1995200020052010Year00270097007Colonies The summary statistics for number of colonies and year are listed below. The correlation between number of colonies and year is -0.4098.Year: mean =2003.5000, standard deviation = 5.3385Colonies: mean =2546.0556, standard deviation = 108.86983. The equation for the regression line is y =+4.Complete the following sentence to interpret the slope of the regression line:In 1 year, the number of colonies is expected to decrease v by5. Use the regression equation to estimate the number of colonies in 2009.6. The actual number of colonies in 2009 was 2498. Complete the following sentence:The residual for yr isThis means the number of colonies in 2009 isA. lower thanB. the same asC. higher thanthe number of colonies predicted by the regression model.

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Answered: The summary statistics for number of…

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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5.34 Please complete both parts. Do Part (a) "by hand": the mean height of the brothers is 69 inches and the standard deviation is 2.72 inches, the mean height of the sisters is 64 inches and the standard deviation is 2.57 inches, and the correlation is 0.558. Draw the scatterplot by hand. The regression line can be approximate
Consider a regression that uses X = # of 60 pound bench press repetitions to predict Y = maximum bench press amount in pounds…..for the ith student among a sample of high school athletes. The estimated predicted sample regression is as follows Maxi= 68.91 + .6885Repsi *the ei is left off for simplicity The table below lists the actual data from which the regression above was estimated. For example, a student in the study who trained by doing 12 “60 pound reps” accomplished a maximum bench press amount of 76 pounds. Xi Yi 12 76 15 82 17 79 3 71 Predict the average maximum bench press for a student that does 17 “60 pound reps” a) a. around 68.85 pounds b) b. around 85.00 pounds c) c. around 80.61 pounds d) d. around 68.91 pounds
For a sample of 306 students in a basic business statistics course, the sample regression liney = 58.813 + 0.2875xwas obtained. Here,y = final student score at the end of the coursex = score on a diagnostic statistics test given at the beginning of the courseThe coefficient of determination was 0.1158, and the estimated standard deviation of the estimator of the slope of the population regression line was 0.04566.a. Interpret the slope of the sample regression line.b. Interpret the coefficient of determination.c. The information given allows the null hypothesis that the slope of the population regression line is 0 to be tested in two different ways against the alternative that it is positive. Carry out these tests and show that they reach the same conclusion.
Use the summary statistics to calculate the regression line for this data. Explain each coefficient. What is the total amount of weight would you expect a weightlifter with a body weight of 84kg to lift? At the 5% significance level, is there evidence that the body weight of the athlete is a useful predictor for the total amount an athlete can lift? Assume the assumptions can be verified.
For a sample of eight bears, researchers measured the distances around the bears' chests and weighed the bears. Minitab was used to find that the value of the linear correlation coefficient is r= 0.846. Using a = 0.05, determine if there is a linear correlation between chest size and weight. What proportion of the variation in weight can be explained by the linear relationship between weight and chest size? Click here to view a table of critical values for the correlation coefficient. .... a. Is there a linear correlation between chest size and weight? A. No, because the absolute value of the test statistic exceeds the critical value of 0.707. B. Yes, because the absolute value of the test statistic exceeds the critical value of 0.707. C. Yes, because the test statistic falls between the critical values of - 0.707 and 0.707. D. The answer cannot be determined from the given information. b. What proportion of the variation in weight can be explained by the linear relationship between…
Find the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 2.1 cm. Can the prediction be correct? What is wrong with predicting the weight in this case? Use a significance level of 0.05. Overhead_Width_(cm) Weight_(kg)8.4 1957.6 1919.7 2769.5 2418.7 2288.3 216 The regression equation is y=---+----x. (Round to one decimal place as needed.) The best predicted weight for an overhead width of 2.1 cm is = ___ kg. Can the prediction be correct? What is wrong with predicting the weight in this case? A. The prediction cannot be correct because a negative weight does not make sense. The regression does not appear to be useful for making predictions. B. The…
mean of x = 2.882, standard deviation = 1.634 Mean of y = 2.588, standard deviation = 0.246 The correlation coefficient is 0.159 slope coefficient for regression line = .0239 y intercept = 2.52 write the regression model
21
Define the term relationship as a perfect positive cross-correlation?
Listed below are altitudes (thousands of feet) and outside air temperatures (°F) recorded during a flight. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 95% confidence level with the altitude of 6327 ft (or 6.327 thousand feet). Altitude 8 15 22 28 31 33 Temperature 56 39 24 - 28 - 41 - 60 a. Find the explained variation. (Round to two decimal places as needed.)
mean of x = 2.882, standard deviation = 1.634 Mean of y = 2.588, standard deviation = 0.246 The correlation coefficient is 0.159 slope coefficient for regression line = .0239 Find the intercept for the regression line.
Listed below are altitudes (thousands of feet) and outside air temperatures (°F) recorded during a flight. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 95% confidence level with the altitude of 6327 ft (or 6.327 thousand feet). Altitude Temperature a. Find the explained variation. (Round to two decimal places as needed.) 2 55 8 40 13 25 20 - 3 28 - 26 31 - 41 34 - 53

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