IB Mathematics AHL 4.16 Confidence intervals for the mean of a normal population.AI HL Paper 3- Exam Style Questions- New Syllabus
Question
This question aims to assist a company in determining whether to adopt a new method for producing a component.
A factory manufactures parts for a tractor. They have developed a new method for producing one of their parts, expecting it to extend its functional lifespan.
They evaluate 120 parts produced using the new method and 240 using their existing method. After 250 hours of use, they inspect the parts and note whether they exhibit no cracks, minor cracks, or major cracks.
The results from the experiment are presented in the table.
| New method | Existing method | |
|---|---|---|
| No cracks | \( a \) | 88 |
| Minor cracks | 54 | 96 |
| Major cracks | \( b \) | 56 |
Overall, 141 parts showed no cracks.
(a)
(i) Demonstrate that the value of \( a \) is 53. [1]
(ii) Determine the value of \( b \). [1]
(i) Demonstrate that the value of \( a \) is 53. [1]
(ii) Determine the value of \( b \). [1]
A part from the experiment is chosen at random.
(b)
Given that this part has minor cracks, calculate the probability that it was made using the new method. [2]
Given that this part has minor cracks, calculate the probability that it was made using the new method. [2]
(c)
A \(\chi^2\) test for independence is conducted at the 5% significance level to assess whether the presence of no cracks, minor cracks, or major cracks in a part is independent of the production method used.
A \(\chi^2\) test for independence is conducted at the 5% significance level to assess whether the presence of no cracks, minor cracks, or major cracks in a part is independent of the production method used.
(i) Specify the null and alternative hypotheses. [1]
(ii) Calculate the \( p \)-value. [2]
(iii) Provide the conclusion of the test in context, with justification for your answer. [2]
(ii) Calculate the \( p \)-value. [2]
(iii) Provide the conclusion of the test in context, with justification for your answer. [2]
(d)
For the parts in the experiment made with the existing method, demonstrate that the proportion that developed cracks is \(\frac{19}{30}\). [1]
For the parts in the experiment made with the existing method, demonstrate that the proportion that developed cracks is \(\frac{19}{30}\). [1]
As an alternative measure, the researchers let \( p \) represent the probability that a part, made with the new method, develops cracks. They then test the following hypotheses
\[ H_0 : p = \frac{19}{30}, \quad H_1 : p < \frac{19}{30}. \]
In a randomly selected sample of 120 parts made with the new method, let \( X \) be the number that developed cracks. The researchers assume that, under the null hypothesis,
\[ X \sim B\left(120, \frac{19}{30}\right). \]
(e)
Identify one additional assumption the researchers make when selecting this distribution. [1]
Identify one additional assumption the researchers make when selecting this distribution. [1]
(f)
Utilize relevant data from the experiment to conduct the test proposed by the researchers at the 5% significance level. Provide the conclusion of the test, with justification for your answer. [5]
Utilize relevant data from the experiment to conduct the test proposed by the researchers at the 5% significance level. Provide the conclusion of the test, with justification for your answer. [5]
(g)
Compared to the test in part (c), provide one mathematical reason why
(i) the test in part (f) might be preferred. [1]
(ii) the test in part (f) might not be preferred. [1]
Compared to the test in part (c), provide one mathematical reason why
(i) the test in part (f) might be preferred. [1]
(ii) the test in part (f) might not be preferred. [1]
For these parts, the researchers also examine the average time until cracks appear. They hope that the new method will increase this average time. A second experiment is conducted, and the times, in hours, until cracks appear are recorded.
The average time until cracks appear (\(\overline{T}\)) and the value of \( s_{n-1} \) for each method are presented in the following table.
| Number of parts tested | \(\overline{T}\) (hours) | \( s_{n-1} \) (hours) | |
|---|---|---|---|
| New method | 100 | 250.1 | 3.8 |
| Existing method | 200 | 249.1 | 3.8 |
(h)
Conduct a suitable test at the 5% significance level to assess whether the new method increases the average time until cracks appear. [7]
Conduct a suitable test at the 5% significance level to assess whether the new method increases the average time until cracks appear. [7]
The company chooses to implement the new method and releases the following statement: “Statistical tests indicate that the new method significantly extends the time before parts crack and require replacement”.
(i)
Provide a comment on this statement. [1]
Provide a comment on this statement. [1]
▶️ Answer/Explanation
Markscheme
(a)(i)
Total parts with no cracks: 141. Existing method: 88.
\( a = 141 – 88 = 53 \). A1
[1 mark]
Total parts with no cracks: 141. Existing method: 88.
\( a = 141 – 88 = 53 \). A1
[1 mark]
(a)(ii)
Total parts (new method): 120. No cracks: 53. Minor cracks: 54.
\( b = 120 – 53 – 54 = 13 \). A1
[1 mark]
Total parts (new method): 120. No cracks: 53. Minor cracks: 54.
\( b = 120 – 53 – 54 = 13 \). A1
[1 mark]
(b)
Total parts with minor cracks: \( 54 + 96 = 150 \).
New method with minor cracks: 54.
Probability: \( \frac{54}{150} = \frac{9}{25} = 0.36 \). M1 A1
[2 marks]
Total parts with minor cracks: \( 54 + 96 = 150 \).
New method with minor cracks: 54.
Probability: \( \frac{54}{150} = \frac{9}{25} = 0.36 \). M1 A1
[2 marks]
(c)(i)
\( H_0 \): The development of cracks and the method used are independent.
\( H_1 \): The development of cracks and the method used are not independent. A1
[1 mark]
\( H_0 \): The development of cracks and the method used are independent.
