Hint
Samples are independent and normally distributed. Weight each sample variance by its residual degrees of freedom.
Worked solution
(a)
\(\displaystyle s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}=\frac{11}{2}\)
Answer: \(\displaystyle \frac{11}{2}\).
Answer: \(\displaystyle \frac{11}{2}\)
(b)
\(\displaystyle \sqrt{s_{p}^{2}(\frac{1}{n_{1}}+ \frac{1}{n_{2}})}= 1.10554159678513\)
Answer: \(1.10554159678513\).
Answer: \(\approx 1.10554159678513\)
(c)
\(\displaystyle t= \frac{\bar{x}_{1}- \bar{x}_{2}}{S}E= 1.80906806746658\)
Answer: \(1.80906806746658\).
Answer: \(\approx 1.80906806746658\)
(d)
With 16 degrees of freedom the upper-tail p-value is \(0.0446340720813295\). Reject \(\displaystyle H_{0}\): the first mean is greater.
Answer:
With 16 degrees of freedom the upper-tail p-value is \(0.0446340720813295\). Reject \(\displaystyle H_{0}\): the first mean is greater.
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: Weight each sample variance by its residual degrees of freedom.
- 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 pool means instead of within-sample variances.
(b) · 2 marks
- 1 mark: Add the independent estimated sampling variances.
- 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: The variance of a difference adds the independent sampling variances.
(c) · 2 marks
- 1 mark: Divide the mean difference by its pooled standard error.
- 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 pool means instead of within-sample variances.
(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.