Question

Please be as detailed as possible and provide solution without using to software

Problem 3 sample of 15 measurements was taken and the sample mean was 3.7867 and the sample variance was 0.94265. Assuming that the measurements represent a random sample from a normal population, (24 points): Measurements were recorded for the aeriation time of a process. A random a) Construct a 95% confidence interval for the mean drying time Construct a 95% prediction interval for the drying time for the next trial of the paint. Construct a tolerance interval with a tolerance of 0.99 and 1-0 0.95 b) c)

Answer 1

n = 15, x = sample mean = 3.7867, sample variance = s2 = 0.94265 95% C.1, for the mean drying time : (1.7043, 5.8691) 14 × 0.025, where, to.025,142.1448 95% P.I. for thedaying time for the next trial of the paint : T- t0.025.14 s1/1-, r + to.025141/1(1.6360, 5.9374) (N -11+2.0453 p= 0.99. k2=E(1+p)/2 where, X2,N-,-upper-tail critical ualue = χ0.05,14-23.6848. (1+p)/220.9952.5758 Tolerance interval: (T - k2s, ks)1.8010, 5.7724)