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*Biology Paper 5 tips*

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those are for immobilised enzymes
i got others for immobilized algae
  • high biomass loading capacity
  • high transparency
  • absence of toxicity to the algal balls
  • suitable for simple immobilization procedures
  • optimum diffusion of nutrients and products
  • resistance to abrasion
  • maximum surface-to-area volume
  • mechanical stability
  • sterilizable
i guess these are enough



Can you explain these points?

· high biomass loading capacity
· absence of toxicity to the algal balls
· suitable for simple immobilization procedures
· mechanical stability
 
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Aoa wr wb!
I wanted to know when is it appropriate to use the chi squared test, and when do we have to use the t-test?
If someone could please elaborate with examples?
JazakAllah khair :)

We use chi squared to test the possibility of our null hypothesis being correct and to see if there is a difference between observed and expected value ...it is used in discrete data values

But
T-test is used in continuous data , It is used to tell you that whether the means of two sets of values ,(each following a normal distribution ) are significantly different from one another ....... In simple words it is used when there are two sets of data present in the question to compare .

If i am wrong please correct me ..!! :)
 
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(1)initially plot the graph.no.of beans per pod in x axis and frequency in y axis.The histogram is usually drawn as a bar graph.Make sure that u select a suitable scale.
i dnt think the next calculation is a big deal for you.

(2)fx is the product of frequency and the no.of beans per pod
sigmafx is the sum of fx values
n is the sum of the frequencies

n = 100
∑fx = 542




(3)substitute the calculated values(sigmafx ,n)in the formula given



the answer is 5.42.


(4)the standard deviation is given as 1.15.
the standard deviation is the spread of the data around the mean.
the mean that we calculated is 5.42
the spread for this data is 5.42 plus or minus 1.15

(5)standard error can be calculated by substituting the s and root n in the equation given

(b) not all the ovules may have been fertilised;
exposed to different environment/example of an environmental difference
and seeds not growing/developing;

hope you understood:unsure::)
 
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(1)initially plot the graph.no.of beans per pod in x axis and frequency in y axis.The histogram is usually drawn as a bar graph.Make sure that u select a suitable scale.
i dnt think the next calculation is a big deal for you.

(2)fx is the product of frequency and the no.of beans per pod
sigmafx is the sum of fx values
n is the sum of the frequencies

n = 100
∑fx = 542




(3)substitute the calculated values(sigmafx ,n)in the formula given



the answer is 5.42.


(4)the standard deviation is given as 1.15.
the standard deviation is the spread of the data around the mean.
the mean that we calculated is 5.42
the spread for this data is 5.42 plus or minus 1.15

(5)standard error can be calculated by substituting the s and root n in the equation given

(b) not all the ovules may have been fertilised;
exposed to different environment/example of an environmental difference
and seeds not growing/developing;

hope you understood:unsure::)



The real problem was the xaxis scale.How will the scale be?
 
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We use chi squared to test the possibility of our null hypothesis being correct and to see if there is a difference between observed and expected value ...it is used in discrete data values

But
T-test is used in continuous data , It is used to tell you that whether the means of two sets of values ,(each following a normal distribution ) are significantly different from one another ....... In simple words it is used when there are two sets of data present in the question to compare .

If i am wrong please correct me ..!! :)

Thank you very much, this is another reply i got, in case you want to confirm:
https://www.xtremepapers.com/commun...ost-your-doubts-here.9858/page-92#post-588598
 
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http://papers.xtremepapers.com/CIE/...and AS Level/Biology (9700)/9700_w08_qp_5.pdf

Ques 2b iv...here my degree of freedom was 19, so should i take the t value for 18 or 20?? according to the ms, the value used was for 20 but they allowed the value for 18 as well. maybe they overlooked it that year, so is there a set rule in situations like this? should i take the value for the higher degree of freedom or something?
 
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w08_qp_5.pdf

Ques 2b iv...here my degree of freedom was 19, so should i take the t value for 18 or 20?? according to the ms, the value used was for 20 but they allowed the value for 18 as well. maybe they overlooked it that year, so is there a set rule in situations like this? should i take the value for the higher degree of freedom or something?

it is better to take 20(always the higer value)
 
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http://papers.xtremepapers.com/CIE/...nd AS Level/Biology (9700)/9700_w09_qp_51.pdf

Ques 3b...i'm quoting the ms here:
3. ref. to no overlap in error bars between values at 60 and 120 (ng per g
lipid) indicates the difference (in damage) is likely to be significant;
4. ref. to above 120–180 (ng per g lipid) error bars have a lot of overlap so
(increase in damage) not likely to be significant;
i don't get it! Could someone please explain what they mean? I'm really weak in stats lol.
 
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it's what you'll assume the difference between the two specimen is: you're gonna assume whther or not the difference is due to chance.
So, if the difference is significant, its nt due to chance
if it's not significant, it's due to chance:)
jazakALLAH!
 
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