IBDP MAI : Topic 4 Statistics and probability - SL 4.11 Formulation of null and alternative hypotheses AI HL Paper 2
Question
A type of generator will only function if a particular switch is working. The generator has a main switch, A, and a ‘back up’ switch, B.
The manufacturer claims the probability of switch A failing within one month of being fitted is 0.1 and the probability of the cheaper switch B failing within one month is 0.3. Whether or not a switch fails is independent of the state of the other switch.
If both switches fail, the generator needs to shut down to replace the switches. Both switches are replaced after a month of use (whether they have failed or not) or whenever the generator needs to be shut down.
The following tree diagram shows the probabilities of a switch failing within one month of them both being replaced, assuming the manufacturer’s claim is correct.
(a) Write down the values of
(i) a.
(ii) b.
(iii) c.
(b) Hence find the probability that the generator needs to shut down within one month of the switches being replaced.
The owner of the generator is suspicious of the switch manufacturer’s claims, so they look
back through the past 200 occasions when the switches were replaced. The records show
whether no switches, one switch or two switches had failed.
The data the owner collected are shown in the following table.
(c) Perform a c 2 goodness of fit test at the 5 % significance level to test whether the manufacturer’s claims are correct using the following hypotheses.
H0: The manufacturer’s claims are correct.
H1: The manufacturer’s claims are not both correct.
▶️Answer/Explanation
Detail Solution: –
Let’s solve this step-by-step. The problem involves a generator that depends on at least one of two switches (A or B) functioning to avoid shutting down. We’re given the probabilities of failure for each switch within one month: P(A fails) = 0.1 and P(B fails) = 0.3. The switches operate independently, and the generator shuts down only if both switches fail. We need to determine the values of a, b, and c from a tree diagram and then calculate the probability of shutdown.
Step 1: From the Tree Diagram
A tree diagram for this scenario typically starts with switch A, which either works or fails, followed by switch B, which also either works or fails. The probabilities are:
P(A fails) = 0.1, so P(A works) = 1 – 0.1 = 0.9
P(B fails) = 0.3, so P(B works) = 1 – 0.3 = 0.7
Since the switches are independent, the joint probabilities for each outcome are calculated by multiplying the individual probabilities along the branches. The possible outcomes are:
1. A works, B works: P(A works) × P(B works) = 0.9 × 0.7
2. A works, B fails: P(A works) × P(B fails) = 0.9 × 0.3
3. A fails, B works: P(A fails) × P(B works) = 0.1 × 0.7
4. A fails, B fails: P(A fails) × P(B fails) = 0.1 × 0.3
The problem asks for the values of a, b, and c, which are likely probabilities on specific branches of the tree diagram, and then the probability of shutdown (both switches failing).
Part (a) – Determine a, b, and c
Without the actual tree diagram, we must infer the meanings of a, b, and c based on standard conventions. Typically:
The first level of the tree represents switch A:
Probability A works = 0.9
Probability A fails = 0.1
The second level represents switch B, conditional on A’s state.
A common assignment in such problems is:
a: Probability A works (first branch) = 0.9
b: Probability B fails given A works (second-level branch) = 0.3
c: Probability B works given A fails (second-level branch) = 0.7
This is consistent with independence, as the probabilities for B don’t change based on A’s state. Let’s calculate the joint probabilities to confirm:
If a = P(A works) = 0.9, then:
b = P(B fails | A works) = 0.3, so P(A works, B fails) = 0.9 × 0.3 = 0.27
If A fails (P = 0.1), then:
c = P(B works | A fails) = 0.7, so P(A fails, B works) = 0.1 × 0.7 = 0.07
These values align with the problem’s structure. Thus:
(i) a = 0.9
(ii) b = 0.3
(iii) c = 0.7
Part (b) – Probability of Shutdown
The generator shuts down if both switches fail. Using the independence of A and B:
P(A fails) = 0.1
P(B fails) = 0.3
P(both fail) = P(A fails) × P(B fails) = 0.1 × 0.3 = 0.03
Alternatively, from the tree diagram, the branch “A fails, B fails” is:
P(A fails) × P(B fails) = 0.1 × 0.3 = 0.03
Part (c): Chi-squared goodness of fit test at 5% significance level
We need to test if the manufacturer’s claims (\( P(A) = 0.1 \), \( P(B) = 0.3 \), independent) match the observed data over 200 replacements. The outcomes are:
Both fail (A and B).
One fails (A fails, B doesn’t; or A doesn’t, B fails).
Neither fails.
Step 1: Expected probabilities
\( P(\text{both fail}) = 0.1 \times 0.3 = 0.03 \).
\( P(\text{A fails, B doesn’t}) = 0.1 \times 0.7 = 0.07 \).
\( P(\text{A doesn’t, B fails}) = 0.9 \times 0.3 = 0.27 \).
\( P(\text{neither fails}) = 0.9 \times 0.7 = 0.63 \).
\( P(\text{one fails}) = 0.07 + 0.27 = 0.34 \).
Categories from the table: “no switches fail,” “one switch fails,” “two switches fail.”
\( P(\text{none}) = 0.63 \).
\( P(\text{one}) = 0.34 \).
\( P(\text{two}) = 0.03 \).
Step 2: Expected frequencies
Total observations = 200.
\( E(\text{none}) = 0.63 \times 200 = 126 \).
\( E(\text{one}) = 0.34 \times 200 = 68 \).
\( E(\text{two}) = 0.03 \times 200 = 6 \).
Step 3: Chi-squared statistic
\[
\chi^2 = \sum \frac{(O – E)^2}{E}
\]
None: \( \frac{(118 – 126)^2}{126} = \frac{64}{126} \approx 0.5079 \).
One: \( \frac{(72-68)^2}{68} = \frac{16}{68} \approx 0.235 \).
Two: \( \frac{(10- 6)^2}{6} = 2.66666\).
\[
\chi^2 = 0.5079 + 0.235 + 2.66666 = 3.4095666
\]
Step 4: Degrees of freedom
Categories = 3, parameters estimated = 0 (probabilities given), so \( df = 3 – 1 = 2 \).
Step 5: Critical value
At 5% significance=0.05 , \( df = 2 \), critical \( \chi^2 = 3.409566 \).
from the above data we can calculate p-value with the help of p value calculator is 0.182
0.18 > 0.05
Hence insufficient evidence to reject $H_{o}$(that the manufacturers claims are correct)
————Markscheme—————–
(a) (i) 0.9 (ii) 0.3 (iii) 0.7
(b) (0.1×0.3=) 0.03
(c) P (no fail) = 0.63
P (one fails) = 0.34
P(two fail) = 0.03
multiplying by 200
degrees of freedom = 2
p-value = 0.182 (0.181781…)
0.182 > 0.05
hence insufficient evidence to reject H₀ (that the manufacturers claims are correct)