chi squared question

http://files.physicsandmathstutor.com/download/Biology/A-level/Past-Papers/Edexcel-A/Paper-3/Specimen%20QP%20-%20Paper%203%20Edexcel%20Biology%20(A)%20A-Level.pdf
Q2b please

Post your working so far.

Question says: "The Chi squared test can be used to determine whether the results of this
investigation indicate a significant difference in the distribution of young larvae between the light and the dark side"

Now your focus is only on the young - and the values in the table are your observed values.

Do you think you can now carry on with the question?

I am so confused

http://files.physicsandmathstutor.com/download/Biology/A-level/Past-Papers/Edexcel-A/Paper-3/Specimen%20QP%20-%20Paper%203%20Edexcel%20Biology%20(A)%20A-Level.pdf
Q2b please

Did you manage to work out the answer to this? I have been trying for the last hour and I don't think the answer is right! I cannot work it out.

Omg thankyou so much this is so helpful!

Right! I know this thread is old but It has been bugging me and I have finally cracked the code!

After looking at the mark scheme and thinking what the actual f.u.q I took it back to basics.

I just followed the steps to chi squared really slowly. So:

As no expected values are given, you must assume an equal division of the larvae --> 5 larvae on each side in each trial. So the total for the expected on the dark side, for both trials is 5 + 5 = 10 and for light side 5 +5 = 10.

So our expected for both dark and light is 10 and 10. Now, we get out observed. tbh, the table of results is confusing and is just a prime example of edexcel's shoddy work but, the way I looked at it is:

In trial 2 flip the results over, as in this trial it was the left side that was light and the right side that was dark, not the way the table has portrayed it. edexcel are sneaky biatches.

so add up the observed values ( light= 1+2, dark 9+8) So our observed for light side is 3 and our observed for dark is 17.

Now, subtract the expected from the observed: 17 - 10 = 7 // 3 - 10 = -7, then square it : 7^2 = 49 // -7^2 = -49

Then divide these answers by the expected for each trial (10) which should give us -4.9 and 4.9.

Where I kept tripping up is me forgetting that negative values become positive at this stage. so all you have to do now, is get rid of that negative sign and add 4.9 and 4.9 together to get 9.8.
Which is the answer on the mark scheme.

EDIT: I noticed that when you manually enter into the calculator -7 x -7 you get a whole number of 49, but when you do -7 and press the square button it comes in a negative form. so to be safe always enter the calculations manually.

thing is i would say since the distribution of young larvae should be equal on both sides so considering the total number of larvae on the first trial is 20 and the total number of young larvae are 10 and the total number of larvae (young and old) on the left side is 11 so the number of larvae in trial 1 on the left side would be
10/20 x 11 around 5.5 then repeat this with the other 4 values and add them up i get around as my value 9.8