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A2 Biology | Post your doubts here

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wht do u mean by :in a t-test, "comparing 2 means obtained from 2 samples in a population"
wht kind of 2 samples....give me an xmple sombody>???!!
 
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wht do u mean by :in a t-test, "comparing 2 means obtained from 2 samples in a population"
wht kind of 2 samples....give me an xmple sombody>???!!
its there in the last page of our book---(mary jhon)
eg u got the mean index finger lenght of ur class...now u have to comapre this with the 4 to 6 other classes in ur school...
al the samples should be from the same population...here school represents the same population of the samples(classes)...:)
 
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its there in the last page of our book---(mary jhon)
eg u got the mean index finger lenght of ur class...now u have to comapre this with the 4 to 6 other classes in ur school...
al the samples should be from the same population...here school represents the same population of the samples(classes)...:)
ohhhh right....how foolish of me??
 
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fo example u measured reaction times of ten people b4 drinking alcohol and after drinking alcohol....so u calculate mean time b4 and after and chek using t test whether difference in mean times is just due to chance or significant
 
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In this experiment a plant has two types of leaves; sun and shade leaves, due to the fact that each leaf grows in different light intensities; light and dark(0 intensity).

A good deal of sampling is used. The mean internode length is found by measuring the internode length between two leaves; for a number of such sets of two leaves sampled from different parts of the plant, in a systematic way, e.g. measuring from every 3rd internode, starting from bottom; and taking the average.

The mean surface area of leaves is calculated in a similar way, by measuring the total surface area (top and bottom side) of a particular no. of leaves and calculating the average. The leaves may be chosen similar to the above method, 3rd leave from bottom.

The dry mass of leaves is measured by drying the leaves in an oven and then measuring mass using mass balance. Sampling is done as before, and the average mass taken.

The rate of water loss is measured using a potometer. The change in volume of water in the capillary tube is measured for every time interval, usually one hour, and the rate of water loss calculated. Several time intervals of 1 hour are sampled and the average taken. The volume may be measured from change in length of water coloumn and the area of capillary tube.
 
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Tomorrow question that will come out is about rate of photosynthesis and chi square. So, dont worry, be happy. From God.
 
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Paper 5 Help

If the degree of freedom is 38 and the table at 0.05 only shows 40 & 30 ... should i pick 40 or 30
 
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Suppose you have 4 observed results for 4 expected results; e.g. in a dihybrid genetic cross. First, subtract each expected result from the corresponding observed result: O - E. Then, square each of the numbers: ((O - E)^2). Then, divide each of these numbers by the corresponding expected results: ((O - E)^2) / E. Then, add all the numbers up. Calculate the degrees of freedom of your results, which is (number of sets of results or data - 1), here 4-1=3. Look up your number in a table of chi-square values, at the corresponding degrees of freedom you found, and at probability of 0.05. If your number is smaller than the value in the table, then there is a probability of greater than 0.05, that the difference between your results is due to chance; such differences occur in more than 5 of every 100 experiments. Therefore you are quite certain that this difference is certainly due to chance. The higher the probability, the more certain the difference is due to chance.

P.S. Biology book- Mary Jones- pg-234.
 
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what are the critical values and how do we know the difference bw 2 means is significant or not??????????
 
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in nov 2010 paper 51 ms says that ''both the results are greater than critical value so they are significant''
can some1 explain this to me
are we comparing the critical value p<0.05 with the 2 values of t(12.6 and 8.88)......or we see the corresponding value of degree of freedom in the given table?and there is no value of 58 in the table so which value to compare to?
thanx in advance
 
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''in nov 2010 paper 51 ms says that ''both the results are greater than critical value so they are significant''
can some1 explain this to me
are we comparing the critical value p<0.05 with the 2 values of t(12.6 and 8.88)......or we see the corresponding value of degree of freedom in the given table?and there is no value of 58 in the table so which value to compare to?
thanx in advance ''

The probability of 0.05 shows that there is 5 percent chance that the difference between the two means is due to chance so the critical value as shown in the the critical value shown in the table is quite less then our both values therefore our probablity is quite less the 0.05 ... due tot his the difference is significant
 
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''in nov 2010 paper 51 ms says that ''both the results are greater than critical value so they are significant''
can some1 explain this to me
are we comparing the critical value p<0.05 with the 2 values of t(12.6 and 8.88)......or we see the corresponding value of degree of freedom in the given table?and there is no value of 58 in the table so which value to compare to?
thanx in advance ''

The probability of 0.05 shows that there is 5 percent chance that the difference between the two means is due to chance so the critical value as shown in the the critical value shown in the table is quite less then our both values therefore our probablity is quite less the 0.05 ... due tot his the difference is significant
The two values of t tests are larger than the critical value at p 0.05 so the differnce is significant and not due to chance :D
 
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