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When growing giant pumpkins for competitions, growers need to keep track of the weights of the pumpkins while they are growing. It is difficult to weigh a large pumpkin before it is harvested, so a method has been developed for estimating the weight. The grower measures around the pumpkin both horizontally and vertically, then adds the results. This is called the OTT (over the top) measurement and is used to predict the weight of the pumpkin. Following are the OTT measurements and actual weights of the 10 largest pumpkins entered into official competitions in a recent year.

OTT and pounds
490.0 / 2433.9
457 / 2079
477/2157.5
456/2077
469/2469
450/2017.5
451/2114
480/2416.5
454/2020.5
490/2528

(a) Compute the least-squares regression line for predicting weight (y) from OTT (x). Round the slope and y-intercept to four decimal places as needed.

11 answers

To compute the least-squares regression line for predicting weight (y) from OTT (x), we need to perform the following calculations:

  1. Organize the data points:

    • \(x\) (OTT) values: 490.0, 457, 477, 456, 469, 450, 451, 480, 454, 490
    • \(y\) (weight) values: 2433.9, 2079, 2157.5, 2077, 2469, 2017.5, 2114, 2416.5, 2020.5, 2528

  2. Calculate the necessary sums:

    • \(n = 10\) (number of data points)
    • \( \sum x \)
    • \( \sum y \)
    • \( \sum x^2 \)
    • \( \sum xy \)

Perform the calculations: \[ \sum x = 490.0 + 457 + 477 + 456 + 469 + 450 + 451 + 480 + 454 + 490 = 4324.0 \] \[ \sum y = 2433.9 + 2079 + 2157.5 + 2077 + 2469 + 2017.5 + 2114 + 2416.5 + 2020.5 + 2528 = 23984.0 \] \[ \sum x^2 = 490.0^2 + 457^2 + 477^2 + 456^2 + 469^2 + 450^2 + 451^2 + 480^2 + 454^2 + 490^2 \] Calculating each term:

  • \(490.0^2 = 240100\)
  • \(457^2 = 208849\)
  • \(477^2 = 227729\)
  • \(456^2 = 207936\)
  • \(469^2 = 219961\)
  • \(450^2 = 202500\)
  • \(451^2 = 203401\)
  • \(480^2 = 230400\)
  • \(454^2 = 206116\)
  • \(490^2 = 240100\)

Adding them up: \[ \sum x^2 = 240100 + 208849 + 227729 + 207936 + 219961 + 202500 + 203401 + 230400 + 206116 + 240100 = 2081731 \]

Now calculate \(\sum xy\): \[ \sum xy = (490.0 \cdot 2433.9) + (457 \cdot 2079) + (477 \cdot 2157.5) + (456 \cdot 2077) + (469 \cdot 2469) + (450 \cdot 2017.5) + (451 \cdot 2114) + (480 \cdot 2416.5) + (454 \cdot 2020.5) + (490 \cdot 2528) \] Calculating each term:

  • \(490.0 \cdot 2433.9 = 1193361\)
  • \(457 \cdot 2079 = 950103\)
  • \(477 \cdot 2157.5 = 1027987.5\)
  • \(456 \cdot 2077 = 947592\)
  • \(469 \cdot 2469 = 1154951\)
  • \(450 \cdot 2017.5 = 907875\)
  • \(451 \cdot 2114 = 953814\)
  • \(480 \cdot 2416.5 = 1165920\)
  • \(454 \cdot 2020.5 = 917870\)
  • \(490 \cdot 2528 = 1238720\)

Adding them up: \[ \sum xy = 1193361 + 950103 + 1027987.5 + 947592 + 1154951 + 907875 + 953814 + 1165920 + 917870 + 1238720 = 11294993.5 \]

  1. Now we can calculate the slope (m) and y-intercept (b) of the regression line using the formulas: \[ m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2} \] \[ b = \frac{\sum y - m(\sum x)}{n} \]

Substituting the sums: \[ m = \frac{10(11294993.5) - (4324)(23984)}{10(2081731) - (4324)^2} \] \[ = \frac{112949935 - 103762656}{20817310 - 18661076} = \frac{9189279}{2156234} \approx 4.26 (rounded to 4.2611) \]

Now calculate \(b\): \[ b = \frac{23984 - (4.2611)(4324)}{10} = \frac{23984 - 18483.26}{10} = \frac{5498.74}{10} \approx 549.874 \]

Thus, the least-squares regression equation is: \[ y = 4.2611x + 549.874 \]

Final result:

  • Slope (m): 4.2611
  • Intercept (b): 549.8740

The least-squares regression line for predicting weight \(y\) from OTT \(x\) is: \[ y \approx 4.2611x + 549.874 \]

No the answer is -2949.8733+11.0851x

Is it possible to interpret the y-intercept?
[yes or no?] because the y-intercept is [positive or negative?] and weights cannot be [positive or negative?]

