Question

4. According to 32 randomly selected customers at a Burger King restaurant, we found the service time for the drive-through has a mean of 181.3sec with a standard deviation of 82.2sec. According to another 45 randomly selected customers at a McDonald's restaurant in the same area, we found the service time for the drive-through has a mean of 167.8sec with a standard deviation of 60.6sec. a) Do we have enough evidence to conclude that the average service time at McDonald is more than 150 sec, at 5% level of significance? b) Test if the average service time at McDonald's is less than Burger King at 2% level of significance. (Use both critical value method and p-value method)

Answer 1

solution n = 32X = 181-3,0 = 82.2₂ = 45 X = 167.8, O2 = 60.6 Since no, we apply z-test. Hypothesis for Mc Donald Hol=150 (Average Service time at Mc Donald is equal to 100) Hi MZ 150 Average service time is more than 150) Test - statistic (for McDonald Alone) Z = X-M = 167.8-150 7.8 olm 60.6 9.0337 Z-1.97 Critical Region = Zx at 5% =Zoros = 1.96 (Table] Since Z calculated =1.97 is greater than 2x=196, we reject to and we infer that avereage service time at Mc Donald is more than iso.

teret of (6 To Test if werage Service time at Me Donald's is less than Burger King at 27 Sigorta cante Hypother's Ho : Average Service time at Me Donald 5 ös not Cantly less than Burger king Hi Average service time at Mc Donald's is significantly less than Burger King. Test statistic z = x - xa 쪽 za +8613--167-8 82.22 32 + 60.6² 45 z = 13.5 13.5 5292.758 211.15 + 81.Gof 2 = 13.5 = 0.789 17:1 220-789

from normal mal) if =0.02 statistics Table 2=2% = Zo.o2 2:3 3 26 ( Decision Since Z - Calculated = 0.789 less than Zx=21326, we Cannot reject Ho and we conclude that the average Service time at Mc Donalds is not significantly less than Burger King.