Question

In: Statistics and Probability

Consider n experiment to test the effect of caffeine on athletic performance. Twenty-five subjects were included...

Consider n experiment to test the effect of caffeine on athletic performance. Twenty-five subjects were included in the study. Each subject was initially given two tests, a swimming, and a running. The subject then consumed caffeine and then repeated the two tests. Score was recorded for each of the four series of tests.

A) Suggest model with appropriate notation.

B) Consider the null hypothesis that consumption of caffeine has no effect on either the mean swim score or the mean running score. State Null and Alternative in your notation and explain how to calculate the test statistic and the distribution of it under the null.

C) Consider the single parameter: the decrease in the mean swim score due to caffeine consumption minus the decrease in the mean running score due to caffeine consumption. Give an expression for this parameter using your notation, and explain how you would construct a 95% confidence interval for the parameter.

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Expert Solution

A)

We can use pairwise t test to each of activities ie swimming, and running and determine we there is effect of caffeine on either of swimming, and running.

B)

Activity: Swimming

Mean of Difference (X_D) = Mean of Swimming with Caffeine - Mean of Swimming without Caffeine

Null and Alternate Hypothesis

H₀: µ_D = 0 (There is no effect of Caffeine on swimming)

H_A: µ_D <> 0 (There is no effect of Caffeine on swimming)

Activity: Running

Mean of Difference (X_D) = Mean of Running with Caffeine - Mean of Running without Caffeine

Also, we will calculate Standard Error of Differences

Alpha = 0.05

Null and Alternate Hypothesis

H₀: µ_D = 0 (There is no effect of Caffeine on Running)

H_A: µ_D <> 0 (There is no effect of Caffeine on Running)

Test Statistic

t = (X_D – 0)/SE

p-value = TDIST(t,25-1,2)

Decision Rule

If p-value is less than 0.05, we reject the null hypothesis.

Result

Based on p-value we either reject or fail to reject the null Hypothesis

C)

We will calculate

Mean of Difference (X_D) = decrease in the mean swim score due to caffeine consumption - decrease in the mean running score due to caffeine consumption

Also, we will calculate Standard Error (SE) of Differences

At alpha = 0.05

Z_Critical = 1.96

Hence 95% CI of Mean of Difference = X_D +/- 1.96 * SE

orchestra answered 2 years ago

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