Yes, it is possible to interpret the y-intercept in the context of this problem. The y-intercept is -2949.8733, which is negative.

Since weights cannot be negative, this negative y-intercept indicates that when the OTT measurement is zero (which doesn't correspond to a realistic scenario, as a pumpkin cannot have an OTT of zero), the predicted weight would be negative. This suggests that while the regression line can fit the data within the range of OTT measurements observed, it does not have a meaningful interpretation for very low values of OTT, as that scenario is outside the domain of the actual data.

In summary:

  • Is it possible to interpret the y-intercept? Yes.
  • The y-intercept is: Negative.
  • Weights cannot be: Negative.
The answer was no.

(c) If two pumpkins differ in OTT by 12 inches, by how much would you predict their weights to differ? Round the answer to two decimal places as needed.

To predict the difference in weights of two pumpkins that differ in OTT by 12 inches, we can use the slope of the least-squares regression line.

The slope of the regression line is given as \( m = 11.0851 \). This slope indicates how much the weight (y) changes for each unit change in OTT (x).

If the OTT values differ by 12 inches, the predicted difference in weight \( \Delta y \) can be calculated as:

\[ \Delta y = m \cdot \Delta x \]

Where:

  • \( m \) is the slope (11.0851),
  • \( \Delta x \) is the difference in OTT (12).

Substituting the values:

\[ \Delta y = 11.0851 \cdot 12 \] \[ \Delta y = 133.0212 \]

Rounding to two decimal places:

\[ \Delta y \approx 133.02 \]

Thus, if two pumpkins differ in OTT by 12 inches, you would predict their weights to differ by approximately 133.02 pounds.

(d) Predict the weight of a pumpkin whose OTT is 460 inches. Round the answer to two decimal places as needed.

To predict the weight of a pumpkin with an OTT of 460 inches using the least-squares regression equation:

\[ y = -2949.8733 + 11.0851x \]

where \(x\) is the OTT measurement.

Substituting \(x = 460\):

\[ y = -2949.8733 + 11.0851 \cdot 460 \]

Calculating the product \(11.0851 \cdot 460\):

\[ 11.0851 \cdot 460 = 5091.846 \]

Now substituting that back into the equation:

\[ y = -2949.8733 + 5091.846 \] \[ y \approx 5091.846 - 2949.8733 \approx 2141.9727 \]

Rounding to two decimal places, the predicted weight is:

\[ y \approx 2141.97 \]

Therefore, the predicted weight of a pumpkin whose OTT is 460 inches is approximately 2141.97 pounds.

that is incorrect

Let's re-evaluate the calculation step by step to predict the weight of a pumpkin with an OTT of 460 inches using the provided regression equation:

The regression equation is:

\[ y = -2949.8733 + 11.0851x \]

Where \(x\) represents the OTT measurement.

Substituting \(x = 460\):

\[ y = -2949.8733 + 11.0851 \cdot 460 \]

Calculating the second term:

First, calculate \(11.0851 \cdot 460\):

\[ 11.0851 \cdot 460 = 5091.846 \]

Now substitute that back into the regression equation:

\[ y = -2949.8733 + 5091.846 \]

Now perform the addition/subtraction:

\[ y = 5091.846 - 2949.8733 \approx 2141.9727 \]

Now, rounding to two decimal places gives:

\[ y \approx 2141.97 \]

If this calculation is still incorrect, I would need to double-check the given slope and intercept values in case they've changed or if inaccuracies arose with previous values.

Therefore, given the regression equation parameters:

  • Intercept: -2949.8733
  • Slope: 11.0851

Predicting weight for \(x = 460\):

\[ y = -2949.8733 + 11.0851(460) = 2141.97 \]

If there's a mistake in these values or assumptions, please provide the accurate regression coefficients or more context to rectify the prediction. Thank you!

the regression line is -2949.8733 + 11.0851x

(d) Predict the weight of a pumpkin whose OTT is 460 inches. Round the answer to two decimal places as needed.

Thank you for your patience. Let’s accurately compute the predicted weight of a pumpkin with an OTT of 460 inches using the provided regression equation:

\[ y = -2949.8733 + 11.0851x \]

Substituting \(x = 460\):

\[ y = -2949.8733 + 11.0851 \cdot 460 \]

Calculating \(11.0851 \cdot 460\):

\[ 11.0851 \cdot 460 = 5091.846 \]

Now substitute this value back into the equation:

\[ y = -2949.8733 + 5091.846 \]

Now perform the calculation:

\[ y = 5091.846 - 2949.8733 \]

Subtracting:

\[ y = 2141.9727 \]

Now, rounding to two decimal places:

\[ y \approx 2141.97 \]

Therefore, the predicted weight of a pumpkin whose OTT is 460 inches is 2141.97 pounds. If there is a specific correction or context needed for a different answer, please let me know!