IB Mathematics SL 4.3 Measures of central tendency AI SL Paper 2- Exam Style Questions- New Syllabus

Question

A Principal would like to compare the students in his school with a national standard. He decides to give a test to eight students made up of four boys and four girls. One of the teachers offers to find the volunteers from his class.

(a) Name the type of sampling that best describes the method used by the Principal. [1]

The marks out of 40, for the students who took the test, are:
25, 29, 38, 37, 12, 18, 27, 31.

(b)
For the eight students find
(i) the mean mark. [2]
(ii) the standard deviation of the marks. [2]

The national standard mark is 25.2 out of 40.

(c) Perform an appropriate test at the 5% significance level to see if the mean marks achieved by the students in the school are higher than the national standard. It can be assumed that the marks come from a normal population. [5]

(d) State one reason why the test might not be valid. [1]

Two additional students take the test at a later date and the mean mark for all ten students is 28.1 and the standard deviation is 8.4. For further analysis, a standardized score out of 100 for the ten students is obtained by multiplying the scores by 2 and adding 20.

(e)
For the ten students, find
(i) their mean standardized score. [2]
(ii) the standard deviation of their standardized score. [2]

▶️ Answer/Explanation

Markscheme

(a)
The method used is quota sampling combined with volunteer sampling (A1)
Result:
Quota/volunteer sampling

(b)(i)
\( \text{Mean} = \frac{25 + 29 + 38 + 37 + 12 + 18 + 27 + 31}{8} = \frac{217}{8} = 27.125 \)
\( \approx 27.1 \) (A1)
Result:
27.1

(b)(ii)
    First calculate the sum of squares: \( \sum x^2 = 25^2 + 29^2 + 38^2 + 37^2 + 12^2 + 18^2 + 27^2 + 31^2 = 6537 \)
    \( \text{Variance} = \frac{6537}{8} – (27.125)^2 \approx 68.875 \)
    \( \text{Standard deviation} = \sqrt{68.875} \approx 8.30 \) (A1)
Result:
8.30

(c)
    We perform a one-sample t-test:
    \( H_0: \mu = 25.2 \)
    \( H_1: \mu > 25.2 \)
    Test statistic: \( t = \frac{27.125 – 25.2}{8.30 / \sqrt{8}} \approx 0.655 \)
    Degrees of freedom = 7
    p-value \( \approx 0.279 > 0.05 \) (A1)
    Conclusion: Fail to reject \( H_0 \) (A1)
    Insufficient evidence that school mean > national standard (A1)
Result:
Fail to reject \( H_0 \) (p-value = 0.279 > 0.05)

(d)
The test may not be valid because the sample was not random (volunteers from one class) (A1)
Result:
Non-random sample/volunteer bias

(e)(i)
\( \text{New mean} = 28.1 \times 2 + 20 = 76.2 \) (A1)
Result:
76.2

(e)(ii)
\( \text{New SD} = 8.4 \times 2 = 16.8 \) (A1)
Result:
16.8

IB Mathematics SL 4.3 Measures of central tendency AI SL Paper 2- Exam Style Questions- New Syllabus

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