Question

Question #4
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.

Before	9.4	9.0	9.2	9.4	9.2	9.3	9.7
After	9.5	9.3	9.5	9.6	9.4	9.5	9.9

Using a 0.10 level of significance and Critical Value method to test the claim that the training technique is effective in raising the gymnasts' scores. (Mean difference is less than zero)

Answer 1

4.

Before	After	d=Before-After
9.4	9.5	9.4 - 9.5 = -0.1
9.0	9.3	9.0 - 9.3 = -0.3
9.2	9.5	9.2 - 9.5 = -0.3
9.4	9.6	9.4 - 9.6 = -0.2
9.2	9.4	9.2 - 9.4 = -0.2
9.3	9.5	9.3 - 9.5 = -0.2
9.7	9.9	9.7 - 9.9 = -0.2

claim : The training technique is effective in raising the gymnasts' scores. (Mean difference is less than zero)

Null and alternative hypothesis are ,

Ho: μd =0

Ha: μd < 0     (claim)

Test Statistic :

d bar : is the sample mean of differences (d)

sd : is the sample standard deviation of differences (d)

n = 7

Using excel function , =AVERAGE( select the differences)

=AVERAGE(C2:C8)

= -0.214286

dbar = -0.214286

Using Excel function , =STDEV( select the differences )

=STDEV(C2:C8)

=0.069007

sd = 0.069007

Now plug the values in the formula ,

t = -8.216

For critical value ,

significance level ( α ) = 0.10

degrees of freedom = n - 1 = 7 -1 = 6

df =6

As the alternative hypothesis (Ha) is less than (<) type , this is one tailed test (left tailed) .

For one tailed test , using Excel function    =TINV( 2 α , df )

=TINV(0.20,6)

=1.440

so critical value for left tailed test is -1.440

t critical value = -1.440

Decision rule :

If t-test statistic is less than the t-critical value we reject the null hypothesis (Ho).

Since t test statistic (-8.216 ) is less than t-critical value (-1.440 ) we reject the null hypothesis (Ho).

Conclusion :

There is sufficient evidence to support the claim that the training technique is effective in raising the gymnasts' scores.