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	54.11	48.55	49.73	24.35	43.16	49.12	59.80	43.05	42.86
Carbohydrate Replacement	23.88	34.40	40.39	9.74	47.42	27.20	37.77	14.99	45.47

Answer 1

for p-value we use spss software and run paired sample t test.

	chocolate_milk	carbohydrate_replacements
1	54.11	23.88
2	48.55	34.40
3	49.73	40.39
4	24.35	9.74
5	43.16	47.42
6	49.12	27.20
7	59.80	37.77
8	43.05	14.99
9	42.86	45.47

Paired Samples Statistics
		Mean	N	Std. Deviation	Std. Error Mean
Pair 1	chocolate_milk	46.0811	9	9.88467	3.29489
	carbohydrate_replacements	31.2511	9	13.21664	4.40555

Paired Samples Test
		Paired Differences					t	df	Sig. (2-tailed)
		Mean	Std. Deviation	Std. Error Mean	95% Confidence Interval of the Difference
					Lower	Upper
Pair 1	chocolate_milk - carbohydrate_replacements	14.83000	12.31632	4.10544	5.36284	24.29716	3.612	8	.007

P-value = 0.007

Since, p value is less than 0.05

So we can say that the difference is significant at 5% level of significance.

Mean of chocolate milk is 46.0811

Mean of carbohydrate replacement is 31.2511

Chocolate milk is more effective than carbohydrate replacement.