Babies typically learn to crawl approximately 6 months after birth. However, it may take longer for babies to learn to crawl in the winter, when they are often bundled in clothes that restrict their movement. Thus, there may be an association between a baby's crawling age and the average temperature during the month they first try to crawl. Below are the average ages (in weeks) at which babies began to crawl for a sample of babies born in each of the 12 months of the year. In addition, the average temperature (in °F) for the month that is 6 months after the birth month is listed.
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$\begin{array} { | l | l | l | } \hline \text { Birth month } & \text { Average crawling age } & \text { Average temperature } \\\hline \text { January } & 29.84 & 66 \\\hline \text { February } & 30.52 & 73 \\\hline \text { March } & 29.70 & 72 \\\hline \text { April } & 31.84 & 63 \\\hline \text { May } & 28.58 & 52 \\\hline \text { June } & 31.44 & 39 \\\hline \text { July } & 33.64 & 33 \\\hline \text { August } & 32.82 & 30 \\\hline \text { September } & 33.83 & 33 \\\hline \text { October } & 33.35 & 37 \\\hline \text { November } & 33.38 & 48 \\\hline \text { December } & 32.32 & 57 \\\hline\end{array}$ We want to investigate whether the average age at which infants begin to crawl (y) can be predicted from the average outdoor temperature (x) 6 months after birth, when the babies are likely to begin crawling. We decide to fit a least-squares regression line to the data, with x as the explanatory variable and y as the response variable. We compute the following quantities:
R = correlation between x and y = -0.7
X?= mean of the values of x = 50.25
?= mean of the values of y = 31.77
S_x = standard deviation of the values of x = 15.85
S_y = standard deviation of the values of y = 1.76
Which of the following values is closest to the intercept of the least-squares regression line?

Question

Babies typically learn to crawl approximately 6 months after birth. However, it may take longer for babies to learn to crawl in the winter, when they are often bundled in clothes that restrict their movement. Thus, there may be an association between a baby's crawling age and the average temperature during the month they first try to crawl. Below are the average ages (in weeks) at which babies began to crawl for a sample of babies born in each of the 12 months of the year. In addition, the average temperature (in °F) for the month that is 6 months after the birth month is listed.
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\begin{array} { | l | l | l | } \hline \text { Birth month } & \text { Average crawling age } & \text { Average temperature } \\\hline \text { January } & 29.84 & 66 \\\hline \text { February } & 30.52 & 73 \\\hline \text { March } & 29.70 & 72 \\\hline \text { April } & 31.84 & 63 \\\hline \text { May } & 28.58 & 52 \\\hline \text { June } & 31.44 & 39 \\\hline \text { July } & 33.64 & 33 \\\hline \text { August } & 32.82 & 30 \\\hline \text { September } & 33.83 & 33 \\\hline \text { October } & 33.35 & 37 \\\hline \text { November } & 33.38 & 48 \\\hline \text { December } & 32.32 & 57 \\\hline\end{array}

We want to investigate whether the average age at which infants begin to crawl (y) can be predicted from the average outdoor temperature (x) 6 months after birth, when the babies are likely to begin crawling. We decide to fit a least-squares regression line to the data, with x as the explanatory variable and y as the response variable. We compute the following quantities:
R = correlation between x and y = -0.7
X?= mean of the values of x = 50.25
?= mean of the values of y = 31.77
S_x = standard deviation of the values of x = 15.85
S_y = standard deviation of the values of y = 1.76
Which of the following values is closest to the intercept of the least-squares regression line?

Quizplus · Accepted Answer

We can use the equation of the least-squares regression line to find the intercept as follows: ŷ = a + bx, where ŷ is the predicted value of y, a is the intercept, b is the slope, and x is the value of the explanatory variable. We know that R = -0.7, which means that there is a negative linear relationship between x and y. Therefore, b must be negative. We can use the formula: b = R(_y/)/(_x/) Substituting in the values, we get: b = (-0.7)(1.76/15.85)/(1/15.85) = -0.138 Next, we can use the formulas: a = Ȳ - bX̄ Substituting in the values, we get: a = 31.77 - (-0.138)(50.25) = 35.68 Therefore, the closest value to the intercept of the least-squares regression line is 35.68, which is answer choice B.

Babies Typically Learn to Crawl Approximately 6 Months After Birth