"Investigating The Distribution Of Streams Invertebrates"

3559 words - 15 pages

Method1) We investigate the distribution of Benthic invertebrates at the Nant Iago stream, which is near Abergaveng in South Wales;2) We measured a 150 m stretch of the stream, each group had a 10 m stretch meter long of stream;3) A surber sample was used, the size of the holes in the net was 1 mm2, while the size of the quadrat was 0.1 m2 (this size prevents the invertebrates from passing through the holes, but small particles of fine sand could);4) We placed the quadrats so it was flat in an area which was firstly slow flowing and then fast flowing (stony and sandy regions);5) We the disturbed the area by turning the rocks and brushed then lightly ...view middle of the document...

Alternative HypothesisThere will be a higher distribution of chironomus in the sandy substrate than in the stony substrate.v I'm going to use the ÷2 (chi) test, because the sample that I have is smaller than 30 and we don't know the exact distribution.Total number of chironomus in sandy region: 10+4+6+1+5+6+7+4+80+8+5 = 136Total number of chironomus in stony region: 5+3 = 8 Sandy Stony TotalObserved 8 136 144Expected 144 = 72 2 144 = 72 2 ÷2 Test ÷2 = Ó (O-E)2E Note:O = ObservedE = Expected Sandy StonyObserved (O) 136 8Expected (E) 144 = 72 2 144 = 72 2O - E 136 - 72 = 64 8 - 72 = -64(O - E)2 642 = 4096 -642 = 4096(O - E)2 E 4096 = 56.89 72 4096 = 56.89 72Ó(O - E)2 E 56.89 = 56.89 = 113.78 56.89 = 56.89 = 113.78 ÷2 = 113.78 Degrees of freedom = 1, because number of samples = 2.Number of data classes - 1 = Degrees of freedomTherefore in our investigation the degree of freedom equals 1, as we have 2 data classes (either sandy or stony).The critical ÷2 value from table is 3.84. That is more than 95% sure that the null hypothesis is wrong. For this we used 1 degree of freedom to measure the value of the ÷2 and also the confidence level. Therefore by looking at the degree of frequency table and at my results of the ÷2 test, which was 113.78, I am more than 95% and even 99.9% certain that the null hypothesis is wrong and that the alternative hypothesis is correct.Therefore I can validate with plenty of confidence the alternative hypothesis, which was that there are higher distributions of chironomus in the sandy substrate than in the stony substrate.Mean calculations Stony region: Number of chironomus = 5+0+0+0+0+0+0+0+0+0+0+0+0+3 = (0.1 m2 quadrat) 14 = 0.6 per 0.1 m2 quadrat Detritus rating = 3+3+2+2+2+3+2+3+3+2+1+2+2+2 =(1 to4) 14 = 2.3 Flow rate =(ms-1)0.25+0.38+0.37+0.50+0.49+0.51+0.25+0.60+0.30+0.68+0.24+0.22+0.23+0.52=14= 0.4 ms-1 pH = 8.0+8.1+8.0+8.0+8.0+8.1+7.7+8.1+8.1+8.0+8.0+8.1+8.0 =13= 8.0 Dissolved Oxygen =(% Saturation)101+102+99+101+103+105+99+103+103+104+101+103+103+102.1 =14= 102.1 % Water depth =(m)0.17+0.12+0.15+0.19+0.21+0.16+0.09+0.12+0.19+0.17+0.19+0.15+0.20+0.18= 14= 0.2 m Temperature =(0C)12.0+11.9+12.1+12.2+12.0+12.0+12.0+11.9+12.1+13.0+12.1+12.2 = 12= 12.1 0C Total dissolved solids = (Parts per million)65+65+66+66+67+67+59+66+66+60+66+63= 12= 64.7 parts per million Sandy substrate: Number of chironomus =...

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