Question

listed below are the overhead widths (in cm) of seals measured from photographs and the weights (in kg) of the seals. construct a scatterplot, find the value of the linear correlation coefficient r, and find the critical values of r using \\( \alpha = 0.01 \\). is there sufficient evidence to conclude that there is a linear correlation between overhead widths of seals from photographs and the weights of the seals?
\\( \

$$\begin{array} { c | c c c c c c } { \\text { overhead width } } & { 7.1 } & { 7.6 } & { 9.8 } & { 9.4 } & { 8.9 } & { 8.3 } \\\\ \\hline \\text { weight } & { 113 } & { 188 } & { 249 } & { 201 } & { 206 } & { 190 } \\end{array}$$

\\)
click here to view a table of critical values for the correlation coefficient.
construct a scatterplot. choose the correct graph below.
o a.
o b.
o c.
o d.

Explanation:

Step1: List paired data points

Let \(x\) = overhead width (cm): [7.1,7.6,9.8,9.4,8.9,8.3]; \(y\) = weight (kg): [113,188,249,201,206,190]

Step2: Calculate sums for \(r\)

\(\sum x = 7.1+7.6+9.8+9.4+8.9+8.3 = 51.1\)
\(\sum y = 113+188+249+201+206+190 = 1147\)
\(\sum xy = (7.1×113)+(7.6×188)+(9.8×249)+(9.4×201)+(8.9×206)+(8.3×190) = 792.3+1428.8+2440.2+1889.4+1833.4+1577 = 9961.1\)
\(\sum x^2 = 7.1^2+7.6^2+9.8^2+9.4^2+8.9^2+8.3^2 = 50.41+57.76+96.04+88.36+79.21+68.89 = 440.67\)
\(\sum y^2 = 113^2+188^2+249^2+201^2+206^2+190^2 = 12769+35344+62001+40401+42436+36100 = 229051\)
\(n = 6\)

Step3: Compute linear correlation coefficient \(r\)

$$ r = \frac{n\sum xy - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} $$

Numerator: \(6×9961.1 - 51.1×1147 = 59766.6 - 58611.7 = 1154.9\)
Denominator part1: \(6×440.67 - 51.1^2 = 2644.02 - 2611.21 = 32.81\)
Denominator part2: \(6×229051 - 1147^2 = 1374306 - 1315609 = 58697\)
Denominator: \(\sqrt{32.81×58697} ≈ \sqrt{1926848.57} ≈ 1388.11\)
\(r ≈ \frac{1154.9}{1388.11} ≈ 0.832\)

Step4: Find critical values (\(\alpha=0.01, n=6\))

Critical values from table: \(\pm 0.917\)

Step5: Check correlation evidence

\(|r| = 0.832 < 0.917\), so no sufficient evidence.

Answer:

Scatterplot: (Assuming correct graph matches the data trend; since graphs are not visible, typical trend is positive association)
Linear correlation coefficient \(r ≈ 0.83\)
Critical values: \(\pm 0.917\)
No sufficient evidence for linear correlation.