SUBMISSION REQUIREMENTS: Please submit a single document (word or PDF) for submission. Your submission should contain a summary of your results (and answers to questions asked on the homework) as well as your R code used to generate your results (please append your R code to the end of your submission).+ An experiment was conducted to determine the effect of children participating in a given meal preparation on calorie intake for that meal. Data are recorded below. Save the data to a format that can be read into R. Read the data in for analysis. Use R to calculate the quantities and generate the visual summaries requested below. You will lose points if you are not utilizing R.+ (1) Summarize the data by whether children participated in the meal preparation or not. Use an appropriately labelled table to show the results. Also include a graphical presentation that shows the distribution of calories for participants vs. non-participants. Describe the shape of each distribution and comment on the similarity ( or lack thereof) between the distributions in each population. (2 points)+ (2) Does the mean calorie consumption for those who participated in the meal preparation differ from 425? Formally test at the alpha = 0.05 level using the 5 steps outlined in the module. (6 points)~ (3) Calculate a 90% confidence interval for the mean calorie intake for participants in the meal preparation. Interpret the confidence interval. (4 points) (4) Formally test whether or not participants consumed more calories than non-participants at the alpha = 0.05 level using the 5 steps outlined in the module. (6 points ) (5) Are the assumptions of the test used in (4) met? How do you know? (2 points)~4. D SIEPS Slgnmcant IESE: 1) Let the participants be the group 1, and let the non-participants be the group 2. H0:u1= 112 ,Ha:u1u2>0 0120.05 2) Use the t test equation as below: x_1 2 312 $22 + 3) g = 02$ =o.025 Find the critical value for it is 1.960, If |z| 31.96, reject H0, if not, do not reject H0. 4) Calculate the pvalue 410.1 374.1 121.5138 3 121.5138 133.1393 :1: 133.1393 25 + 22 4) Calculate the pvalue 410.1 374.1 121.5138 3 121.5138 133.1393 3 133.1393 25 + 22 = 36/373678303333 #036339551102 Use R to calculate the p- value by using the code as below: 2*(1pnorm(0.9633955)) Get the output of the pvalue as: 0.3353491 5) Because p as 0.34, p