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 α = 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?
overhead width | 7.0 | 7.5 | 9.8 | 9.4 | 8.6 | 8.4
weight | 109 | 194 | 250 | 204 | 193 | 192
click here to view a table of critical values for the correlation coefficient.

construct a scatterplot. choose the correct graph below.
○ a. ○ b. ○ c. ○ d.
(graphs of scatterplots with weight (kg) on y - axis and width (cm) on x - axis, with different point distributions)

the linear correlation coefficient is r =
(round to three decimal places as needed.)

the critical values are r =

Explanation:

Step1: Identify Data Points

Let \( x \) be overhead width (cm) and \( y \) be weight (kg). The data points are: \((7.0, 109)\), \((7.5, 194)\), \((9.8, 250)\), \((9.4, 204)\), \((8.6, 193)\), \((8.4, 192)\).

Step2: Calculate Necessary Sums

• \( n = 6 \)
• \( \sum x = 7.0 + 7.5 + 9.8 + 9.4 + 8.6 + 8.4 = 50.7 \)
• \( \sum y = 109 + 194 + 250 + 204 + 193 + 192 = 1142 \)
• \( \sum xy = (7.0×109) + (7.5×194) + (9.8×250) + (9.4×204) + (8.6×193) + (8.4×192) \)

\( = 763 + 1455 + 2450 + 1917.6 + 1659.8 + 1612.8 = 9858.2 \)

• \( \sum x^2 = 7.0^2 + 7.5^2 + 9.8^2 + 9.4^2 + 8.6^2 + 8.4^2 \)

\( = 49 + 56.25 + 96.04 + 88.36 + 73.96 + 70.56 = 434.17 \)

• \( \sum y^2 = 109^2 + 194^2 + 250^2 + 204^2 + 193^2 + 192^2 \)

\( = 11881 + 37636 + 62500 + 41616 + 37249 + 36864 = 227746 \)

Step3: Compute Correlation Coefficient \( r \)

The formula for \( r \) is:

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

Substitute values:

• Numerator: \( 6×9858.2 - 50.7×1142 = 59149.2 - 57909.4 = 1239.8 \)
• Denominator part 1: \( 6×434.17 - 50.7^2 = 2605.02 - 2570.49 = 34.53 \)
• Denominator part 2: \( 6×227746 - 1142^2 = 1366476 - 1304164 = 62312 \)
• Denominator: \( \sqrt{34.53×62312} \approx \sqrt{2151733.36} \approx 1466.88 \)
• \( r = \frac{1239.8}{1466.88} \approx 0.845 \)

Step4: Find Critical Values

For \( n = 6 \) and \( \alpha = 0.01 \), from critical values table, critical \( r \) values are \( \pm 0.917 \).

Step5: Scatterplot Analysis

Plot \( x \) (width) on x - axis and \( y \) (weight) on y - axis. The points should show a positive trend. Looking at the options, the correct scatterplot (e.g., option with points increasing as width increases) – assuming option C (or the one with positive trend) is correct (since data has positive correlation).

Answer:

Scatterplot: (Assuming correct is option C, e.g., "C. [Graph with positive trend]")
Linear correlation coefficient \( r \approx \boxed{0.845} \)
Critical values \( r = \pm 0.917 \)
Since \( |r| = 0.845 < 0.917 \), there is not sufficient evidence at \( \alpha = 0.01 \) to conclude a linear correlation.