Question

A computer system administrator notices that computers running a particular operating system seem to freeze up more often as the installation of the operating system ages. She measures the time (in minutes) before freeze-up for seven computers one month after installation, and for nine computers seven months after installation. The results are as follows:

One month after install:      207.4    233.1    215.9    235.1    225.6    244.4    245.3
Seven months after install: 84.3      53.2      127.3    201.3    174.2    246.2    149.4    156.4    104.3

Let μXμX represent the population mean for the first month and let μYμY represent the population mean for the seventh month. Find a 95% confidence interval for the difference μX−μYμX−μY. Round down the degrees of freedom to the nearest integer and round the answers to three decimal places.

Answer 1

Solution Giver that: for one Month Mean . 807.47 233:1421597'.---+2453 7 7 - 160648 Telli = 229.5431 Standard deviation (01) . (2014-029543)*- ./ +(245.3-229.543) [C1=16:17] ni = 7 Peven months after Ug = 84.3+5312+12 7.3720113 te----+1043 J 7-1 . 199516 Ilo = 143.981 om borá boriste Standard deviation (2) 1843-143.807 ---#104.3-1987 - Q-

192=59,67 ng = 9 At 95% confidence level X=1-9506 = 1-0.95 = 0.055 4/2=0.045 dr. Certain that niel - (14:17 + 59,67 gr (1917%open df = 9] tary Of = 60025,9 = 3,262 (From t table] = Hi-telax lay game = (229.543-143.96)2.26214489 59677 = 89.58 46.594 -|(381986, 132,174)] V 7