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A clothes shop manager records the weekly sales figures, £ S, and the average weekly temperature, t °C, for 6 weeks during the summer - Edexcel - A-Level Maths Statistics - Question 1 - 2017 - Paper 1

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A clothes shop manager records the weekly sales figures, £ S, and the average weekly temperature, t °C, for 6 weeks during the summer. The sales figures were coded s... show full transcript

Worked Solution & Example Answer:A clothes shop manager records the weekly sales figures, £ S, and the average weekly temperature, t °C, for 6 weeks during the summer - Edexcel - A-Level Maths Statistics - Question 1 - 2017 - Paper 1

Step 1

Find $S_{w}$ and $S_{u}$

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Answer

Sw=784119426=78449=735.S_{w} = 784 - 119 \cdot \frac{42}{6} = 784 - 49 = 735. Su=50×106(42)26=50×106294=50,000,000294=49,999,706.S_{u} = 50 \times 10^6 - \frac{(42)^2}{6}= 50 \times 10^6 - 294 = 50,000,000 - 294 = 49,999,706.

Step 2

Write down the value of $S_{s}$ and the value of $S_{t}$

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Ss=7846=130.67S_{s} = \frac{784}{6} = 130.67, St=1196=19.83S_{t} = \frac{119}{6} = 19.83.

Step 3

Find the product moment correlation coefficient between $s$ and $t$

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Answer

The product moment correlation coefficient, rr, is calculated as: r=(wiwˉ)(titˉ)(wiwˉ)2(titˉ)2r = \frac{\sum (w_i - \bar{w})(t_i - \bar{t})}{\sqrt{\sum (w_i - \bar{w})^2 \sum (t_i - \bar{t})^2}} Substituting the values, we find that r0.801.r \approx -0.801.

Step 4

State, giving a reason, whether or not your value of the correlation coefficient supports the manager's belief

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Answer

The value of r0.801r \approx -0.801 indicates a strong negative correlation between sales and temperature, suggesting that as the temperature increases, sales tend to decrease. This supports the manager's belief about the relationship between the two variables.

Step 5

Find the equation of the regression line of $w$ on $t$, giving your answer in the form $w = a + bt$

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Answer

To find the regression line, we use: w=a+btw = a + bt Where b=wtnwˉtˉt2ntˉ2b = \frac{\sum wt - n \cdot \bar{w} \cdot \bar{t}}{\sum t^{2} - n \cdot \bar{t}^{2}} and a=wˉbtˉa = \bar{w} - b \cdot \bar{t}. Using the calculated values, we find: b=0.065 and a=20.0000.065(19.83)=20.0001.289=18.711.b = -0.065\ and \ a = 20.000 - 0.065(19.83) = 20.000 - 1.289 = 18.711. Thus, w=18.7110.065tw = 18.711 - 0.065t.

Step 6

Hence find the equation of the regression line of $s$ on $t$, giving your answer in the form $s = c + dt$

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Answer

Using the previous result, we find: s=c+dts = c + d \cdot t where d=0.065×1000=65d = 0.065 \times 1000 = 65 and c=18.711×1000=18711.c = 18.711 \times 1000 = 18711. So, the regression equation is: s=1871165t.s = 18711 - 65t.

Step 7

Using your equation in part (f), interpret the effect of a 1°C increase in average weekly temperature on weekly sales during the summer

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Answer

From the equation s=1871165ts = 18711 - 65t, a 1°C increase in average weekly temperature corresponds to a decrease of £65 in weekly sales, indicating that higher temperatures tend to reduce sales significantly.

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