Question

1. (50%) Suppose that the manager of a car taxi fleet in a large city wishes to reevaluate the maintenance contract on its vehicles. A major part of the analysis considers the "wear and tear on the vehicles, i.e., the daily usage as represented by miles logged per taxi per day. Upon examination of his contract, the manager feels that if taxis are averaging more than 70 miles per day, he wants to renegotiate or change the contract. A random sample of 16 taxis is selected at random from the feet. The following table records their mileage for a particular day 107.1 1210 71.2 76. 1 95.7 92.8 74.8 92.1 94.4 42.5 82.3 56.5 74.6 91.7 63.7 62.9 The manager knows you are studying statistics and asks you to completely analyze these data At the 0.05 level of significance, what would you conclude about average daily mileage? Are the taxis averaging more than 70 miles per day? State managerial conclusion. What would Jon recommend to the manager? (18%)

Answer 1

Solution 1: Here the manager only intuested ko wntinue the contract only 4 tari are audliging miles pesday is less than 70 miles olhéiciise herunger will sinoectate or change the the contract. then let theet cur claim for the test i emerge nunnig peimilesen a day is less than to miles aquunst sellernattie eleito greater thin to miles en lag. . Hypothesis Null hypothesis H. U = 70 egallstelle altre este Hp : M < 70.2=0.05) D NOTE traat here hypothesized mean , Mo = 70 and sample of 16-18 güun, as follows: 107.1 121.0 71.2 76.1 95.7 92.8 74.5 92.1 9404 42.5 82.3 56.5 7406 91.7 6307 6209 let x be samples and t be the mean and s be the standard deuiation of x and spibe the sample size. = 107.1 +121 + 71.2 + 16.1+ 95.7+92.8+"74-8 +92.1794048 42.5+ 82.37 56.577406 + 91.7+63.7+62.9 16 T = 1299.4 16 8 = 81.2125

= 2.24933 Ita2.249 Rejection Regums Based on the information provided, the Eleppige cance level, d=0.05 and the evitical ualue for a lefftciled (ie th: M270) test the ualue is te= -1.753. And rejectům regies for the lept toiled Test is equal t: 13-1.753) Decision 1st method Since it is observed that the calculated turlue, t = 2.249. 80 then the t= 0.2499 te= -1.753. there fore due feuled to reject the nuli hypothesis Ho. 204 roethod By pualue approaeb method. the pCtc 2.249). at d = 0,05 iš p= 0.98. and then P = 0.98>2=0.05 ,80 ue feiled to rgeet nul hypothesů.

Conclusün Since we observed that the test slatésűer is guiled to leef the null hypothesis that means lered the alternative hypothesis (Ha). une leih unclude the ende cunning of tani's is more than Fomiles per decy. thue ů slyfterent endence for this. . 80 eve ucommend to the manager to change the contract Because evuerege running of lari is more than to mēles en aday.