Assume that a is set to 05 for a two-tailed test and the estimated p-value is 0.0012, What do you conclude?

Use the  data set provided, answer the following questions:Q1- Q4

A study was conducted to identify the changes in triglyceride levels of subjects followed by a specific diet. The diet consists primarily of protein and animal fat, restricting carbohydrate intake. The study participants   followed the diet with length of time varying from 3 to 17 weeks (Datasheet is provided  in  excel  attachment  )

 

Q1. Write a statistical summary for   participant’s demographic details that includes variable age, gender, weeks   on diet & physical activity?

                                                                                                         Mark -2                   Q2. Compute most appropriate measure of central tendency & dispersion to report the variable “changes in triglyceride level” and address its worthiness?

                                                                                                       Mark – 2

Q3. What is the appropriate statistical test for identifying significance difference in the “changes in   triglyceride levels” in male and females &   Why

                                                                                                       Mark – 1                                                   

Q4. What are the assumptions underlying the test you identified in Question-3?

Mark -1

Q5: In a cross sectional study on coronary heart disease (CHD), the smoking and CHD status are summarized below

        CHD Total
Smoking Present Absent  
Yes 55 84 139
No 552 1927 2479
Total 607 2011 2618
  1. What is the appropriate statistical test for identify the association between   smoking   and   CHD ?                                   Mark -1
  2. What are the assumptions and application of the proposed test identified for 5a                                                        Mark – 2

 

  1. Assume that a is set to 05 for a two-tailed test and the estimated p-value is 0.0012, What do you conclude?               Mark – 1