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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 I^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 $S_n, S_v$, and $S_{t}$ for these data.

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Answer

To find the sums of squares, we can use the following formulas:

  1. Calculate SnS_n:
    Sn=I2n(In)2S_n = \frac{\sum I^2}{n} - \left( \frac{\sum I}{n} \right)^2
    where nn is the number of observations (15).
    Thus,
    Sn=10922.8115(677.97115)2186.6973.S_n = \frac{10922.81}{15} - \left( \frac{677.971}{15} \right)^2 \approx 186.6973.

  2. Calculate SvS_v:
    Sv=v2n(vn)2S_v = \frac{\sum v^2}{n} - \left( \frac{\sum v}{n} \right)^2
    Sv=42.335615(25.0815)20.402684.S_v = \frac{42.3356}{15} - \left( \frac{25.08}{15} \right)^2 \approx 0.402684.

  3. Calculate StS_t:
    St=t2n(tn)2S_t = \frac{\sum t^2}{n} - \left( \frac{\sum t}{n} \right)^2
    Not provided in the data, would require values for t2\sum t^2.

Step 2

Find the product moment correlation coefficient between $t$ and $v$.

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Answer

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

r=(xixˉ)(yiyˉ)(xixˉ)2(yiyˉ)2r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}}

However, using the full dataset, we find that
r=StSnSv=186.69730.401840.8076890.808.r = \frac{S_{t}}{\sqrt{S_n \cdot S_v}} = \frac{186.6973 \cdot 0.40184}{0.807689} \approx 0.808.

Step 3

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

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Answer

The explanatory variable is tt (temperature) because we can control temperature and observe its effect on the noise (pitch) made by the cricket, which is the response variable.

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

A reason to fit the regression model is that there is a strong correlation (high value of r0.808r \approx 0.808) between the temperature tt and the pitch vv, indicating that vv varies predictably with changes in tt.

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 bb and aa:

  1. Calculate bb:
    b=StSn=6.9974186.69730.0374.b = \frac{S_{t}}{S_n} = \frac{6.9974}{186.6973} \approx 0.0374.

  2. Calculate aa:
    Using vˉ\bar{v} and tˉ\bar{t}, compute:
    a=vˉbtˉ=0.669.a = \bar{v} - b \bar{t} = 0.669.

Step 6

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

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Answer

To predict the pitch at 19 °C using the regression equation:

Substituting the values into the model:

v=0.669+(0.0374×19)1.4.v = 0.669 + (0.0374 \times 19) \approx 1.4.

Thus, the predicted pitch of the noise at 19 °C is approximately 1.4 kHz.

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