Question

Some students experience extreme anxiety when taking tests, and this can have a negative impact on their performance. Dr. Rogers has designed a set of simple relaxation and focusing exercises that students can employ while seated at their desks. In order to test her theory that these exercises will help students combat their anxiety and perform well on tests, Dr. Rogers recruited six students from a local middle school. Three of the students took their first math test as they would have normally, but were taught to employ the relaxation techniques before taking their second test. The other three students used the relaxation techniques before taking their first math test, but then did not use them before taking their second test. The following are the test scores for the six students, with and without use of relaxation techniques.

No Relax Tech Relax Tech

72 82

81 90

87 89

75 80

79 83

82 90

1.) Compute the required analysis (one-tailed at the .05 level of significance) to determine if there is a statistically significant difference between the two sets of scores above.  

2.) Calculate the confidence interval for the above analysis.

3.) Calculate the effect size for the above analysis

SHOW WORK

Answer 1

(1)

H0: Null Hypothesis: $\mu _{d}\geq$ 0 (Exercise will not help students combat their anxiety)

HA: Alternative Hypothesis: $\mu _{d}<$ 0 (Exercise will help students combat their anxiety) (Claim)

From the given data, values of d = No Relax Tech - Relex Tech are calculated as follows:

d = No Relax Tech - Relex Tech = - 10, -9, - 2, - 5, - 4, - 8

From d values, the following statistics are calculated:

n = 6

$\bar{d}$ = - 6,333

s_d = 3.141

$SE=\frac{s_{d}}{\sqrt{n}}=\frac{3.141}{\sqrt{6}}=1.2823$

Test Statistic is given by:

$t=\frac{\bar{d}}{SE}=\frac{-6.333}{1.2823}=-4.939$

$\alpha$ = 0.05

df = 6- 1 = 5

From Table, critical value of t = - 2.015

Since calculated value of t = - 4.939 is less than critical value of t = - 2.015, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that exercise will help students combat their anxiety.

(2)

n = 6

$\bar{d}$ = - 6,333

s_d = 3.141

$SE=\frac{s_{d}}{\sqrt{n}}=\frac{3.141}{\sqrt{6}}=1.2823$

$\alpha$ = 0.05

df = 6- 1 = 5

From Table, critical values of t = $\pm$ 2.5706

Confidence Interval:

$\bar{d}\pm (t \times SE)=-6.333\pm (2.5706\times 1.2823)=-6.333\pm 3.296=(-9.629,-3.037)$

So,

Answer is:

(-9.629, - 3.037)

(3)

Effect Size is given by Cohen's d as follows:

$d=\frac{6.333}{3.141}=2.0162$