Y(Mean)=44.085; Standard Deviation(s.d2)=5.243
Number(n2)=26
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (28*92.4675 + 25*27.489) / (55- 2 )
S^2 = 61.8173
The Value of |t | with (n1+n2-2) i.e 53 d.f is 1.674
ME = t a/2 * S * Sqrt( 1 /29+ 1/26 )]
= [ 1.674 * Sqrt(61.8173) * Sqrt( 1 /29+ 1/26 ) ]
= 3.5547