Question

Eleven employees were put under the care of the company nurse because of high cholesterol readings. The nurse lectured them on the dangers of this condition and put them on a new diet. You have the cholesterol readings of the 11 employees both before the new diet and one month after use of the diet began.

(2). Use R command to construct a 98% confidence interval to estimate the population mean difference of cholesterol readings before and after for people who are involved in this program. Assume differences in cholesterol readings are normally distributed in the population.

(3). Import dataset cholesterolReadings.csv to R script file, and write the correct R commands to solve this problem. What is the p value? What is the statistical decision?

Employee_ID	Before	After
1	255	197
2	230	225
3	290	215
4	242	215
5	300	240
6	250	235
7	215	190
8	230	240
9	225	200
10	219	203
11	236	223

Answer 1

The R output is ,

> x=c(255,230,290,242,300,250,215,230,225,219,236)
> x
[1] 255 230 290 242 300 250 215 230 225 219 236
> y=c(197,225,215,215,240,235,190,240,200,203,223)
> y
[1] 197 225 215 215 240 235 190 240 200 203 223
> t.test(x,y,paired=T,conf.level=0.98)

Paired t-test

data: x and y
t = 3.6094, df = 10, p-value = 0.004773
alternative hypothesis: true difference in means is not equal to 0
98 percent confidence interval:
6.581057 49.600761
sample estimates:
mean of the differences
28.0909

(2) The 98% confidence interval to estimate the population mean difference of cholesterol readings before and after for people who are involved in this program is ,

( 6.581057 , 49.600761 )

(3) P value = 0.004773

Decision : Here , P value = 0.004773 <

Therefore , reject Ho at significance level.