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 16.73. The cell means in milligrams for the experiment are given below.

Temperature 25 C 30 C Row Mean Salinity 10% 45.85 67.65 56.75 Salinity 25% 57.71 88.62|73.16 Salinity 40% 71.06 92.29 81.67 Column Mean 58.21 82.85

Answer 1

Number teplications - x = 3 Total sample standard deviation - 16.73 overall Mean - Y = 58.21 + 8 2.85 - 70.53 2 SSAB ** 3:3 ( Yps - Y; - Y> + 7)? 2 = 1 J=1 2 3 (45.85 - 56.75 - 58.21 + 70.53) + 167.65 -56.75 – 82.85 + 70.53) + (57.71 – 73.16 –58.21 + 70-53)?+ (88.62 – 73.16-82.85+ 70-53]$+ (71.06 - 81.67 -58.21+ 70-53)2 + (92.99 - 81•67- 82.85 + 70:53) 2 3 [2.0 2.0164+ 2.0164 +9.7969 + 9.8596 +2.924 1+1 82.8402 df Ca-1) (6-1) = (3-1) (2-1) IAB - 2 So MSAB- SSAB 82.8402 ISAB -141.4201 2

we are given that is 2. ( 16.73)? N-1 2 Ź ( Yijk - 9) = overall Then sample Standard deviation S = 16.73 & (Tisk- 7² S = AS N= 2x3x3 = 18 obseruations, 18 (0)2 = 17 (16.73) Ź Yisk 18(70.53) +17(16.73) 94298.8355 Then sluitwin = (1-7,,) (1 x + (Ts) 27 (155k) 2 - h Y; JK 2 -> 3 کے د 2 = 1 = 1 kal

(94298.8355) - 3( 45.85' +67.65+57.71 +88.622 + 71.062 +92-2014) (94298.8355) - (9428 8.9836) 9.8519 ff mitan = (x-1) (a) (b) (3-1) (3) (2) 12 Smithin 9.8519 Thus, Ms within It within 12 MSAB So = W 41.420) X12 9.8519 ms mitan : 50.4513

So, test statistic = 50.4513