Question

consider the bevariate data below about Advanced Mathematics and English results for a 2015 examination scored by 14 students in a particular school.The raw score of the examination was out of 100 marks.
Questions:
a)Draw a scatter graph
b)Draw a line of Best Fit
c)Predict the Advance Mathematics mark of a student who scores 30 of of 100 in English.
d)calculate the correlation using the Pearson's Correlation Coefficient Formula
e) Determine the strength of the correlation

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Answer 1

Answer:

Explained below.

Step-by-step explanation:

Enter the data in an Excel sheet.

(a)

Go to Insert → Chart → Scatter.

Select the first type of Scatter chart.

The scatter plot is attached below.

(b)

The scatter plot with the line of best fit is attached below.

The line of best fit is:

$y=-0.8046x+103.56$

(c)

Compute the value of x for y = 30 as follows:

$y=-0.8046x+103.56$

$30=-0.8046x+103.56\\\\0.8046x=103.56-30\\\\x=\frac{73.56}{0.8046}\\\\x\approx 91.42$

Thus, the Advance Mathematics mark of a student who scores 30 out of 100 in English is 91.42.

(d)

The Pearson's Correlation Coefficient is:

$r=\frac{n\cdot \sum XY-\sum X\cdot \sum Y}{\sqrt{[n\cdot \sum X^{2}-(\sum X)^{2}][n\cdot \sum Y^{2}-(\sum Y)^{2}]}}$

  $=\frac{14\cdot 44010-835\cdot 778}{\sqrt{[14\cdot52775-(825)^{2}][14\cdot 47094-(778)^{2}]}}\\\\= -0.7062\\\\\approx -0.71$

Thus, the Pearson's Correlation Coefficient is -0.71.

(e)

A correlation coefficient between ± 0.50 and ±1.00 is considered as a strong correlation.

The correlation between Advanced Mathematics and English results is -0.71.

This implies that there is a strong negative correlation.