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Crickets make a noise - Edexcel - A-Level Maths Statistics - Question 4 - 2008 - Paper 2

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Crickets make a noise. The pitch, v kHz, of the noise made by a cricket was recorded at 15 different temperatures, t °C. These data are summarised below. $$\sum t^2... show full transcript

Worked Solution & Example Answer:Crickets make a noise - Edexcel - A-Level Maths Statistics - Question 4 - 2008 - Paper 2

Step 1

Find $S_{t}$, $S_{v}$, and $S_{tv}$ for these data.

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Answer

To find the needed sums:

  1. Calculate StS_{t}:

    St=t=401.3S_{t} = \sum t = 401.3

  2. Calculate SvS_{v}:

    Sv=v=25.08S_{v} = \sum v = 25.08

  3. Calculate StvS_{tv}:

    Stv=tv=vtn=677.971401.315=18267.169S_{tv} = \sum tv = \frac{\sum v \cdot \sum t}{n} = \frac{677.971 \cdot 401.3}{15} = 18 267.169

Step 2

Find the product moment correlation coefficient between t and v.

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Answer

The formula for the product moment correlation coefficient, rr, is given by:

r=n(tv)tv(nt2(t)2)(nv2(v)2)r = \frac{n \sum (tv) - \sum t \cdot \sum v}{\sqrt{(n \sum t^2 - (\sum t)^2)(n \sum v^2 - (\sum v)^2)}}

Substituting the values:

r=1518267.169401.325.08(1510922.81(401.3)2)(1542.3356(25.08)2)r = \frac{15 \cdot 18 267.169 - 401.3 \cdot 25.08}{\sqrt{(15 \cdot 10 922.81 - (401.3)^2)(15 \cdot 42.3356 - (25.08)^2)}}

Calculating gives: r0.808r \approx 0.808

Step 3

State, with a reason, which variable is the explanatory variable.

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Answer

tt is the explanatory variable because we can control temperature, but not the frequency of noise or any equivalent comment.

Step 4

Give a reason to support fitting a regression model of the form v = a + bt to these data.

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Answer

The relationship is expected to be linear; as temperature increases, we expect the pitch, vv, to change systematically.

Step 5

Find the value of a and the value of b. Give your answers to 3 significant figures.

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Answer

To find the coefficients aa and bb:

  1. Calculate bb using the formula:

    b=n(tv)tvnt2(t)2b = \frac{n\sum(tv) - \sum t \cdot \sum v}{n\sum t^2 - (\sum t)^2}

    Which gives: b0.669b \approx 0.669

  2. Then find aa:

    a=vˉbtˉa = \bar{v} - b \bar{t}

    Using our averages we find: a1.4a \approx 1.4

Step 6

Using this model, predict the pitch of the noise at 19 °C.

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Answer

Plugging t=19t = 19 into the regression equation:

v=a+btv = a + bt

v1.4+0.66919v \approx 1.4 + 0.669 \cdot 19

This results in:

v1.649v \approx 1.649

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