Question

1. Suppose we are interested in studying flight delays. We know that on average there are 100 delays in a month with a population standard deviation of 238 delays. Suppose we take a sample of 30 flight delays from American airlines only and we find that the average from that sample for that month is 306 delays. We are interested in seeing if the number of flight delays are increasing.
   1. What is the standard error?
   2. What is the margin of error at 90% confidence?
   3. Using my sample of 30, what would be the 90% confidence interval for the population mean?
   4. If I wanted to control my margin of error and set it to 50 at 90% confidence, what sample size would I need to take instead of the 30?
   5. What are the null and alternative hypotheses?
   6. What is the critical value at 90% confidence?
   7. Calculate the test statistic (using the sample of 30 and NOT the answer from part d).
   8. Find the p-value.
   1. What conclusion would be made here at the 90% confidence level?

Answer 1

a) SE =

          = 238/

          = 43.453

b) At 90% confidence interval the critical value is z_0.05 = 1.645

Margin of error = z_0.05 *

                        = 1.645 * 238/

                        = 71.4796

c) The 90% confidence interval for the population mean is

+/- ME

= 306 +/- 71.4796

= 234.5204, 377.4796

d) Margin of error = 50

or, z_0.05 * = 50

or, 1.645 * 238/ = 50

or, n = (1.645 * 238/50)^2

or, n = 62

e) H₀: = 100

   H₁: > 100

f) At 90% confidence interval the critical value is z_0.05 = 1.645

g) The test statistic z = ()/()

                                  = (306 - 100)/(238/)

                                  = 4.74

h) P-value = P(Z > 4.74)

                 = 1 - P(Z < 4.74)

                 = 1 - 1 = 0

i) Since the test statistic value is greater than the critical value (4.74 > 1.645), so we should reject the null hypothesis.

So at 90% confidence level there is sufficient evidence to conclude that the number of flight delays are increasing.