Question

The university data center has two main computers. The center wants to examine whether computer 1 is receiving tasks that require processing times comparable to those of computer 2. A random sample of 9 processing times from computer I showed a mean of 52 seconds with a standard deviation of 19 seconds, while a random sample of 14 processing times from computer 2 (chosen independently of those for computer 1) showed a mean of 56 seconds with a standard deviation of 17 seconds. Assume that the populations of processing times are normally distributed for each of the two computers and that the variances are equal. Construct a 99% confidence interval for the difference Hy between the mean processing time of computer 1. Hy, and the mean processing time of computer 2. z. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your responses to at least two decimal places. What is the lower limit of the 99% confidence interval? What is the upper limit of the 99% confidence interval? $

Answer 1

Given :  

1
= 52 ,
S1
= 19 , n₁ = 9 and  $//img.homeworklib.com/questions/9971bba0-df8b-11ea-841a-c53dc4db816b.png?x-oss-process=image/resize,w_560$ = 56,
S2
= 17 ,  n₂ = 14

degrees of freedom (df) = n₁+ n₂ - 2 = 9+14-2 = 21

confidence level (c) = 0.99 , so α = 1- 0.99 = 0.01

Therefore critical value t( 0.01, 21 ) = 2.831 ---- ( from t distribution table )

Sp =

1) si(ni 1) + s(n2 ni + n2 - 2

= $\sqrt{\frac{19^{2}(9-1)+17^{2}(14-1)}{9+14-2}}$

Sp = 17.7884

SE = $S_p*\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

SE =$17.7884*\sqrt{\frac{1}{9}+\frac{1}{14}}$

SE = 7.6001

Margin of error (E) = t*SE = 2.831*7.6001

E = 21.5185

(
1
- $1596234160797_blob.png$ ) = 52 - 56 = -4

Lower bound = (
1
- $1596234160797_blob.png$ ) - E = -4 - 21.5185

Lower bound = -25.5185

Upper bound = (
1
- $1596234160803_blob.png$ ) + E =-4 + 21.5185

Upper bound = 17.5185