Question

Question 18 3 pts An SRS of 300 flights of a large airline (called this Airline 1) showed that 246 were on time. An SRS of 300 flights of another large airline (called this Airline 2) showed 231 were on time. Let Pi and pa be the proportions of all flights that are on time for Airlines 1 and 2, respectively. Step 1: Find the test statistic to test Hop - P = 0. Hz:P1 - P2 +0. Give your answer to 2 decimal places. Question 19 2 pts Continued from the previous question: Step 2: Ignore your answer in Step 1. Suppose the test statistic is 1.65. Find the P-value of the test. Give your answer to 3 decimal places. Question 20 2 pts Continued from the previous question: Step 3: Ignore answer in Steps 1 and 2. Suppose the P-value of the test is 0.075. We conclude ata= that there is no significant difference in the proportion of on-time flights between the two airlines since the P-value is greater than a Othere is significant difference in the proportion of on-time flights between the two airlines since the P-value is greater than a Othere is no significant difference in the proportion of on-time flights between the two airlines since the P-value is less than 2a. there is significant difference in the proportion of on-time flights between the two airlines since the P-value is less than 20.

Answer 1

Solution:(18 Data given mi = 300 flights of a large airline (Airline 1) 246_fights were on time, 12 = 300 flights of another lange airline (Airline 2) X2 = 231 flights were on time. Xia Sample propor tion of Airline 1 pi= Xi 246 0.82 ni 300 P=0.82 Sample proportion of Airline 2 P2 = X2 231 300 =0.77 12 P = 0.77 Hypothesis Ho: P.-P2 = 0 Ha PR-P2 0 (Two-tailed test) where be the population that are on time for Airline 1 and 2 . Pl and P2 proportions of all flights

test statistic, under Ho 1 z = Р. P2 후한 where PE X1 + X2 nitha 246+231 300+300 477 -0.795 600 q = 1-2 요 1-0.795 q = 0.205 = z = 0.82 – 0.77 0-795*0-2057 300 300 0.05 0.05 D-162975% 2 300 0.0 0 10865 0.05 0.032962 = 1.517 Z=1.517 ZE).52 Rounded 2 decimal places.

Solution (19) Ignoring answer in last step Suppose the test statistic iš 1.65 , the p-value of the test. Since the test is two tailed, fina P-value = P(Z4-1.65) + P(Z > 1.65) = 0.04947 +0.04947 P-value = 06 09894 IP-value 0.099 (Rounded 3 decimal places) [using normal tables 0004947 0104947 -1.65 +1.65 Solution (20) Ignoring answers in Step 1 and 2 Suppose P Value = 0.075 Significance level = 5% de 0.05 conclusion: There is no significant difference in the proportion of on time flights between the two airlines, since, p-value is greater than a.