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(answered) – 1 STA100A and STA100B Homework 10 (last one!) Due at theDescriptionSolution downloadThe QuestionHi, I have this statistics homework (attached below) that involves using the program R. I would appreciate your help. Thank you!1STA100A and STA100B Homework 10 (last one!)Due at the beginning of class on March 14, 2016Write your name and student IDon the left side at the top of the ?rst page.Write the homework number in the middle at the top of the ?rst page.Write your section number on the right side at the top of the ?rst page.Staple all pages.If your homework is on paper pulled out of a notebook,cut o? all of the fringes (from the torn horizontal threadsthat attached the paper to the notebook).Percentage lost if youdon?t follow the rule2%2%2%2%2%For example, for homework 10 if your name is John Smith, your student ID is 123456789,and you are in section A01, then the top of your ?rst page should look like thisJohn Smith123456789Homework 10A01and if you are in section B01, then the top of your ?rst page should look like thisJohn Smith123456789Homework 10B01Be kind to the grader.? make sure you write your name clearly (so it is easy to read)? write neatly? circle all ?nal answers (so they are easy to ?nd)1. Consider the following data.x61325y673214(a) Make a scatter plot of the data. Look at the plot and answer the followingquestions. (You do not have to print this graph.) Does there there appear to bea relationship between X and Y? Is it linear or nonlinear? Is it weak or strong?x=c(6,1,3,2,5)y=c(6,7,3,2,14)plot(x,y)(b) Compute the sample correlation coe?cient.cov(x,y)/(sd(x)*sd(y))2(c) Is there evidence of a signi?cant linear relationship? Answer this question bytesting the null hypothesis H0 : ? = 0 against the alternative hypothesis H0 : ? = 0with ? = 0.05.2. Data were obtained on the reduction in cholesterol count (in mg per 100 ml of bloodserum) for a sample of 15 male subject participating in a study to test if a lowcholesterol diet reduces serum cholesterol. Each subject participated in the studyfor four weeks. The response variable was the reduction in cholesterol. It was alsoanticipated that the reduction in cholesterol might depend on the subject?s age. Youcan cut and paste the data into R. In case you have di?culty cutting from this pdf ?leI also put the data into a text ?le on SmartSitex=c(45,43,46,49,50,37,34,30,31,26,22,58,60,52,27)y=c(30,52,45,38,62,55,25,30,40,17,27,44,61,58,45)(a) Make a scatter plot of the data. Don?t print the graph until you have added theregression line (part e).plot(x,y,xlab=”age”,ylab=”cholesterol reduction”)(b) Calculate the slope of the regression line.slope=cov(x,y)/var(x)slope(c) Calculate the intercept of the regression line.intercept=mean(y)-slope*mean(x)intercept(d) You can verify the formulas used above to calculate the slope and intercept withthe following R command.lm(y~x)No idea why the ? prints as if it is a superscript. It is not.You don?t need to provide any answer to this question. I just want you to seefrom (b) and (c) how the slope and intercept are calculated and also show youthe lm() function in R that does linear regression.(e) Add the regression line to the scatter plot and turn in the graph with your homework. We add the line by calculating the predicted values of y at the minimumvalue of x (which is 22) and the maximum value of x (which is 60). We then usethese two pairs of x and y as the end points for a straight line that is added tothe plot with the R command lines() shown below.#you need to have already run the commands above to set the# intercept and slope and produce the graph#FIND THE

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