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

Answer:

For one month

= 229.543, s1 = 14.17, n1 = 7

For seven month

= 143.733 = 144 , s2 = 59.93, n2 = 9

df = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

    = ((14.17)^2/7 + (59.93)^2/9)^2/(((14.17)^2/7)^2/6 + ((59.93)^2/9)^2/8) = 9

At 95% confidence interval the critical value is t* = 2.262

The 95% confidence interval for the difference in population means is

() +/- t* * sqrt(s1^2/n1 + s2^2/n2)

= (229.543 - 143.733) +/- 2.262 * sqrt((14.17)^2/7 + (59.93)^2/9)

= 85.81 +/- 46.783

= 39.027, 132.593

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