Predicted 0 0756 pieces s 0 636 yes they have changed

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Predicted price = 15 .03 + 0 .0756 (pieces), ࠵? S = 0 .636 ; yes, they have changed quite a bit. Chapter 10.3 Exercise Question 14 Crickets Consider the following two scatterplots based on data gathered in a study of 30 crickets, with temperature measured in degrees Fahrenheit and chirp frequency measured in chirps per minute. If the goal is to predict temperature based on a cricket's chirps per minute, which is the appropriate scatterplot to examine—the one on the left or the one on the right? Explain briefly. One of the following is the correct equation of the least squares line for predicting temperature from chirps per minute: A. predicted temperature = 35.78 + 0.25 chirps per minute B. predicted temperature = −131.23 + 3.81 chirps per minute C. predicted temperature = 83.54 − 0.25 chirps per minute Which is the correct equation? Circle your answer and explain briefly. Use the correct equation to predict the temperature when the cricket is chirping at 100 chirps per minute. Interpret the value of the slope coefficient, in this context, for whichever equation you think is the correct one.
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Chapter 10.3 Exercise Question 15 Cat jumping Harris and Steudel (2002) studied factors that might be associated with the jumping performance of domestic cats. They studied 18 cats, using takeoff velocity (in centimeters per second) as the response variable. They used body mass (in grams), hind limb length (in centimeters), muscle mass (in grams), and percent body fat in addition to sex as potential explanatory variables. The data can be found in the CatJumping data file. A scatterplot of takeoff velocity vs. body mass is shown in the figure for Exercise 10.3.15. Describe the association between these variables. Moderately strong negative linear association Use the Corr/Regression applet to determine the equation of the least squares line for predicting a cat's takeoff velocity from its mass. Predicted takeoff velocity = 394 .47 − 0 .0122 (bodymass) Interpret the value of the slope coefficient in this context. For each additional gram of body mass the cat's predicted takeoff velocity decreases 0.0122 cm/sec. Interpret the value of the intercept coefficient. Is this a context in which the intercept coefficient is meaningful? For a cat whose body mass is zero the takeoff velocity is 394.47 cm/sec; this doesn't make sense in the context of the study because a cat wouldn't have a body mass of zero. Determine the proportion of variability in takeoff velocity that is explained by the least squares line with mass. 24.6% of the variability seen in takeoff velocity is explained by the linear association with the body mass of the cat. Chapter 10.3 Exercise Question 16 Reconsider the previous exercise and the CatJumping data file.
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Determine the predicted takeoff velocity for a cat with a mass of 5,000 grams (which is about 11 pounds). 333.47 cm/sec Determine the predicted takeoff velocity for a cat with a mass of 10,000 grams. Also explain why it's not advisable to have much confidence in this prediction.
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  • Summer '18
  • Null hypothesis, Statistical hypothesis testing, Correlation and dependence, Pearson product-moment correlation coefficient

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