Question 23M.1.SL.TZ2.21
| Date | May 2023 | Marks available | [Maximum mark: 1] | Reference code | 23M.1.SL.TZ2.21 |
| Level | SL | Paper | 1 | Time zone | TZ2 |
| Command term | Question number | 21 | Adapted from | N/A |
A student wanted to know whether the density of buttercup (Ranunculus) flowers in two fields was the same. He used a quadrat to estimate the number of flowers in equal-sized areas of each field. The table shows the results.
| Number of flowers | Field 1 | Field 2 |
|
Observed |
75 | 51 |
| Expected | 63 |
63 |
A chi-squared test was carried out to determine whether the density of buttercups was the same in both areas. The chi-squared calculated value was 4.571.
| Probability level | ||||||
| Degrees of freedom (df) | 0.5 | 0.10 | 0.05 | 0.02 | 0.01 | 0.001 |
| 1 | 0.455 | 2.706 | 3.841 | 5.412 | 6.635 | 10.827 |
| 2 | 1.386 | 4.605 |
5.991 |
7.824 | 9.210 | 13.815 |
With reference to the probability table, what conclusion can be drawn about the null hypothesis with 95 % confidence?
A. It is rejected as 4.571 is less than 5.991.
B. It is rejected as 4.571 is greater than 3.841.
C. It is rejected as 4.571 is greater than 0.455.
D. It is rejected as 4.571 is greater than 1.386.
[1]
B
This question asking students to draw conclusions from the results of a chi-squared test discriminated quite well.