Question

Babies typically learn to crawl approximately six 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. We want to investigate if the average age at which infants begin to crawl (Y) can be predicted from the average outdoor temperature (X) six months after birth when they 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 = mean of the values of X = 50.25 = mean of the values of Y = 31.77 = standard deviation of the values of X = 15.85 = standard deviation of the values of Y = 1.76

in a study for each month of the year research sampled babies born during that month and measured the average age in weeks of first crawl they also recorded the average temperature 6 months later when babies usually start crawling for example the babies born in january on average had their first crawl at age of 29.84 weeks and the average temperature 6 months after january that july was 66 degree we should record this data point as 66 degree 29.84 weeks this was done for all 12 month so there are 12 data points
a)what are the slope and intercept of the least square regression line
b)what percent of the variation in y explanied by the regression on x
c)make a prediction for the baby first crawl when the baby was born in january at the average temp of 66 degrees what is the residual corresponding to the obsevation referred above 66 degree 29.84 week
d) one of the data points is 72 weeks y the residual for the obsevation y is -0.81 find y

Answer 1

Given:

$\bar {X} =$ 50.25; $\bar {Y} =$ 31.77; s_x =15.85; s_y =1.76 and Correlation(X, Y) =r =0.7

a)

The slope of the least square regression line is:

Slope, b1 =r(s_y/s_x) =0.7(1.76/15.85) =0.07773

​​​​​​The intercept of the least square regression line is:

Intercept, b0 = $\bar {Y} - (b1*\bar {X})$ =31.77 - (0.07773*50.25) =27.8641

b)

R-squared value =Co-efficient of determination =r² =0.7² =0.49 =0.49 =49%. So, 49% of the variation in Y was explained by the regression on X.

c)

X =66 degrees.

Predicted age, $\hat {Y}$ =b0 + b1(X) =27.8641 + 0.07773(66) = 32.99 weeks.

For X =66, the observed age is Y =29.84 weeks

Residual, e = $Y - \hat {Y}$ =29.84 - 32.99= -3.15 weeks.

d)

Given: one of the data points is Y =72 weeks, i.e., observed value is Y =72 and the residual is e = -0.81. We need to find predicted Y, i.e., $\hat {Y}$

e = $Y - \hat {Y} \implies$ $\hat {Y} =$ Y - e =72 - (-0.81) =72+0.81 =72.81 weeks.

Thus, predicted Y =72.81 weeks.

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