A random sample of 50 salmon was caught by a scientist - Edexcel - A-Level Maths Statistics - Question 1 - 2011 - Paper 1
Question 1
A random sample of 50 salmon was caught by a scientist. He recorded the length l cm and weight w kg of each salmon.
The following summary statistics were calculated... show full transcript
Worked Solution & Example Answer:A random sample of 50 salmon was caught by a scientist - Edexcel - A-Level Maths Statistics - Question 1 - 2011 - Paper 1
Step 1
Find \(S_{l} \) and \(S_{w} \)
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Answer
To find (S_{l} ) and (S_{w} ), we use the following formulas:
The formula for (S_{l}) is:
Sl=∑l2−n(∑l)2
Substituting the known values:
Sl=327754.5−50(4027)2Sl=327754.5−5016216029Sl=327754.5−324320.58≈3433.92
The formula for (S_{w}) is:
Sw=∑w2−n(∑w)2
We have (S_{ww} = 289.6), thus:
Sw=289.6−50(357.1)2Sw=289.6−50127056.41Sw=289.6−2541.128≈289.6−2451.13≈2863.46
Step 2
Calculate, to 3 significant figures, the product moment correlation coefficient between \(l \) and \(w \)
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Answer
The product moment correlation coefficient (r) is calculated using:
r=(n∑l2−(∑l)2)(n∑w2−(∑w)2)n∑lw−∑l∑w
Where (n = 50):
Calculating the numerator:
Numerator=50×29330.5−4027×357.1=1466525−1437067.7=2947.3
Calculating the denominator:
Denominator=(50×327754.5−(4027)2)(50×289.6−(357.1)2)=(16387725−16216029)(14480−127056.41)=171696×1781.59
Calculate:
≈550668.8764≈740.36
Substituting back:
r=740.362947.3≈0.398≈0.57 to 3 significant figures
Step 3
Give an interpretation of your coefficient
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Answer
The calculated correlation coefficient of approximately 0.57 indicates a moderate positive correlation between the length and weight of the salmon. This suggests that, generally, as the length of the salmon increases, their weight also tends to increase.