Question

The measurements given below were recorded for the drying time, in hours, of a certain brand of latex paint. Assuming that the measurements represent a random sample from a normal population, find a 99% prediction interval for the drying time for the next trial of the paint. 4.5 2.6 3.9 4.2 3.8 2.9 5.7 3.2 5.5 3.6 3.3 5.7 4.4 4.5 3.1 Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. Click here to view page 1 of the table of critical values of the t-distribution. Click here to view page 2 of the table of critical values of the t-distribution. The prediction interval is (Round to two decimal places as needed.) <xo <

Answer 1

5.7 4.5 3.1 The 99% prediction interval for the drying Answer: given data 4.5 2.6 3.9 4.2 3.8 2.9 s.7 312 sis 3.6 3.3 from the given data = 4:06 S = 0.99986 n= 15 given 99%. I d.f= n-1=15-1514 2.927 from t-value 0.005, 14 time for the next trial of the paint as X-te xs/1th a rolet KE I-0.99 = 0.01 к 0.005 t-value +

4.06- 2.97780.999864 namang IAL Lhoc 4.06+29778 15 0.99986 +1 15 4.06-2.9724 0.99986X1.033 LooL 4.06+2.9774 0.99986 X 1.033 4.06- 3.075 L 4.06 + 3.015 0.985 LoL 7.135 0.99 LXL 2.14 no L7.14 : The prediction interval is 10.99 NOTE: Please likeccecce