Question

Daniel believes that people perform better in the barbell curl, on average, if they are encouraged by a coach. He recruited 29 subjects to participate in an experiment and randomly assigned them into two groups. Daniel gave one group verbal encouragement during the exercise and was quiet the exercise for the other group. He recorded the total number of barbell curls each subject was able to complete before setting the bar down.

(a) Explain the purpose of random assignment in this experiment.

(b) Compare these distributions,

(c) State the hypotheses Daniel should use to test his belief about receiving encouragement during exercise. Make sure to define any parameters you use.

(d) Identify the significance test Daniel should use to analyze the results of his experiment and show that the conditions for this test are met.

(e) The P-value for Daniel's test is 0.107. What conclusion should Daniel make at the a=0.05 significance level?

Answer 1

a) to remove bias in the study

b) The number of barbell curls performed by group who was encourged was higher than the group who was not encouraged

c) H0: mean of barbell curls for group encouraged= mean of barbell curls for group not encouraged

Ha: mean of barbell curls for group encouraged≠ mean of barbell curls for group not encouraged

d) two sample t-test.

Requirement has been met as there are two separate groups whose means have to be compared. Median of box plot is equal to mean. Standard Deviation is equal to range divided by four. Sample size will have to be assumed

e) Encouragement affects the number of barbell curl performed by an individua

Explanation:

a) random assignemnt removes bias in any study.

b) box plot is attached

c) null hypthesis and alternate hypothesis would compare means of the two distributions

d) As we are comparing two groups, we will use two sample t-test.

The median of box plot is equal to mean for both groups.

The rnge/4 is equal to standard deviation.

mean of encouraged group= 30

mean of not encoouraged group=19

SD of encouraged group= (75-5)/4= 15

SD of not encouraged group= (60-0)/4= 15

Assume number of inidivudals in encouraged group =15

assume number of individuals in not encouraged group = 14

e) test statistic= (mean1-mean2)/√(SD1²/(N1-1)+ SD2²/(N2-1))

= (30-19)/(√(15²/14+ 15²/13)

= 1.9734

Since test statistic is greater than p-value, null hypothesis is rejected.

Encouragement affects the number of barbell curl performed by an individual