Question

Investigators working on commericially cultivating shrimp are interested in determining the best conditions in which for them to grow. They are interested in knowing what impact water temperature (either 25 C, 30 C) and the salinity level (10 % 25 % and 40 %) have on the growth of shrimp. Three replicates were used for each of the 6 possible treatment combinations. Post-larval shrimp were placed in each of the 18 tanks and fed the same amount of food. After one month the overall weight of all the shrimp in a tank was measured and compared to how much they weighed previously. The total sample standard deviation was found to be 18.96. The cell means in milligrams for the experiment are given below. Temperature 25 C 30 C Row Mean Salinity 10% 50.24 81.96 71.10 Selinity 25% 79.33 109.70 4.76 Salinity 40% 98.37 114.32 101.34 Column Mean 76.14 101.99 The investigators want to know whether there is an interaction between temperature and salinity. Which of the following would be the correct null hypothesis? My for all (0) - for all :) H;= w; for all i (03)y for some J [1 pts PSA Tries 0/1 The investigators want to know whether there is an interaction between temperature and salinity. Which of the following would be the correct alternative hypothesis? (cy - for some Hwy for all W (0) - for all (2) for some [1 pis1 Trics / What is the grand mean? [1 pers! Subre Are Tries 0/3 Calculate the degrees of freedom Salinity Temperature Subre A Tries 0/3 Interaction Emor [1 pt) Calculate MSAB [2 posil Rabat RISA Tries 0/3 Calculate the test statistic [2 pts. Pub Tries 0/3 Do we reject the null hypothesis of their being an interaction at significance level a = 0.057 No, we fail to reject the null hypothesis Yes, reject the null hypothesis [1 ptis)]

Answer 1

Row Mean T6.14 101.99 an interaction betwee to know whether there is 9 Wountions Sol five data kmperature 25°C 30°C Sabinily 10*)60-24 $4.96 7110 Salinity 25/79-83 104.-10 94.76 Salinity 40% 88-37 114.32 101:34 Column Mean (xp); -o for at j ( Null hypothesis) Option B is Correct 2) The investigators want to know whether there is on interaction between temperature 8 Salinity Cha) (Altemate hypothesis) Option D is correct 3) firand Mean - Row Mean (Total) (xp); to for Some ij 71.10+ · 76+101.34 3 267.2 3 89.06

Degree of freedom for PAGE 2 Salinity 3-1 2 2-1 — 12 2 ss 3 AB ij =3 60.24-71-10 84.0612 Temperature Pnteraction = (3-1)(2-1) Error 3X2 (3-1) SE ( Gigo - Tigo - Yojo - Joco) [(60. 76.14 + 89.06)+(81.96 (81.96 - 71.10 - 101.99 +84-06)2+( 79-83 - 94.76 – 76:14 + +(109.10 –9 4.76 – 101.99 +89-06)2+(88.37–101. 34 76-14 + 89-06) ( 114-32 – 101.34 –101.99 + 89-06) 3[(3:06) + (-2.07)+(-2-01)+(2.0172+60.05) ? + (0-05) -31 0.0025 .. (1053) 2 2 4.2436 + 4.2849 + 4.0401 + 4.0401 +0.0025 +0.0025 : ] SS AB 49.8411 SSAB 49.8411 49.8411 where 3-1 MS AB na1 Df AB 49-8411 2 24.92055

SSA 3x2 PAGE 3 2 - 3.X2 (11:10 - 89.06)2 + ( 94.16 = 89.06)2+ (101.34.-89.06)?] [(-13-2) 2 + (*-+)2+(13:24) 322. 5616 +32.49 + 150.7984] 6( 505.85) 6 C SSA 3035:1 SSB 3x3 (176-14 -89.06) ? + (101.99 -89.06) 89.06)2] aſ (-12.92)2 + (12-43)* ac ( 334 166. 9264 + 167. 1849 49 9 3007. 0017 SST - Total Sample standard deviation Das 12+ 2+2+1=17 (18.967** 17 359. 4816 * * 6111. 1872

PAGE 4 SSE SST SSA SSB.- SSAB 6111:1872 3035:1 300 F. 0017 49.8411 So, SSE 19.2444 Test statistics MS(AB) . MSE of 12.46027 19.2444 12 12.46027 oll 1-6037 Il To 1698

level of Significance (X). 0.054 PAGE 5 Foost, 2,12 from Fritical Value Calculator Then we get f 3.612 critical fo.ost - 3.672 s fo fobservations 12,12 So we Reject the Noll hypothesis, RATE MY WORK THANK YOU.....IF ANY ERRORS I WILL DO AGAIN BUT DON'T DISLIKE OK...
please do like it will take 2 hours to calculate in easy method those students whom are understand easily.thank u