Question

question 3 and 4 with work please

cient on the assumpoon that phone surveys can be completed with The estate hypotheses in this station are 1 1.l .co n ces of making a Type I error 2 0 vs H an average time of 20 minutes or less > 20 more time is required a A The company changes the customer the premium rate when they should B. The company does not change the customer the premium when they should not c. The company does not change the customer the premium rate when they should D. The company charges the customer the premium rate when they should not 2. let the consequences of marang a Type 11 error A The company does not change the customer the premiumbe when they should not 8. The company changes the customer the premium when they should se. The company does not change the customer the premium rate when they should D. The company change the customer the premium rate when they should not To monitor the bang will change the per the manager is going to take a random sample of 20 Surveys each and calculate the average survey time in the sample They make a decision rule f orat o Assume the popuso standard deviation is 3 minutes 2. What is the pro t othe company will make a type i morsing this decision rule Round your answer to four decimal places 20.5 they 4. Ung this decision w 2075 Round your ith power of the best the actual meanime locomote the survey 2075m four decimal aces ? Thats what is the pray they wirect He when the actual arage meis

Answer 1

(3)

Given : u = 20,0= 3 and n= 20 By Central Limit Theorem, ă follows a Normal distribution with mean = u = 20 standard error = 0/Vn = (3/720) = 0.6708 Then z=(x-mean)/SD = (x-20)/0.6708 follows standard Normal.

When 7=20.5, z=(20.5-20)/0.6708 = 0.75 P(Z<0.75) =0.7734 [using Normal table or Excel : NORM.S.DIST(0.75, TRUE)] Pã> 20.5 ) = 1-Pā< 20.5) = 1-P(Z <0.75 ) = 1-0.7734 = 0.2266

(4)

Given : u = 20.75, 0= 3 and n= 20 By Central Limit Theorem, & follows a Normal distribution with mean = u = 20.75 standard error = o/Vn = (3/120) = 0.6708 Then z=(x-mean)/SD = (x-20.75)/0.6708 follows standard Normal.

When 7=20.5, z=(20.5-20.75)/0.6708 = -0.37 P(Z<-0.37) =0.3557 [using Normal table or Excel : NORM.S.DIST(-0.37, TRUE)] Põ > 20.5 ) = 1-P& < 20.5) - 1-P(Z <-0.37 ) = 1-0.3557 = 0.6443