In the simple linear regression Y = 20 5X, if the coefficient of determination is 0.8, then the sample correlation coefficient is Select one: O a. 0.89 Ob. 0.64 OC. -0.89 od. -0.64

Porphyrin is a pigment in blood protoplasm and other body fluids that is significant in body energy and storage. Let � be a random variable that represents the number of milligrams of porphyrin per deciliter of blood. In healthy circles, � is normally distributed with mean �=44 and standard deviation �=8. Find the following probabilities.A) � is greater than or equal 16.7 Keep your answer in 4 decimal places.B) � is greater than 54.3 Keep your answer in 4 decimal places.

If you have a correlation (r=.90), the coefficient of determination would be:

In a regression study, relating Price/unit (x) to Weekly Sales (in Kg.), with the scatter plot showing a strong negative direction, 63% of the variability in sales could be accounted for by the variation in the Unit Price. The correlation coefficient in this study is: 0.7 -0.4 -0.79 0.4

Suppose the correlation coefficient r=0.16 Find:a. The coefficient of determination r2=  (round to 3 decimal places)b. The percentage of explained variation = % (round to 1 decimal place)c. The percentage of unexplained variation = % (round to 1 decimal place)

Run a regression analysis on the following bivariate set of data with y as the response variable. y 62.8 75.2 60 70.4 59.8 63 59.3 68.5 59 70.3 52.4 50.8 63.4 61.9 70.3 68.1 65.6 69.3 56.8 60.9 67.5 77.5 55.5 57.2 Find the correlation coefficient and report it accurate to three decimal places. What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place. (If the answer is 0.84471, then it would be 84.5%...you would enter 84.5 without the percent symbol.) r2 = Based on the data, calculate the regression line (each value to three decimal places) y = Predict what value (on average) for the response variable will be obtained from a value of 52.5 as the explanatory variable. Use a significance level of a = 0.05 to assess the strength of the linear correlation. What is the predicted response value? (Report answer accurate to one decimal place.) y = Submit Question IO 01 100 O Details

The regression equation is:       ŷ = 67.16 + 8.417x where ŷ is the miles traveled, and x is the MPG. The sample size used was all 110 MPG records. The correlation coefficient r = 0.620. Use the information to obtain an estimate of my mileage if my MPG is 22.   Is it option:  a.) cannot estimate ŷ rcrit = 0.195; the correlation IS NOT significant b.) ŷ = 252.33 rcrit = 0.195; the correlation IS significant c.) ŷ = 252.33 rcrit = 0.187; the correlation IS significant d.) cannot estimate ŷ rcrit = 0.187; the correlation IS NOT significant

A random sample of college students was surveyed about how they spend their time each week.  The scatterplot below displays the relationship between the number of hours each student typically works per week at a part- or full-time job and the number of hours of television each student typically watches per week.  The correlation between these variables is r = –0.63, and the equation we would use to predict hours spent watching TV based on hours spent working is as follows:             Predicted hours spent watching TV = 17.21 – 0.23(hours spent working) Since we are using hours spent working to help us predict hours spent watching TV, we’d call hours spent working a(n) __________________ variable and hours spent watching TV a(n) __________________ variable. The correlation coefficient, along with what we see in the scatterplot, tells us that the relationship between the variables has a direction that is _________________ and a strength that is ______________________. According to the…

Two statistically independent random have variances of Variables X and 0² and new ran- dom T variables Y 54 are + = 25. TWO defined by U = 3x + 4 Y V = 5X-2Y

A veterinarian collects data about puppies in her practice. She makes a scatterplot with their age in days on the horizontal axis and weight in ounces (oz) on the vertical axis and sees that the scatterplot is football shaped. The average age of the puppies is 42 day, with an SD of 7 days. The average weight is 25 oz, with an SD of 5 oz. The correlation is moderately strong, with r = 0.7. 4. (a) The age of two puppies differ by 20 days. The difference in their weight is predicted to be oz. (b) The baseline prediction for the weight of puppies ignores their age and just uses the information from all puppies. The baseline predicted weight is oz and the RMS error for this prediction is o. (c) Suppose we converted the units of age from days to weeks. The average age is now weeks, with an SD of weeks. The correlation between age in weeks and weight in oz is now r =