Question

a study was done to determine if daily maximal-squat training had a positive effect on a 3 rep squat max compared to once per week high-intensity training men in their 20s. Twenty-four individuals were randomly selected from a large list of competitive weightlifters, all between 20 and 25 years old. Each of the men had been competing and training in weightlifting since the age of 16 and each could back-squat more than double their own bodyweight. The men each performed a 3-repetition maximal effort in the back squat before their group's training began.
The 24 individuals were randomly put into one of two groups:
Group 1 used daily maximal squatting
- squatted 6 sessions per week and rested on sundays
- each session consisted of a dynamic warm-up and stretchinf followed by successive sets of 3, adding 20 pounds each set, until they reached a maximum-effort set of 3 reps.
-they rested for 90 seconds between sets
Group 2 used once per week high-intensity squatting
-squatted once per week on sundays
-each squat session consisted of a dynamic warm-up and stretching followed by successive sets of 3, adding 20 pounds each set, until they reached a maximum effort set of 3 reps. they then completed two "back-off" sets as many reps as possible, using 60% of the 3 rep-max they previously worked up to in that session.
- they rested for 90 seconds between the successive sets of 3 and rested for 5 minures before and after the first "back-off" max repetition set.

Each group went through the specified training for 6 weeks
the following table gives the percent-change in the 3 rep max back-squat after 6 weeks of training for each of the 24 individuals
Group 1:
1.80%, 1.90%, -1.60%, -0.5%, 1.22%, 1.23%, 0.89%, 0.10%, 2.5%, -2.1%, 1.09%, 1.10%
Group 2:
1.17%, -1.09%, -1.01%, 3.56%, 1.13%, -1.11%, -0.90%, 1.00%, 1.05%, 2.7%, 3.20%, 2.35%

Questions:
(a) conduct a hypothesis test at the 5% significance level to determine if Group 1 saw a larger percent-change, on average, than Group 2 did. State the null and alternative hypothesis here:
H0:
Ha:
(b) what distribution should you utilize for this test? state the distribution using words and symbols.
(c) what is the test statistic? what does this mean in this context?
(d) what is the p-value? explain what the p-value means for this problem. be precise here
(e) sketch the picture of the situation. make sure to label the axis, the p-value, and the level of significance, a. also, label the locations of the sample difference and claimed population difference.
(f) should we accept or reject the null hypothesis? state the conclusion of this study with respect to the claim, in simple clear language

Answer 1

Solution

[Note: Final answers are given first. Back-up Theory and Detailed Working follow at the end.]

Part (a)

Hypotheses:

Null: H₀: µ1 = µ2 Vs Alternative: H_A: µ1 > µ2 Answer 1

Part (b)

Distribution: t-distribution with degrees of freedom 22 Answer 2

In symbols: t₂₂Answer 3

Part (c)

Test statistic = - 0.5743 Answer 4

Part (d)

p-value = 0.7142 Answer 5

Interpretation: high p-value implies that the null hypothesis is only very likely to be true and hence needs to be accepted. Answer 6

This, in turn, leads to the conclusion that the claim that Group 1 on an average recorded a higher percentage change is not tenable. Answer 7    

DONE

Back-up Theory and Detailed Working

Let X = percentage change in Group 1

      Y = percentage change in Group 2

Then, X ~ N(µ₁, σ₁²) and Y ~ N(µ₂, σ₂²), where σ₁² = σ₂² = σ², say and σ² is unknown.

Claim:

Group 1 saw a larger percent-change, on average, than Group 2 did.

Hypotheses:

Null: H₀: µ1 = µ2 Vs Alternative: H_A: µ1 > µ2 [claim]

Test Statistic:

t = (Xbar - Ybar)/{s√(2/n)} where

s² = (s₁² + s₂²)/2;

Xbar and Ybar are sample averages and s₁,s₂ are sample standard deviations based on n observations each on X and Y.

Calculations

Summary of Excel calculations is given below:

n =	12
Xbar =	0.635833
Ybar =	1.004167
s1 =	1.40357
s2 =	1.722168
s^2 =	2.467936
s =	1.570966
tcal =	-0.57431
α =	0.05
tcrit =	1.717144
p-value =	0.7142

Distribution, Significance Level, α , Critical Value and p-value:

Under H₀, t ~ t_{2n - 2}. Hence, for level of significance α%, Critical Value = upper α% point of t_{2n - 2} and p-value = P(t_{2n - 2} > tcal).

Using Excel Functions, and given α = 0.05,

Using Excel Function: Statistical TINV TDIST the above are found to be as given in the above table.

Decision:

Since tcal < tcrit, or, equivalently, since p-value > α, H₀ is accepted.

Conclusion:

There is not sufficient evidence to suggest that the claim is valid.

Complete