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Edexcel IAL Biology 6P.5 Uncertainty, errors and conclusions

Quote the relevant values or overlap, compare the calculated statistic with its critical value and distinguish random variability from systematic error before concluding.

Syllabus
First assessment 2019
Course
Biology YBI11
Level
A2

Exam points

  • Compare a calculated statistic with its critical value to decide significance.
  • Interpret error-bar overlap and variability when comparing treatment means.
  • Support conclusions with graph or table values while identifying uncertainty or error.

6P.5—Evaluating results, uncertainty and errors question 1

[Maximum number: 2]

The photograph shows a young locust eating the leaves of a plant.

Figure for Question 6P.5—Evaluating results, uncertainty and errors question 1 — Edexcel A-Level Biology A2

Locusts can breed in large numbers and destroy crops, such as sorghum, in East Africa. A scientist observed that some varieties of sorghum were more likely to be eaten by young locusts. The scientist collected the seeds of two varieties of sorghum (A and B) and grew them in trays. Locust eggs were collected and hatched into young locusts. Twenty young locusts were placed in a cage containing 100 g of fresh sorghum leaves of variety A. The cage was kept at 30C30^{\circ} \mathrm{C}. After 24 hours, the leaves were removed and the mass of leaves eaten was calculated. This method was repeated with fresh sorghum leaves of variety B. The method was repeated six times for these two varieties of sorghum leaves. The results: Mass of leaves eaten of variety A 17.317.217.517.016.716.9 Mass of leaves eaten of variety B 17.817.917.717.617.817.4

The student analysed the data using the t test formula:

t=(xˉAxˉB)(SA)2nA+(SB)2nBt=\frac{\left(\bar{x}_{A}-\bar{x}_{B}\right)}{\sqrt{\frac{\left(S_{A}\right)^{2}}{n_{A}}+\frac{\left(S_{B}\right)^{2}}{n_{B}}}}

Where: x\overline{\mathrm{x}} is the mean value for each treatment n is the number of samples for each treatment (SA)2=0.084\left(\mathrm{S}_{\mathrm{A}}\right)^{2}=0.084 and (SB)2=0.032\left(\mathrm{S}_{\mathrm{B}}\right)^{2}=0.032

The table shows the critical values of t for different degrees of freedom. The number of degrees of freedom =(n11)+(n21)=\left(n_{1}-1\right)+\left(n_{2}-1\right). Describe the conclusions that can be drawn from this investigation. Use the information in the table to support your answer.

Table for Question 6P.5—Evaluating results, uncertainty and errors question 1 — Edexcel A-Level Biology A2
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