Question

A coach uses a new technique to train gymnasts. 7 gymnasts were randomly selected and their competition scores were recorded before and after the training. The results are shown below.
Subject A B C D E F G

Before 9.5 ,9.4, 9.6, 9.5, 9.5, 9.6, 9.7,

After 9.6, 9.6, 9.6, 9.4, 9.6, 9.9, 9.
Using a 0.01 level of significance, test the claim that the training technique is effective in raising the gymnasts' scores. Include your null and alternative hypotheses, the test statistic, P-value or critical value(s), conclusion about the null hypothesis, and conclusion about the claim in your answer.
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Answer 1

Here, we have to use paired t test.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: the training technique is not effective in raising the gymnasts' scores.

Alternative hypothesis: Ha: the training technique is effective in raising the gymnasts' scores.

H0: µd = 0 versus Ha: µd > 0

This is a right tailed test.

We take difference as after minus before.

Test statistic for paired t test is given as below:

t = (Dbar - µd)/[Sd/sqrt(n)]

From given data, we have

Dbar = -0.0143

Sd = 0.3288

n = 7

df = n – 1 = 6

α = 0.01

t = (Dbar - µd)/[Sd/sqrt(n)]

t = (-0.0143 – 0)/[ 0.3288/sqrt(7)]

t = -0.1150

The p-value by using t-table is given as below:

P-value = 0.5439

P-value > α = 0.01

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that the training technique is effective in raising the gymnasts' scores.