We considered the differences between the temperature readings in january 1 of 1968 and 2008 at 51 locations in the continental us in exercise 5.19. the mean and standard deviation of the reported differences are 1.1 degrees and 4.9 degrees respectively.

Actually this problem has the following three questions:

(a) Calculate a 90% confidence interval for the average difference between the temperature measurements between 1968 and 2008.

(b) Interpret this interval in context.

(c) Does the confidence interval provide convincing evidence that the temperature was higher in 2008 than in 1968 in the continental US? Explain.

Solutions:

(a) We solve for the confidence interval using the formula:

confidence interval = x ± z s / sqrt(n)

where x is the mean value = 1.1, z is taken from the standard normal tables at 90% confidence level, s is standard deviation = 4.9, and n is number of sample locations = 51

From the tables at 90% confidence interval, the z score is:

z = 1.645

confidence interval = 1.1 ± [1.645 * 4.9 / sqrt(51)]

confidence interval = 1.1 ± 1.129

confidence interval = -0.029, 2.229

lower bound: -0.03 degrees

upper bound: 2.23 degrees

(b) There is a 90% chance that the difference in temperatures in a city from year to year will be between the lower bound and upper bound

(c) No, because the confidence interval contains 0. There is also a great change that it fall at 0 degrees hence they could just be the same.

inequality:

620-37.4p ≥ 450

the minimum amount of packages needed to be removed from the elevator is 5.

step-by-step explanation:

step-by-step explanation: