Sunday, October 6, 2019
Production of EPA by the Diatom Coursework Example | Topics and Well Written Essays - 1500 words
Production of EPA by the Diatom - Coursework Example The experiment will use the diatom, Nitzschialaevis. Treatments to be used in experiment PI metal solution (Cepà ¡k, PÃ
â¢ibyl, Kohoutkovà ¡, & KaÃ
¡tà ¡nek, 2014). à The source of variation in experiment one is the treatment (PI solutions) used in the experiment. A PI solution of 4.5m/L and 3.5m/L used in the experiment will yield different yields in EPA. The treatment used in the experiment will account for the variation in the experiment. à From table 1.1 EPA 13.5ml/l had the highest EPA yield (mean = 219.5) compared to EPA 4.5 (mean=217.2). It implies that PI of the concentration of 13.5ml/l had a higher EPA yield. There was a big disparity yield of PI EPA 13.5ml\l (standard deviation =5.396) compared to EPA 4.5ml/l (standard deviation =3.821) this implies that yield of 13.5 PI was less spread compared to yield in 4.5 PI, which was more spread. It could be assumed to the way the experiment was handled; the time was taken to record the yield in the two experiments. The data of the two experiments were the same, most of the data points were to the left of the mean (skewness, -0.238, and 0.0377). However, the yield of 13.5ml/l was more skewed compared to the yield of 4.5ml/l. à Data of the treatment 4.5ml/l was more reliable to make inferences and conclusions (cv =1.759), compared to data of treatment 13.5ml/l (cv=2.464) which was less reliable to make a justification for the claim. à From the summary statistics, the yield of the two treatments was different this implies that level of PI concentration had a different effect on the diatom Nitzschialaevis (UTEX 2047). Metal PI was therefore important since different concentration yield different results. à Non-parametric test to be used is a chi-square while a parametric test to be used is a t-test. A chi-square test tests for the independence of the samples while a t-test test for the difference on the means of the two samples. Ã
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