Answer to Question #192975 in Statistics and Probability for sagun poudel

Question

A company manufactures two types of bulbs, (A and B). The manager of the company tests a
random sample of 50 bulbs of type A and 60 bulbs of type B and obtains the following information:
Mean Life
(in hours)
Standard Deviation
(in hours)
Type A 1300 50
Type B 1200 60
Obtain 99% confidence interval for the difference of the average life of the two types of
bulbs.(Given that Z0.005 = 2.58)

Answer 1

Solution:

Given, $\bar{X}_{A}=1300,\bar{X}_{B}=1200,\sigma_A=50, \sigma_B=60,n_A=50,n_B=60$

Now, z-value of 99% confidence interval=2.58

Then, 99% confidence interval$=(\bar{X}_{A}-\bar{X}_{B}\pm z\sqrt{\dfrac{\sigma^2_A}{n_A}+\dfrac{\sigma^2_B}{n_B}})$

$=(1300-1200\pm 2.58\sqrt{\dfrac{50^2}{50}+\dfrac{60^2}{60}}) \\=(100\pm 2.58\sqrt{110}) \\=(72.941,127.059)$