| Date | May 2009 | Marks available | 3 | Reference code | 09M.2.sl.TZ2.2 |
| Level | SL only | Paper | 2 | Time zone | TZ2 |
| Command term | Show that | Question number | 2 | Adapted from | N/A |
Question
A manufacturer claims that fertilizer has an effect on the height of rice plants. He measures the height of fertilized and unfertilized plants. The results are given in the following table.
A chi-squared test is performed to decide if the manufacturer’s claim is justified at the 1 % level of significance.
The population of fleas on a dog after t days, is modelled by
\[N = 4 \times {(2)^{\frac{t}{4}}},{\text{ }}t \geqslant 0\]
Some values of N are shown in the table below.
Write down the null and alternative hypotheses for this test.
For the number of fertilized plants with height greater than 75 cm, show that the expected value is 97.5.
Write down the value of \(\chi_{calc}^2\).
Write down the number of degrees of freedom.
Is the manufacturer’s claim justified? Give a reason for your answer.
Write down the value of p.
Write down the value of q.
Using the values in the table above, draw the graph of N for 0 ≤ t ≤ 20. Use 1 cm to represent 2 days on the horizontal axis and 1 cm to represent 10 fleas on the vertical axis.
Use your graph to estimate the number of days for the population of fleas to reach 55.
Markscheme
H0: The height of the rice plants is independent of the use of a fertilizer. (A1)
Notes: For independent accept “not associated”, can accept “the use of a fertilizer has no effect on the height of the plants”.
Do not accept “not correlated”.
H1: The height of the rice plants is not independent (dependent) of the use of fertilizer. (A1)(ft)
Note: If H0 and H1 are reversed award (A0)(A1)(ft).
[2 marks]
\(\frac{{180 \times 195}}{{360}}\) or \(\frac{{180}}{{360}} \times \frac{{195}}{{360}} \times 360\) (A1)(A1)(M1)
= 97.5 (AG)
Notes: Award (A1) for numerator, (A1) for denominator (M1) for division.
If final 97.5 is not seen award at most (A1)(A0)(M1).
[3 marks]
\( \chi_{calc}^2 = 14.01 (14.0, 14)\) (G2)
OR
If worked out by hand award (M1) for correct substituted formula with correct values, (A1) for correct answer. (M1)(A1)
[2 marks]
2 (A1)
[1 mark]
\( \chi_{calc}^2 > \chi_{crit}^2\) (R1)
The manufacturer's claim is justified. (or equivalent statement) (A1)
Note: Do not accept (R0)(A1).
[2 marks]
\(p = 4\) (G1)
[1 mark]
\(q = 4(2)^{\frac{16}{4}}\) (M1)
\(= 64\) (A1)(G2)
[2 marks]
(A1)(A1)(A1) (A3)
Notes: Award (A1) for x axis with correct scale and label, (A1) for y axis with correct scale and label.
Accept x and y for labels.
If x and y axis reversed award at most (A0)(A1)(ft).
(A1) for smooth curve.
Award (A3) for all 6 points correct, (A2) for 4 or 5 points correct, (A1) for 2 or 3 points correct, (A0) otherwise.
[6 marks]
15 (±0.8) (M1)(A1)(ft)(G2)
Note: Award (M1) for line drawn shown on graph, (A1)(ft) from candidate's graph.
[2 marks]
Examiners report
It was clear that the candidates who performed poorly in part (i) lacked the basic knowledge of chi-squared analysis. Some mixed up the null and alternate hypotheses and also were not able to correctly demonstrate the way of finding the expected value. There were many errors in finding the critical value of \(\chi^2\) at the 1% level of significance.
It was clear that the candidates who performed poorly in part (i) lacked the basic knowledge of chi-squared analysis. Some mixed up the null and alternate hypotheses and also were not able to correctly demonstrate the way of finding the expected value. There were many errors in finding the critical value of \(\chi^2\) at the 1% level of significance.
It was clear that the candidates who performed poorly in part (i) lacked the basic knowledge of chi-squared analysis. Some mixed up the null and alternate hypotheses and also were not able to correctly demonstrate the way of finding the expected value. There were many errors in finding the critical value of \(\chi^2\) at the 1% level of significance.
It was clear that the candidates who performed poorly in part (i) lacked the basic knowledge of chi-squared analysis. Some mixed up the null and alternate hypotheses and also were not able to correctly demonstrate the way of finding the expected value. There were many errors in finding the critical value of \(\chi^2\) at the 1% level of significance.
It was clear that the candidates who performed poorly in part (i) lacked the basic knowledge of chi-squared analysis. Some mixed up the null and alternate hypotheses and also were not able to correctly demonstrate the way of finding the expected value. There were many errors in finding the critical value of \(\chi^2\) at the 1% level of significance.
Candidates found this part rather easy, with some making arithmetic mistakes and thus losing one or more marks. The graph was well done with a high percentage scoring full marks. Some candidates did not label the axes, others had an incorrect scale and a few lost one mark for not drawing a smooth curve.
Candidates found this part rather easy, with some making arithmetic mistakes and thus losing one or more marks. The graph was well done with a high percentage scoring full marks. Some candidates did not label the axes, others had an incorrect scale and a few lost one mark for not drawing a smooth curve.
Candidates found this part rather easy, with some making arithmetic mistakes and thus losing one or more marks. The graph was well done with a high percentage scoring full marks. Some candidates did not label the axes, others had an incorrect scale and a few lost one mark for not drawing a smooth curve.
Candidates found this part rather easy, with some making arithmetic mistakes and thus losing one or more marks. The graph was well done with a high percentage scoring full marks. Some candidates did not label the axes, others had an incorrect scale and a few lost one mark for not drawing a smooth curve.
