Question

7. The average number of planes that arrive at O'Hare Airport (Chicago) per hour was 195 with a historical (known) standard deviation of 13. To increase arrivals, a new air traffic control procedure was implemented. The number of planes that arrived per hour was collected for each of the 30 hours immediately following implementation. Those data are below. 210 193 199 215 200 209215 188 174 1 89 192 176 200 213 179196 2 02 195 202 225 195 181 199 208 197 196 222 199 210 221 (a) Using symbols and not words, state the null and alternate hypotheses for a right-tailed hypothesis test, using the historical average as the null hypothesis about the mean. (b) What is the sample mean? (c) Calculate the z-value test statistic. Round it to 2 decimal places. (d) Calculate the p-value for this hypothesis test. Round it to 4 decimal places. (e) At the 90% confidence level, do you Reject Ho or Fail to Reject Ho? (f) At the 95% confidence level, do you Reject Ho or Fail to Reject Ho? (g) At the 99% confidence level, do you Reject Ho or Fail to Reject H ? (h) Did the new procedure work?

Answer 1

given : n = 30 = 13 Ho : A = 195 H:p> 195 (u is average no of planes )b) Let the guven data be X sample mean = ī= Zri - 0000000 n 30 = 200 test statistics is : T- co/ putting all values we get = 2.106625 – 2.11 13/30 (under HoH = 195) z_200 - 195 p-value fot this test is:P(Z > 2.11) = 0.0174 at 90%, a = 0.1 Since p-value(0.0174) is less than alpha(0.1) hence we have enough evidences to reject Null Hypothesis. f)at 95% a = 0.05 Since p-value(0.0174) is less than alpha(0.05) hence we have enough evidences to reject Null Hypothesis. at 99% a=0.01 Since p-value(0.0174) is greater than alpha(0.01) hence we fail to reject Null Hypothesis.

at 90% and 95% confidence levels we have enough evidence to reject Null Hypothesis that mean on these confidence level the new procedure worked but at 99% it did not worked as we failed to reject Null Hypothesis at 99%