Question

A sample of 100 women is​ obtained, and their heights​ (in inches) and pulse rates​ (in beats per​ minute) are measured. The linear correlation coefficient is 0.239 and the equation of the regression line is y = 18.5 + 0.890x​, where x represents height. The mean of the 100 heights is 62.9 in and the mean of the 100 pulse rates is 73.4 beats per minute. Find the best predicted pulse rate of a woman who is 65 in tall. Use a significance level of α=0.01

Answer 1

Answer:

$\hat y= 18.5 +0.890(65)=76.35$

Step-by-step explanation:

For this case we have a linear model given by:

$\hat y= 18.5 +0.890 X$

Where X= represent the height.

Y= represent the beats per minute

And the slope is given by $\beta_1 =0.890$ and the intercept $\beta_0 =18.5$

We have the mean for x and y given also $\bar X= 62.9$ $\bar Y= 73.4$

And in order to find the best predicted value we just need to replace the value of X= 65, and we got this:

$\hat y= 18.5 +0.890(65)=76.35$

If we want a confidence interval for the response variable when the predictor variable is x, is given by this formula:

$\hat y \pm \sqrt{1+\frac{1}{n} +\frac{(x-\bar X)^2}{(n-1)s^2_x}}$

And if we replace we got this:

$76.35 \pm \sqrt{1+\frac{1}{100} +\frac{(65-62.9)^2}{(100-1)s^2_x}}$

But for this case since we don't have the deviation we can just express it like this.