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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 suggests a negative relationship between the number of minutes exercising (m) and the resting heart rate (h). As the number of minutes spent exercising increases, the resting heart rate tends to decrease. This indicates that exercise has a diminishing effect on heart rate, implying that additional minutes of exercise may lead to a slower resting heart rate.

Step 2

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

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Answer

  1. State your hypotheses clearly:

    • Null Hypothesis (H₀): ρ = 0 (no correlation)
    • Alternative Hypothesis (H₁): ρ < 0 (negative correlation)
  2. Use a 5% level of significance: Using the significance level, the critical value for this test can be found from the correlation coefficient table for n=19, yielding a critical value of -0.387.

  3. State the critical value used: The correlation coefficient calculated is -0.897. Since -0.897 < -0.387, we reject the null hypothesis and conclude that there is evidence of a significant 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

Using the equation provided:

y = -0.05 x + 1.92

We can express this equation in log form:

  1. By substituting back we find: log10h=log10(mk)+log10(a)log_{10} h = log_{10}(m^{k}) + log_{10}(a)
    where k=0.05k = -0.05 and a=101.92a = 10^{1.92}. Therefore, we have:

    h=amkh = a m^{k}

    Now calculating values: a can be evaluated by: aext(approx.=83.2)a ext{ (approx. } = 83.2) Hence, the model for h on m is: hext(approx.=83.2m0.05)h ext{ (approx. } = 83.2 m^{-0.05})

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