The answer is not 76.56. That being said, what else could it be? A sample o 80 w

Question

The answer is not 76.56. That being said, what else could it be?

A sample o 80 women is obtained, and their heights (in inches and pulse rates in beats per minute are measure The in ear co elation coeffice tS·218 and equation of the regression line is y = 17.6 + 0.880x, where x represents height. The mean of the 80 heights is 63.1 in and the mean of the 80 pulse rates is 74.4 beats per minute. Find the best predicted pulse rate of a woman who is 67 in tall. Use a significance level of = 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted pulse rate of a woman who is 67 in tall is beats per minute. (Type an integer or decimal rounded to two decimal places as needed.)

Explanation / Answer

The given regression line is: y = 17.6 + 0.880*x where y = pulse rate, x = height

The mean values of x and y are 63.1 and 74.4 respectively.

Any regression line always passes through the mean of x and y. Putting the value of x-mean and y-mean and subtracting it will give a new updated and accurate intercept.

Intercept = y - 0.880*x = 74.4 - 0.880*63.1 = 74.4 - 55.5 = 18.9

So, the new regression line is:  y = 18.9 + 0.880*x

when x = 67 then y = 18.9 + 0.880*67 = 18.9 + 58.96 = 77.86

Best predicted pulse rate = 77.86

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