A sociologist expects the life expectancy of people in Afric

A sociologist expects the life expectancy of people in Africa is different than the life expectancy of people in Asia. The data obtained is shown in the table below. Determine the 95% confidence interval for the difference in the population means.

Africa Asia

x₁= 55.3 yr.                  x₂=65.2 yr.

₁ = 8.1 yr.                   ₂=9.3 yr.

n₁ =53                          n₂ = 42

A. -11.4 < µ₁- µ₂< -7.6

B. -16.3< µ₁- µ₂<-6.0

C.-13.5< µ₁- µ₂<-6.3

D.-12.2< µ₁- µ₂<-6.9

Solution

CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /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)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x1)=55.3
Standard deviation( sd1 )=8.1
Sample Size(n1)=53
Mean(x2)=65.2
Standard deviation( sd2 )=9.3
Sample Size(n12=42
CI = [ ( 55.3-65.2) ±Z a/2 * Sqrt( 65.61/53+86.49/42)]
= [ (-9.9) ± Z a/2 * Sqrt( 3.2972) ]
= [ (-9.9) ± 1.96 * Sqrt( 3.2972) ]
= [-13.459 , -6.341]

ANS: C.-13.5< µ1- µ2<-6.3