Question

An experiment was conducted to compare the alcohol content of soy sauce on two different production lines. Production was monitored eight times a day. The data are shown below. Assume both populations are normal. It is suspected that production line 1 is not producing as consistently as production line 2 in terms of alcohol content Test the hypothesis that 1 = 02 against the alternative that 67 762. Use a P-value. Production line 1: Production line 2: 0.47 0.39 0.41 0.53 0.39 0.49 0.51 0.53 0.39 0.37 0.39 0.41 0.38 0.39 0.39 0.39 Calculate the test statistic. f= (Type an integer or a decimal. Round to two decimal places as needed.)

Answer 1

172 Solincon Let xi Production line 1 Y; production line 2 we have X2 Y? 0152) X 10.47 10.39 0.4) lo 53 0:39 10.49 0-5) 10.53 Y 0 39 0:37 0-39 041 0 38 0:39 0:39 0:39 02209 0152) 0 1681 0.2809 0.1521 0.240) 260) 0.1369 0152 01681 6:1444 0152) 01521 0.152) 02809 Sum=3072 3211 13755 1.211 Nowe Sample kahance to production line p L, 52 = Exi2.4(Exi)? -1 h +755 ت 2009. f 3.722 8-1 528 0.0036 for 2 production line 5.2= (E7,2 ({4.) 2) n-) CS sdanned with CamScanner

522 - 1 Il 1:21- (3:11)2 (3:1132) 2 S2 0.000141 * Now, we have, 52 -0.0036 522= 0.000141 h2=8 Since & is not given, we take ni = 8 x=0.05 The hypothesis is, Ho: 0=62 Hoo €62. The test Statistic is, F = = 25.5319 S2 S22 = 0.0036 0.000141 le F= 25.5319 15. Now, the p-value alue = 0.000 17975 Here p-value < d (0.05). Hence we reject Ho. Therekse it is concluded that there is enough exidence to claim that to the production line 1 not produung consislantly as production line 2. tale accept that JE62. 15 CS scanned with CamScanner