Question

You may need to use the appropriate appendix table or technology to answer this question.

The following results come from two independent random samples taken of two populations.

Sample 1	Sample 2
n1 = 50	n2 = 25
x1 = 13.6	x2 = 11.6
σ1 = 2.4	σ2 = 3

(a)

What is the point estimate of the difference between the two population means? (Use

x₁ − x₂.)

(b)

Provide a 90% confidence interval for the difference between the two population means. (Use

x₁ − x₂.

Round your answers to two decimal places.)

to

(c)

Provide a 95% confidence interval for the difference between the two population means. (Use

x₁ − x₂.

Round your answers to two decimal places.)

to

Answer 1

a)

the point estimate of the difference between the two population means = 13.6 - 11.6 = 2

b)

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(5.76/50 + 9/25)
sp = 0.6893

Given CI level is 0.9, hence α = 1 - 0.9 = 0.1                  
α/2 = 0.1/2 = 0.05, zc = z(α/2, df) = 1.64                  
                  
Margin of Error                  
ME = zc * sp                  
ME = 1.64 * 0.6893                  
ME = 1.13                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (13.6 - 11.6 - 1.64 * 0.6893 , 13.6 - 11.6 - 1.64 * 0.6893                  
CI = (0.87 , 3.13)                  

c)

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(5.76/50 + 9/25)
sp = 0.6893

Given CI level is 0.95, hence α = 1 - 0.95 = 0.05                  
α/2 = 0.05/2 = 0.025, zc = z(α/2, df) = 1.96                  
                  
Margin of Error                  
ME = zc * sp                  
ME = 1.96 * 0.6893                  
ME = 1.35                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (13.6 - 11.6 - 1.96 * 0.6893 , 13.6 - 11.6 - 1.96 * 0.6893                  
CI = (0.65 , 3.35)