Did the estimated regression equation provide a good fit Sup
Solution
A) ANNUAL MAINTENACE = 10.5279568755945 + (0.95344043969982 * WEAKLY USAGE)
B) SINCE P-VALUE < 0.05..THE TEST IS NOT SIGNIFICANT...
C) WEAKLY USAGE = 30.. T-SCORE FOR DF (10-2) =8 AND ALPHA = 0.05 IS = 2.306004...
standard error of weekly usage = 0.138159709382233.....
confidence interval for weekly usage = 30 + or - ( 0.138159709382233*2.306004)... = [ 29.6814 , 30.3186 ]
now, Annual maintenance when weekly usage = 30.3186 is...= 39.435..
and when weekly usage = 29.6814... annual maintenance = 38.827...
confidence interval for annual maintenance when weekly usage is 30 = [ 38.827 , 39.435 ]...
d) maintenace contract = 3000$ i.e, 30 in hundred of dollars..
30 = a + b*(weekly usage)..
we know a and b..so, usage = (30-a) / b = 20.42293..
mean of weekly usage recored here is = 25.3....
but for a maintenance cost $3000, the ideal weekly usage should be 20.42293..
so, as the mean weekly usage is more than that , it is likely to cost more than $3000..so, I would not recommend purchasing it,
