Suppose a doctor measures the height, x, and head circumference, y, of 8 children and obtains the data below. The correlation coefficient is 0.860 and the least squares regressionline is y = 0.196x + 12.045. Complete parts (a) and (b) below.Beight, x27.5 25 26 25.5 27.75 26.5 26.25 26.75 26.75 27 27.25Head Circumference, y 17.3 17.0 17.2 16.9 17.5 17.4 17.1 17.4 17.3 17.4 17.4(a) Compute the coefficient of determination, R².R² =% (Round to one decimal place as needed.)

Given info: The correlation coefficient is given as r = 0.860.

Answered: Suppose a doctor measures the height,…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The data from the table below gives a regression that is a) reliable. b) unreliable. c) unable to determine the reliability.
Research was conducted on the amount of training for 5K and the time a contestant took to run the race. The researcher recorded the number of miles during training ( a 1 month period) and the time to complete the 5K. The results are below. Miles Trained 45 77 37 25 60 64 Time (Minutes) 32.9 28 36.2 27.5 19.5 18.3 a) Give the correlation coefficient. r = b) Give the equation of the regression line. Round the values to 4 deimal places.This is in the form y = ax + b, so the slope goes in the first box and the y-inercept in the 2nd box.y = x + c) Interpret the y-intercept (Explain it's meaning, do not just give its value.)d) Predict the time in the 5K if someone trained 37 miles.Use the rounded values entered for the slope and intercept of the regression line to compute the predicted y-value.y = e) Give the residual for 37 miles trained.This is the actual y value for the runner who trained for miles minus the predicted value.Residual=
The data is given as follow. xi 2 6 9 13 20 yi 7 18 9 26 23 The estimated regression equation for these data is = 7.6 + .9x.Compute SSE, SST, and SSR (to 1 decimal). SSE SST SSR What percentage of the total sum of squares can be accounted for by the estimated regression equation (to 1 decimal)? %What is the value of the sample correlation coefficient (to 3 decimals)?
ANSWER D ONLY
The fish in my pond have mean length 10 inches and mean weight 4 pounds. The correlation coefficient of length and weight is .9. If the length of a particular randomly selected fish is reported to be 10 inches, then what should we predict for the weight of that fish using simple linear regression? 4 10 2.5 NOT ENOUGH INFORMATION GIVEN NONE OF THE OTHERS
The statistic used to test whether individual regression coefficients are different from zero in the population is: Select one: a. F O b. b Oc. R2 O d. t Clear my choice
5
Suppose a doctor measures the height, x, and head circumference, y, of 8 children and obtains the data below. The correlation coefficient is 0.858 and the least squares regression line is y = 0.228x +11.187. Complete parts (a) and (b) below. Height, x 27.5 25.75 26.5 25.5 27.25 26.25 25.75 27.25 27 27.25 27 Head Circumference, y 17.4 17.2 17.2 16.9 17.6 17.1 17.1 17.4 17.4 17.3 17.3 (a) Compute the coefficient of determination, R². R² =% (Round to one decimal place as needed.) (b) Interpret the coefficient of determination and comment on the adequacy of the linear model. Approximately % of the variation in (Round to one decimal place as needed.) is explained by the least-squares regression model. According to the residual plot, the linear model appears to be
11. For temperature (x) and number of ice cream cones sold per hour (y). (65, 8), (70, 10), (75, 11), (80,13), (85, 12), (90, 16). Interpret the coefficient of determination. Optional Answers: 1. 88.2% of the variability in the number of cones sold is explained by the least-squares regression model. 2. 93.9% of the variability in the number of cones sold is explained by the least-squares regression model. 3. 88.2% of the variability in the temperature is explained by the least-squares regression model. 4. 93.9% of the variability in the temperature is explained by the least-squares regression model.
10.2.13 Question Help The data show the chest size and weight of several bears. Find the regression equation, letting chest size be the independent (x) variable. Then find the best predicted weight of a bear with a chest size of 63 inches. Is the result close to the actual weight of 562 pounds? Use a significance level of 0.05. Chest size (inches) Weight (pounds) 58 50 65 59 59 48 414 312 499 450 456 260 Click the icon to view the critical values of the Pearson correlation coefficient r. What is the regression equation? y =+x (Round to one decimal place as needed.)
Prev The table below gives the age and bone density for five randomly selected women. Using this data, consider the equation of the regression line, y = bo + b₁x, for predicting a woman's bone density based on her age. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, In practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant. Age Bone Density 61 62 68 69 40 357 350 343 340 315 Step 1 of 6: Find the estimated slope. Round your answer to three decimal places. Table Copy Data Next

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