Hint
Independent differences come from a continuous symmetric distribution under the null. Rank absolute values and add only ranks attached to positive differences.
Worked solution
(a)
\(W^+=16\)
Answer: \(16\).
Answer: \(16\)
(b)
\(W^-=5\)
Answer: \(5\).
Answer: \(5\)
(c)
\(\displaystyle P(W^+\ge 16)=\frac{5}{32}\) by all 64 equally likely sign allocations.
Answer: \(\displaystyle \frac{5}{32}\).
Answer: \(\displaystyle \frac{5}{32}\)
(d)
The p-value is \(0.15625\). Since it exceeds 0.05, do not reject \(\displaystyle H_{0}\): the sample gives insufficient evidence that the population has a positive location shift. This does not give the probability that \(\displaystyle H_{0}\) is true.
Answer:
The p-value is \(0.15625\). Since it exceeds 0.05, do not reject \(\displaystyle H_{0}\): the sample gives insufficient evidence that the population has a positive location shift. This does not give the probability that \(\displaystyle H_{0}\) is true.
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.
(a) · 2 marks
- 1 mark: Rank absolute values and add only ranks attached to positive differences.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Rank absolute differences, not the signed numbers.
(b) · 2 marks
- 1 mark: Add ranks attached to negative differences.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not use the sign-test binomial distribution for a sum of ranks.
(c) · 2 marks
- 1 mark: Enumerate the null sign allocations of the ranks.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Rank absolute differences, not the signed numbers.
(d) · 1 marks
- 1 mark: Compare the relevant p-value with 0.05 and state the contextual conclusion without claiming the null is proved.
Alternative methods: A comparison with the correct critical value and the same valid contextual conclusion is acceptable.
Common error: A p-value is not the probability that the null hypothesis is true.
Total: 7 marks.