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Barbara is investigating the relationship between average income (GDP per capita), x US dollars, and average annual carbon dioxide (CO₂) emissions, y tonnes, for different countries - Edexcel - A-Level Maths Statistics - Question 3 - 2019 - Paper 1
Question 3
Barbara is investigating the relationship between average income (GDP per capita), x US dollars, and average annual carbon dioxide (CO₂) emissions, y tonnes, for dif... show full transcript
Worked Solution & Example Answer:Barbara is investigating the relationship between average income (GDP per capita), x US dollars, and average annual carbon dioxide (CO₂) emissions, y tonnes, for different countries - Edexcel - A-Level Maths Statistics - Question 3 - 2019 - Paper 1
Step 1
Stating your hypotheses clearly, test, at the 5% level of significance, whether or not the product moment correlation coefficient for all countries is greater than zero.
Answer
To test the hypothesis, we begin with:
- Null hypothesis (): (no correlation)
- Alternative hypothesis (): (positive correlation)
Given the product moment correlation coefficient (pmcc) of 0.446 and the critical value for a one-tailed test at the 5% level of significance for 22 degrees of freedom (n - 2, where n = 24) is 0.3438, we conclude:
Since 0.446 > 0.3438, we reject the null hypothesis and accept that there is a significant positive correlation between average CO₂ emissions and average income at the 5% level.
Step 2
Explain how this value supports Barbara’s belief.
Answer
The correlation coefficient between c and m is found to be 0.882, which is quite high. This strong positive correlation suggests a closer relationship between the logged values of CO₂ emissions and income. Therefore, it supports Barbara's belief that a non-linear model is better fitted to the data than a linear model.
Step 3
Show that the relationship between y and x can be written in the form y = axⁿ where a and n are constants to be found.
Answer
Starting from the coded equation:
ext{log}_{10} y &= -1.82 + 0.89 ext{log}_{10} x \\ ext{log}_{10} y &= ext{log}_{10} igg(10^{-1.82} igg) + ext{log}_{10} igg( x^{0.89} igg) \\ ext{log}_{10} y &= ext{log}_{10} igg( 10^{-1.82} x^{0.89} igg)\\ herefore y &= 10^{-1.82} x^{0.89} \\ herefore a = 10^{-1.82} ext{ and } n = 0.89. ext{Thus, } y = 0.01542 x^{0.89}. ext{This confirms the form } y = ax^{n}. ext{where } a ext{ and } n ext{ are constants to be found.} ext{Final equation: } y = 0.01542 x^{0.89}. ewline ext{(rounded values)}