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Anna is investigating the relationship between exercise and resting heart rate - Edexcel - A-Level Maths Statistics - Question 6 - 2022 - Paper 1

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Anna is investigating the relationship between exercise and resting heart rate. She takes a random sample of 19 people in her year at school and records for each per... show full transcript

Worked Solution & Example Answer:Anna is investigating the relationship between exercise and resting heart rate - Edexcel - A-Level Maths Statistics - Question 6 - 2022 - Paper 1

Step 1

Interpret the nature of the relationship between h and m

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Answer

The scatter diagram indicates a negative relationship between the number of minutes of exercise (m) and resting heart rate (h). As the number of minutes spent exercising increases, the resting heart rate tends to decrease. This suggests that higher levels of exercise are associated with lower resting heart rates.

Step 2

Test whether or not there is significant evidence of a negative correlation between x and y

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Answer

Let the null hypothesis be:

  • H₀: ρ = 0 (No correlation) The alternative hypothesis is:
  • H₁: ρ < 0 (There is a negative correlation)

Using the product moment correlation coefficient, which is -0.897, we will compare it with the critical value at a 5% significance level. The critical value for n = 19 (degrees of freedom = 17) is approximately -0.387. Since -0.897 < -0.387, we reject the null hypothesis and conclude that there is significant evidence of a negative correlation between x and y.

Step 3

The equation of the line of best fit of y on x is

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Answer

We can express the equation as:

y=0.05imesextlog10(m)+1.92y = -0.05 imes ext{log}_{10}(m) + 1.92

To find the model for h in terms of m, we start with

y=extlog10(h)y = ext{log}_{10}(h) Substituting the equation of the line of best fit gives us:

extlog10(h)=0.05imesextlog10(m)+1.92 ext{log}_{10}(h) = -0.05 imes ext{log}_{10}(m) + 1.92 Taking the inverse logarithm:

h=10(0.05imesextlog10(m)+1.92)h = 10^{(-0.05 imes ext{log}_{10}(m) + 1.92)} Using the properties of logarithms, we have:

h=101.92imesm0.05h = 10^{1.92} imes m^{-0.05} Thus, we can express it in the form:

h=aimesmkh = a imes m^k where

  • a = 10^{1.92} ext{ (approximately 83.21)}
  • k = -0.05.

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