Question

A study of working actors looked at age and gender. One sample of 65 male actors had a mean age of 28 and a standard deviation of 3. The other sample included 65 female actors with a mean age of 39 and a standard deviation of 2, Estimate with 98% confidence the difference between the average ages of working male (M1) and female (H2) actors. Round answers to the nearest hundredth.

Answer 1

here as sample size is high we may use z distribution

	male		female
x1            =	28.000	x2            =	39.000
n1           =	65.000	n2           =	65.000
σ1           =	3.000	σ2           =	2.000
std error of difference σ1-σ2=				√(σ21/n1+σ22/n2)    =	0.4472

for 98 % CI value of z=					2.3263
margin of error E=z*std error                            =					1.0404
lower confidence bound=estimated mean diff.-margin of error =					-12.04
Upper confidence bound=estimated mean diff+margin of error=					-9.96

hence 98% CI =-12.04<$\mu$₁-$\mu$₂ <-9.96

(Note from t - distributioon this interval can be (-12.06 ; -9.94) or (-12.05 ; -9.95)