let x = 78.3
y = 87.2
σₓ = 5.6 and σ₍y₎ = 6.3
number of pieces n = 50
z = z-score value =1.96 in this case
A 100(1-α)%
confidence interval for the difference in large sample means is:
(x – y) ± z x √( σₓ² / n+ σ₍y₎²/n)
by putting the values,
=(78.3-87.2) ± 1.96 x √(5.6² /50 + 6.3² /50)
= -8.9 ± 2.34
= [-11.24, -6.56]
