Question

A consumer electronics company is comparing the bright-
ness of two different types of picture tubes for use in its television
sets. Tube type A has mean brightness of 100 and standard devia-
tion of 16, and tube type B has unknown mean brightness, but the
standard deviation is assumed to be identical to that for type A
. A random sample of n = 25 tubes of each type is selected, and
X B − X A is computed. If μ B equals or exceeds μ A , the manufac-
turer would like to adopt type B for use. The observed difference
is x B − x A = 3 . 5 . What decision would you make, and why?

Answer 1

We have

n₁=n₂=25

hypothesis are:

H₀:$\mu_B-\mu_A≥0$

H_A:$\mu_B-\mu_A≤0$

Test statistics will be

z=(($\bar x_B-\bar x_A$)-($\mu_B-\mu_A$ ))/$\sqrt{\sigma^2_B/n_1+\sigma^2_A/n_2}$ =$(3.5-0)/\sqrt{16^2/25+16^2/25}$

z=0.77

Test is left tailed so P-value of the test is P-value=P(z<0.77)=0.7794

Since P-value is large so manager should adopt type B.