IB Mathematics SL 4.11 Formulation of null and alternative hypotheses AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Sophie is investigating whether a six-sided die is fair. She rolls the die 60 times and records the observed frequencies in the following table:
| Number on die | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Observed frequency | 8 | 7 | 6 | 15 | 12 | 12 |
Sophie carries out a \( \chi^2 \) goodness-of-fit test at a 5% significance level.
(a) Write down the null and alternative hypotheses. [1]
(b) Write down the degrees of freedom. [1]
(c) Write down the expected frequency of rolling a 1. [1]
(d) Find the \( p \)-value for the test. [2]
(e) State the conclusion of the test. Give a reason for your answer. [2]
▶️ Answer/Explanation
Markscheme
(a)
Null hypothesis: \( H_0 \): The die is fair (\( P(\text{any number}) = \frac{1}{6} \)).
Alternative hypothesis: \( H_1 \): The die is not fair (\( P(\text{any number}) \neq \frac{1}{6} \)). A1
[1 mark]
Null hypothesis: \( H_0 \): The die is fair (\( P(\text{any number}) = \frac{1}{6} \)).
Alternative hypothesis: \( H_1 \): The die is not fair (\( P(\text{any number}) \neq \frac{1}{6} \)). A1
[1 mark]
(b)
Degrees of freedom: \( 6 – 1 = 5 \). A1
[1 mark]
Degrees of freedom: \( 6 – 1 = 5 \). A1
[1 mark]
(c)
Expected frequency for rolling a 1:
\( \frac{60}{6} = 10 \) A1
[1 mark]
Expected frequency for rolling a 1:
\( \frac{60}{6} = 10 \) A1
[1 mark]
(d)
Calculate the chi-squared test statistic:
\[ \begin{aligned} \chi^2 &= \sum \frac{(O_i – E_i)^2}{E_i} \\ &= \frac{(8 – 10)^2}{10} + \frac{(7 – 10)^2}{10} + \frac{(6 – 10)^2}{10} + \frac{(15 – 10)^2}{10} + \frac{(12 – 10)^2}{10} + \frac{(12 – 10)^2}{10} \\ &= \frac{4}{10} + \frac{9}{10} + \frac{16}{10} + \frac{25}{10} + \frac{4}{10} + \frac{4}{10} = 0.4 + 0.9 + 1.6 + 2.5 + 0.4 + 0.4 = 5.6 \end{aligned} \]
For \( \chi^2 = 5.6 \) with 5 degrees of freedom, the \( p \)-value is approximately 0.287. A2
[2 marks]
Calculate the chi-squared test statistic:
\[ \begin{aligned} \chi^2 &= \sum \frac{(O_i – E_i)^2}{E_i} \\ &= \frac{(8 – 10)^2}{10} + \frac{(7 – 10)^2}{10} + \frac{(6 – 10)^2}{10} + \frac{(15 – 10)^2}{10} + \frac{(12 – 10)^2}{10} + \frac{(12 – 10)^2}{10} \\ &= \frac{4}{10} + \frac{9}{10} + \frac{16}{10} + \frac{25}{10} + \frac{4}{10} + \frac{4}{10} = 0.4 + 0.9 + 1.6 + 2.5 + 0.4 + 0.4 = 5.6 \end{aligned} \]
For \( \chi^2 = 5.6 \) with 5 degrees of freedom, the \( p \)-value is approximately 0.287. A2
[2 marks]
(e)
Compare \( p \)-value to significance level: \( 0.287 > 0.05 \). R1
Conclusion: Insufficient evidence to reject the null hypothesis; the die is fair. A1
[2 marks]
Compare \( p \)-value to significance level: \( 0.287 > 0.05 \). R1
Conclusion: Insufficient evidence to reject the null hypothesis; the die is fair. A1
[2 marks]
Total Marks: 7
Question
Amelia has geese on their farm. They claim the mean weight of eggs from their black geese is less than the mean weight of eggs from their white geese. The weights of eggs, in grams, from a random selection of geese are shown in the table:
| Type | Weight 1 | Weight 2 | Weight 3 | Weight 4 | Weight 5 | Weight 6 |
|---|---|---|---|---|---|---|
| Black geese | 136 | 134 | 142 | 141 | 128 | 126 |
| White geese | 135 | 138 | 141 | 140 | 136 | 134 |
To test their claim, Amelia performs a t-test at a 10% significance level. The weights of eggs are assumed to be normally distributed with equal variances.
