Question

The R# Medical Centre undertook a survey to determine the average waiting time of patients in the surgery before they saw a doctor. From 12 responses, the average and standard deviation of the waiting times was found to be 14.1 and 11.84 minutes, respectively. Assuming the information was collected through a random sample and that the waiting time is approximately normally distributed, calculate a 98% confidence interval for the mean waiting time at the medical centre. State the lower bound correct to two decimal places.

Answer 1

soin: here for 12 responses D = 22 8 0 = 0.02 f average waiting time of Standasd deviation for response was found to be 14.18 11.84 minutes ie ž = #8 14.1 min - & 5o = 14.84 min here waiting time is approximately normally distributed here we use t test for one sample - The pivotal quantity is - t = šllo here we have to calculate 981. confidence interval for the mean waiting time at medical centre (1-0) 981. confidence interval is given as a DULE

pl - talq2t Leo/2) = (1-4) 981. PL-tale L x-lo Ltd 12 ) =(1-2J987 510 P/x-stal Leol 7 +5 tal =(1-d98%. ... Ci-a) 981 for the mean waiting time a medical centre is 1 ž - 5 tal xts tol tal=2.7180 here | 14.1 - 11.84 72.718, 14.1 +11.84 X2.718 √12 - J 12 [ 4.81, 23.39 ) (1-0) 981. confidence intervall Here for the mean waiting time at medical centre is. (4.81 , 23.39) e lower bound = 4. 81