Question

A study is being conducted to determine whether a particular exercise program is effective at reducing the mean blood cholesterol level on males between the ages of 35 and 50. Six subjects participated in the program, and their cholesterol levels before and after the program are as follows. Blood Cholesterol Level Subject Before After 265 229 2 240 231 258 227 4 295 240 5 251 238 6 225 226 Using this data, we wish to test the following hypothesis using a level of significance of 0.05, where y, denotes the mean blood cholesterol level before the program, and Mz denotes the mean blood cholesterol level after the program. H:H1 - Mz = 0 H:M1 -Uz > 0 Answer questions a through e assuming that you have chosen to use a paired t-test.

Answer 1

First we conduct paired t- test on Excel.

Path to conduct paired t tet is as follows:

Step-1 : Enter given data in Excel sheet.

Step-2 : Go to Data ---> Data Analysis .

Step-3:  Select T-test :paired two Sample for means. And see the output as follows:

B12 for 2.01504837208812 A 1 t-Test: Paired Two Sample for Means B с 2 Variable 1 Variable 2 255.6666667 231.8333333 569.4666667 34.16666667 6 6 0.640439456 0 3 4 Mean 5 Variance 6 Observations 7 Pearson Correlation 8 Hypothesized Mean Difference 9 df 10 t Stat 11 P(T<=t) one-tail 12 t Critical one-tail 13 P(T<=t) two-tail 14 t Critical two-tail 5 2.831933031 0.018295751 2.015048372! 0.036591503 2.570581835

Here we can see that p- value for one tail is = 0.018295. As p-value is less than 0.05 we reject H0.

Part d: Now to construct confidence interval following are steps.

Step -1 : Formula:   
d
  
(n-1,0)
  
Sa in

Step-2 : We can do this in excel also. as follows.

Enter the data and calculate difference. Calculate the mean by using average command and standard deviation by using Stdev command.

19 A F G H B с Subject Before 265 D after 1 2 1 E difference 229 36 231 9 31 240 55 3 2 240 4 3 258 227 5 4 295 Mean difference = 23.83333 Std Deviation of Difference = 20.61472 Std. Error of Difference = 8.415924 T stat (95% C.I.) = 2.015048 Lower confidence level = 6.874839 Upper Confidence level = 65.37299 6 5 251 238 13 7 6 225 226 -1 00 9

In this Std. error of difference =

Std deviation of difference % no. of observation

Lower confidence leve= Mean difference - Std.Error of Difference *T stat (95% C.I.) = 6.874839

Upper confidence level = Mean difference + Std.Error of Difference *T stat (95% C.I.) = 65.37299.

Therefore the confidence interval is given by  

( 16.95849).

+ | 23.8333