(a) State the null hypothesis in words. [1]
(b) Determine the p-value for this test. [2]
(c) State whether the test result supports Amelia’s claim and justify the reasoning. [2]
▶️ Answer/Explanation
Markscheme
(a)
The population mean weight of eggs from black geese is equal to the population mean weight of eggs from white geese (\( \mu_B = \mu_W \)). A1
[1 mark]
The population mean weight of eggs from black geese is equal to the population mean weight of eggs from white geese (\( \mu_B = \mu_W \)). A1
[1 mark]
(b)
Black mean:
\[ \begin{aligned} \bar{x}_B &= \frac{136 + 134 + 142 + 141 + 128 + 126}{6} = 134.5 \end{aligned} \]
Black variance:
\[ \begin{aligned} s_B^2 &= \frac{(136 – 134.5)^2 + (134 – 134.5)^2 + (142 – 134.5)^2 + (141 – 134.5)^2 + (128 – 134.5)^2 + (126 – 134.5)^2}{5} \\ &= \frac{2.25 + 0.25 + 56.25 + 42.25 + 42.25 + 72.25}{5} = 34.3 \end{aligned} \]
White mean:
\[ \begin{aligned} \bar{x}_W &= \frac{135 + 138 + 141 + 140 + 136 + 134}{6} \approx 137.333 \end{aligned} \]
White variance:
\[ \begin{aligned} s_W^2 &= \frac{(135 – 137.333)^2 + (138 – 137.333)^2 + (141 – 137.333)^2 + (140 – 137.333)^2 + (136 – 137.333)^2 + (134 – 137.333)^2}{5} \\ &= \frac{5.428 + 0.444 + 13.444 + 7.111 + 1.777 + 11.111}{5} \approx 7.467 \end{aligned} \] M1
Pooled variance:
\[ \begin{aligned} s_p^2 &= \frac{(5 \times 34.3 + 5 \times 7.467)}{10} \approx \frac{171.5 + 37.335}{10} \approx 20.885 \end{aligned} \]
t-statistic:
\[ \begin{aligned} t &= \frac{134.5 – 137.333}{\sqrt{20.885 \times \left( \frac{1}{6} + \frac{1}{6} \right)}} \\ &= \frac{-2.833}{\sqrt{20.885 \times \frac{2}{6}}} \approx \frac{-2.833}{\sqrt{6.961667}} \approx -0.963 \end{aligned} \]
Degrees of freedom: \( 6 + 6 – 2 = 10 \).
p-value (one-tailed, \( t \approx -0.963 \), df = 10): \( \approx 0.177 \). A1
[2 marks]
Black mean:
\[ \begin{aligned} \bar{x}_B &= \frac{136 + 134 + 142 + 141 + 128 + 126}{6} = 134.5 \end{aligned} \]
Black variance:
\[ \begin{aligned} s_B^2 &= \frac{(136 – 134.5)^2 + (134 – 134.5)^2 + (142 – 134.5)^2 + (141 – 134.5)^2 + (128 – 134.5)^2 + (126 – 134.5)^2}{5} \\ &= \frac{2.25 + 0.25 + 56.25 + 42.25 + 42.25 + 72.25}{5} = 34.3 \end{aligned} \]
White mean:
\[ \begin{aligned} \bar{x}_W &= \frac{135 + 138 + 141 + 140 + 136 + 134}{6} \approx 137.333 \end{aligned} \]
White variance:
\[ \begin{aligned} s_W^2 &= \frac{(135 – 137.333)^2 + (138 – 137.333)^2 + (141 – 137.333)^2 + (140 – 137.333)^2 + (136 – 137.333)^2 + (134 – 137.333)^2}{5} \\ &= \frac{5.428 + 0.444 + 13.444 + 7.111 + 1.777 + 11.111}{5} \approx 7.467 \end{aligned} \] M1
Pooled variance:
\[ \begin{aligned} s_p^2 &= \frac{(5 \times 34.3 + 5 \times 7.467)}{10} \approx \frac{171.5 + 37.335}{10} \approx 20.885 \end{aligned} \]
t-statistic:
\[ \begin{aligned} t &= \frac{134.5 – 137.333}{\sqrt{20.885 \times \left( \frac{1}{6} + \frac{1}{6} \right)}} \\ &= \frac{-2.833}{\sqrt{20.885 \times \frac{2}{6}}} \approx \frac{-2.833}{\sqrt{6.961667}} \approx -0.963 \end{aligned} \]
Degrees of freedom: \( 6 + 6 – 2 = 10 \).
p-value (one-tailed, \( t \approx -0.963 \), df = 10): \( \approx 0.177 \). A1
[2 marks]
(c)
Significance level: \( 0.1 \).
p-value: \( 0.177 \).
Since \( 0.177 > 0.1 \), fail to reject \( H_0 \). A1
Insufficient evidence to support \( \mu_B < \mu_W \). Amelia’s claim is not supported. A1
[2 marks]
Significance level: \( 0.1 \).
p-value: \( 0.177 \).
Since \( 0.177 > 0.1 \), fail to reject \( H_0 \). A1
Insufficient evidence to support \( \mu_B < \mu_W \). Amelia’s claim is not supported. A1
[2 marks]
Total Marks: 5