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.
