Question

A mobile computer network consists of a number of computers (called nodes) that communicate with each other while moving throughout a region. A node that is out of transmission range of the other nodes, so that it is unable to communicate, is said to be partitioned. In studies carried out at the Colorado School of Mines, S. Kurkowski found that in a network containing 186 nodes in which destinations were chosen at random from a uniform distribution, 19 nodes were partitioned. J. Norman found that in a network containing 167 nodes whose movements mimicked those on a college campus, 30 nodes were partitioned. Let p_X represent the population proportion of partitioned nodes in the network that mimicked those on a college campus and let p_Y represent the population proportion of partitioned nodes in the network whose destinations were chosen at random. Find a 99% confidence interval for the difference p_X – p_Y. Round the answers to four decimal places.

The 99% confidence interval is ( , ).

Explanation from a similar problem with different input values:

A mobile computer network consists of a number of computers (called nodes) that communicate with each other while moving throughout a region. A node that is out of transmission range of the other nodes, so that it is unable to communicate, is said to be partitioned. In studies carried out at the Colorado School of Mines, S. Kurkows ki found that in a network containing 187 nodes in which destinations were chosen at random from a uniform distribution, 19 nodes were partitioned. J. Norman found that in a network containing 165 nodes whose movements mimicked those on a college campus, 30 nodes were partitioned. Let pxrepresent the population proportion of partitioned nodes in the network that mimicked those on a college campus and let pyrepresent the population proportion of partitioned nodes in the network whose destinations were chosen at random. Find a 99% confidence interval for the difference PX-Py. Round the answers to four decimal places. The 99% confidence interval is ( Explanation X 30, nx 165, px = (30 + 1)/(165 2) 0.185629 Y= 19, ny 187, py = (19 1)/ (187 2) 010582, z0.005 = 2.58. The confidence interval is 0.185629(1-0.185629) 0.10582(1-0.10582) + 0.185629-0.10582 ± 2.58, 187+2 165+2 (-0.0169, 0.1765). or

Answer 1

Sample 1:

n1 = 167

x1 = 30

p̂1 = (x1+1)/(n1+2) = 0.1834

Sample 2:

n2 = 186

x2 = 19

p̂2 = (x2+1)/(n2+2) = 0.1064

α = 0.01

99% confidence interval for the difference p_X – p_Y :

At α = 0.01, two tailed critical value, z_c = NORM.S.INV(0.01/ 2) = 2.576

(1-P (1-p) (pi -P2)te + ni n2 (0.1834- 0.1064)土2.576。,/0.1834* 0.8166 167 2 0.1064 0.8936 + 186 2 -0.0191, 0.1732)