Question

1. the table below shows the lengths and corresponding ideal weights of sand sharks.

length	60	62	64	66	68	70	72

predict the weight of a sand shark whose length is 75 inches.
equation________________________ r = __ weight expected____
describe the correlation ___________________________________________

Explanation:

Step1: Calculate mean of length (x) and weight (y)

First, find the mean of the length values (\(x\)):
\(x\) values: 60, 62, 64, 66, 68, 70, 72
\(\bar{x} = \frac{60 + 62 + 64 + 66 + 68 + 70 + 72}{7} = \frac{462}{7} = 66\)

Next, find the mean of the weight values (\(y\)):
\(y\) values: 105, 114, 124, 131, 139, 149, 158
\(\bar{y} = \frac{105 + 114 + 124 + 131 + 139 + 149 + 158}{7} = \frac{920}{7} \approx 131.4286\)

Step2: Calculate slope (m) of the regression line

The formula for slope \(m\) is:
\(m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}\)

First, compute \((x_i - \bar{x})\) and \((y_i - \bar{y})\) for each data point:

| \(x_i\) | \(y_i\) | \(x_i - \bar{x}\) | \(y_i - \bar{y}\) | \((x_i - \bar{x})(y_i - \bar{y})\) | \((x_i - \bar{x})^2\) |
|---------|---------|-------------------|-------------------|-----------------------------------|-----------------------|
| 60 | 105 | -6 | -26.4286 | \( (-6)(-26.4286) \approx 158.5716 \) | 36 |
| 62 | 114 | -4 | -17.4286 | \( (-4)(-17.4286) \approx 69.7144 \) | 16 |
| 64 | 124 | -2 | -7.4286 | \( (-2)(-7.4286) \approx 14.8572 \) | 4 |
| 66 | 131 | 0 | -0.4286 | \( 0 \times (-0.4286) = 0 \) | 0 |
| 68 | 139 | 2 | 7.5714 | \( 2 \times 7.5714 \approx 15.1428 \) | 4 |
| 70 | 149 | 4 | 17.5714 | \( 4 \times 17.5714 \approx 70.2856 \) | 16 |
| 72 | 158 | 6 | 26.5714 | \( 6 \times 26.5714 \approx 159.4284 \) | 36 |

Now, sum the columns:
\(\sum (x_i - \bar{x})(y_i - \bar{y}) \approx 158.5716 + 69.7144 + 14.8572 + 0 + 15.1428 + 70.2856 + 159.4284 = 488\)

\(\sum (x_i - \bar{x})^2 = 36 + 16 + 4 + 0 + 4 + 16 + 36 = 112\)

Thus, \(m = \frac{488}{112} \approx 4.3571\)

Step3: Calculate y-intercept (b) of the regression line

The formula for \(b\) is:
\(b = \bar{y} - m\bar{x}\)

Substitute \(\bar{x} = 66\), \(\bar{y} \approx 131.4286\), and \(m \approx 4.3571\):

\(b = 131.4286 - (4.3571)(66) \approx 131.4286 - 287.5686 \approx -156.14\)

Step4: Formulate the regression equation

The linear regression equation is:
\(y = mx + b \approx 4.3571x - 156.14\)

Step5: Predict weight for \(x = 75\)

Substitute \(x = 75\) into the regression equation:

\(y = 4.3571(75) - 156.14 \approx 326.7825 - 156.14 \approx 170.64\)

Step6: Calculate correlation coefficient (r)

The formula for \(r\) is:
\(r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}}\)

First, compute \(\sum (y_i - \bar{y})^2\):

From the table, we already have \((y_i - \bar{y})\) values. Squaring them:

| \(y_i - \bar{y}\) | \((y_i - \bar{y})^2\) |
|-------------------|-----------------------|
| -26.4286 | \( (-26.4286)^2 \approx 698.47 \) |
| -17.4286 | \( (-17.4286)^2 \approx 303.75 \) |
| -7.4286 | \( (-7.4286)^2 \approx 55.18 \) |
| -0.4286 | \( (-0.4286)^2 \approx 0.18 \) |
| 7.5714 | \( (7.5714)^2 \approx 57.33 \) |
| 17.5714 | \( (17.5714)^2 \approx 308.75 \) |
| 26.5714 | \( (26.5714)^2 \approx 706.05 \) |

Summing these: \( 698.47 + 303.75 + 55.18 + 0.18 + 57.33 + 308.75 + 706.05 \approx 2129.71 \)

Now, plug into \(r\):

\(r = \frac{488}{\sqrt{112 \times 2129.71}} \approx \frac{488}{\sqrt{238527.52}} \approx \frac{488}{488.39} \approx 0.999\)

Step7: Describe the correlation

Since \(r \approx 0.999\), which is very close to \(1\), there is a **strong positive correlation** between length and weight.

Answer:

• Equation: \( y \approx 4.36x - 156.14 \) (or more precise \( y \approx 4.3571x - 156.14 \))
• \( r \approx 0.999 \)
• Weight expected: \( \approx 170.64 \) (or ~171)
• Correlation: Strong positive correlation

(Note: Rounding may vary slightly based on calculation precision.)