Question

To determine if chocolate milk was as effective as other carbohydrate replacement drinks, nine male cyclists performed an intense workout followed by a drink and a rest period. At the end of the rest period, each cyclist performed an endurance trial where he exercised until exhausted and time to exhaustion was measured. Each cyclist completed the entire regimen on two different days. On one day the drink provided was chocolate milk and on the other day the drink provided was a carbohydrate replacement drink. Data consistent with summary quantities appear in the table below. (Use a statistical computer package to calculate the P-value. Subtract the carbohydrate replacement times from the chocolate milk times. Round your test statistic to two decimal places, your df down to the nearest whole number, and the P-value to three decimal places.) Cyclist Time to Exhaustion (minutes) 1 2 3 4 5 6 7 8 9 Chocolate Milk 48.96 38.32 53.20 38.00 24.50 44.05 55.30 51.45 30.99 Carbohydrate Replacement 25.03 19.33 31.90 12.89 38.63 30.70 17.13 14.09 20.74

t =

df =

P =

Answer 1

Now we do this using a statistical package named R Studio. The R code used is,

x<-c(48.96,38.32,53.20,38.00,24.50,44.05,55.30,51.45,30.99)
y<-c(25.03,19.33,31.90,12.89,38.63,30.70,17.13,14.09,20.74)
t.test(x, y, paired = TRUE, alternative = "greater")

and the output we got is

Paired t-test

data: x and y
t = 3.6971, df = 8, p-value = 0.003034
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
9.627427 Inf
sample estimates:
mean of the differences
19.37
So test statistic, t= 3.70

degrees of freedom, df=9-1=8

p-value= 0.003