To determine if chocolate milk was as effective as other carbohydrate replacemen

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.)

Is there sufficient evidence to suggest that the mean time to exhaustion is greater after chocolate milk than after carbohydrate replacement drink? Use a significance level of 0.05.

Yes

No   

Cyclist Time to Exhaustion (minutes) 1 2 3 4 5 6 7 8 9 Chocolate
Milk 45.44 47.84 29.15 52.90 37.92 28.12 35.20 38.50 51.06 Carbohydrate
Replacement 53.79 46.57 7.26 46.21 9.55 10.81 22.67 15.88 12.48

Explanation / Answer

Solution:-

Working with two given data we get,

Chocolate Milk :-

Total Numbers, N = 9
Sum = 366.13
Mean (Average) = 40.681111111111
Sample Standard Deviation, s = 9.1110570797856

Carbohydrate Replacement :-

Total Numbers, N = 9
Sum = 225.22
Mean (Average) = 25.024444444444
Sample Standard Deviation, s   = 18.520622364753

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 1 - 2 = 0 , i.e., the mean time to exhaustion is same after chocolate milk than after carbohydrate replacement drink
Alternative hypothesis: 1 - 2 0, i.e., the mean time to exhaustion is greater after chocolate milk than after carbohydrate replacement drink

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(9.12/9) + (18.52/9] = 6.872

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (9.12/9 + 18.52/9)2 / { [ (9.12 / 9)2 / (8) ] + [ (18.52 / 9)2 / (8) ] }
DF = 2230.5679 / { 10.58 + 180.76 } = 12.30

t = [ (x1 - x2) - d ] / SE = [ (40.68 - 25.042 ) - 0 ] / 6.872 = 2.2756

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is

The P-Value is 0.041514.
The result is significant at p < 0.05.

Interpret results. Since the P-value is less than the significance level (0.05), we cannot accept the null hypothesis.

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