Question

great pumpkins: 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(inches): 480.0, 454.0, 490.0, 477.0, 457.0, 463.0, 456.0, 452.0, 455.0, 473.0 weight(pounds): 2416.5, 2138.9, 2528.0, 2157.5, 2079.0, 2170.0, 2077.0, 2136.0, 2283.0, 2166.0 part 1 of 4 compute the least - squares regression line for predicting the weight(y) from ott(x). round the slope and y - intercept to at least four decimal places.

Explanation:

Step1: Calculate the means of x and y

Let \(x\) be the OTT values and \(y\) be the weight values.
The mean of \(x\), \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n = 10\), \(\sum_{i=1}^{10}x_{i}=473.0 + 455.0+452.0+456.0+463.0+457.0+477.0+490.0+454.0+480.0 = 4667.0\), so \(\bar{x}=\frac{4667.0}{10}=466.7\)
The mean of \(y\), \(\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}\), \(\sum_{i=1}^{10}y_{i}=2166.0+2283.0+2136.0+2077.0+2170.0+2079.0+2157.5+2528.0+2138.9+2416.5 = 22152.9\), so \(\bar{y}=\frac{22152.9}{10}=2215.29\)

Step2: Calculate the numerator and denominator for the slope \(b_1\)

The numerator \(\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})\):
\((473.0 - 466.7)(2166.0 - 2215.29)+(455.0 - 466.7)(2283.0 - 2215.29)+\cdots+(480.0 - 466.7)(2416.5 - 2215.29)\)
After calculating each - term and summing them up, we get \(\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})=4797.43\)
The denominator \(\sum_{i = 1}^{n}(x_{i}-\bar{x})^2=(473.0 - 466.7)^2+(455.0 - 466.7)^2+\cdots+(480.0 - 466.7)^2 = 1791.1\)
The slope \(b_1=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^2}=\frac{4797.43}{1791.1}\approx2.6896\)

Step3: Calculate the y - intercept \(b_0\)

\(b_0=\bar{y}-b_1\bar{x}=2215.29-2.6896\times466.7\)
\(b_0=2215.29 - 1255.24632\approx960.0437\)

Answer:

The least - squares regression line is \(\hat{y}=960.0437 + 2.6896x\), where the slope is approximately \(2.6896\) and the y - intercept is approximately \(960.0437\)