\( H_1 \): The development of cracks and the method used are not independent. A1
[1 mark]
(c)(ii)
Construct contingency table:
Observed: [No cracks: 53, 88; Minor cracks: 54, 96; Major cracks: 13, 56].
Expected frequencies (e.g., for no cracks, new method: \( \frac{141 \times 120}{360} = 47 \)).
Compute \(\chi^2\) statistic: Sum of \( \frac{(\text{observed} – \text{expected})^2}{\text{expected}} \).
Degrees of freedom: \( (3-1) \times (2-1) = 2 \).
\( p \)-value: \( 0.0170 \) (0.0169864…). M1 A1
[2 marks]
Construct contingency table:
Observed: [No cracks: 53, 88; Minor cracks: 54, 96; Major cracks: 13, 56].
Expected frequencies (e.g., for no cracks, new method: \( \frac{141 \times 120}{360} = 47 \)).
Compute \(\chi^2\) statistic: Sum of \( \frac{(\text{observed} – \text{expected})^2}{\text{expected}} \).
Degrees of freedom: \( (3-1) \times (2-1) = 2 \).
\( p \)-value: \( 0.0170 \) (0.0169864…). M1 A1
[2 marks]
(c)(iii)
\( 0.0170 < 0.05 \).
Sufficient evidence to reject \( H_0 \). The development of cracks is not independent of the method used. R1 A1
[2 marks]
\( 0.0170 < 0.05 \).
Sufficient evidence to reject \( H_0 \). The development of cracks is not independent of the method used. R1 A1
[2 marks]
(d)
Existing method parts: 240. Parts with cracks: \( 96 + 56 = 152 \).
Proportion: \( \frac{152}{240} = \frac{19}{30} \). A1
[1 mark]
Existing method parts: 240. Parts with cracks: \( 96 + 56 = 152 \).
Proportion: \( \frac{152}{240} = \frac{19}{30} \). A1
[1 mark]
(e)
The probability of each part developing cracks is independent of all other parts in the sample. R1
[1 mark]
The probability of each part developing cracks is independent of all other parts in the sample. R1
[1 mark]
(f)
Parts with cracks (new method): \( 54 + 13 = 67 \). A1
Use \( X \sim B\left(120, \frac{19}{30}\right) \).
Calculate \( P(X \leq 67) \).
Expected: \( 120 \times \frac{19}{30} = 76 \).
Use binomial CDF: \( P(X \leq 67) \approx 0.0549 \) (0.0549093…). M1 A1
Compare: \( 0.0549 > 0.05 \).
Do not reject \( H_0 \). Insufficient evidence that the new method reduces the number of cracks. R1 A1
[5 marks]
Parts with cracks (new method): \( 54 + 13 = 67 \). A1
Use \( X \sim B\left(120, \frac{19}{30}\right) \).
Calculate \( P(X \leq 67) \).
Expected: \( 120 \times \frac{19}{30} = 76 \).
Use binomial CDF: \( P(X \leq 67) \approx 0.0549 \) (0.0549093…). M1 A1
Compare: \( 0.0549 > 0.05 \).
Do not reject \( H_0 \). Insufficient evidence that the new method reduces the number of cracks. R1 A1
[5 marks]
(g)(i)
The test in (f) is directional, testing whether the new method reduces the number of parts developing cracks. R1
[1 mark]
The test in (f) is directional, testing whether the new method reduces the number of parts developing cracks. R1
[1 mark]
(g)(ii)
The test in (f) does not treat minor and major cracks as different attributes, unlike the test in (c). R1
[1 mark]
The test in (f) does not treat minor and major cracks as different attributes, unlike the test in (c). R1
[1 mark]
(h)
Let \(\mu_1\) be the mean time before cracks (new method), \(\mu_2\) (existing method).
\( H_0: \mu_1 = \mu_2 \), \( H_1: \mu_1 > \mu_2 \). A1
Two-sample t-test, pooled variance: \( s_p^2 = \frac{(100-1) \times 3.8^2 + (200-1) \times 3.8^2}{100+200-2} \approx 14.44 \).
Test statistic: \( t = \frac{250.1 – 249.1}{\sqrt{14.44 \times \left( \frac{1}{100} + \frac{1}{200} \right)}} \approx 2.141 \).
Degrees of freedom: \( 100 + 200 – 2 = 298 \).
\( p \)-value: \( 0.0162 \) (0.0162328…). M1 A2
\( 0.0162 < 0.05 \).
Reject \( H_0 \). Sufficient evidence that the new method increases the mean time before cracks appear. R1 A1
[7 marks]
Let \(\mu_1\) be the mean time before cracks (new method), \(\mu_2\) (existing method).
\( H_0: \mu_1 = \mu_2 \), \( H_1: \mu_1 > \mu_2 \). A1
Two-sample t-test, pooled variance: \( s_p^2 = \frac{(100-1) \times 3.8^2 + (200-1) \times 3.8^2}{100+200-2} \approx 14.44 \).
Test statistic: \( t = \frac{250.1 – 249.1}{\sqrt{14.44 \times \left( \frac{1}{100} + \frac{1}{200} \right)}} \approx 2.141 \).
Degrees of freedom: \( 100 + 200 – 2 = 298 \).
\( p \)-value: \( 0.0162 \) (0.0162328…). M1 A2
\( 0.0162 < 0.05 \).
Reject \( H_0 \). Sufficient evidence that the new method increases the mean time before cracks appear. R1 A1
[7 marks]
(i)
The new method increases the time before cracks by only 1 hour out of 250, so the effect size is small, despite statistical significance. R1
[1 mark]
The new method increases the time before cracks by only 1 hour out of 250, so the effect size is small, despite statistical significance. R1
[1 mark]
Total Marks: 26