banned substances list in 2004 by the World Anti-Doping Agency, caffeine has bee
ID: 3326246 • Letter: B
Question
banned substances list in 2004 by the World Anti-Doping Agency, caffeine has been used by athletes with the expectancy that it enhances their workout and performance. However, few studies look at the role caffeine plays in sedentary females. Researchers at the University of Western Australia conducted a test in which they determined the rate of energy expenditure (kilojoules) on 10 healthy, sedentary females who were nonregular caffeine users. Each female was randomly assigned either a placebo or caffeine pill (6 mg/kg) 60 minutes prior to exercise. The subject rode an exercise bicycle for 15 minutes at 65% of their maximum heart rate, and the energy expenditure was measured. The process was repeated on a separate day for the remaining treatment. The mean difference in energy expenditure (caffeine- placebo) was 18 kJ with a standard deviation of 19 kJ. Source: Wallman, Karen E. "Effect of Caffeine orn Exercise Performance in Sedentary Females," Journal of Sports Science and Medicine (2010) 9, 183-189 (a) State the null and alternative hypotheses to determine if caffeine increases energy expenditure. (b) Assuming the differences are normally distributed, determine if caffeine appears to increase energy expenditure at the 0.05 level of significance.Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: d< 0
Alternative hypothesis: d > 0
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs z-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, z statistic test statistic (z).
s = 19
SE = s / sqrt(n)
S.E = 6.008
z = [ (x1 - x2) - D ] / SE
z = 2.996
where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.
Since we have a two-tailed test, the P-value is the probability that a z statistic greater than 2.996.
Thus, the P-value = 0.0012
Interpret results. Since the P-value (0.0012) is less than the significance level (0.05), we have to reject the null hypothesis.
Reject H0. We can conclude that caffience appears to increase energy expenditure.
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