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gizmo_the_mogwai [7]
3 years ago
8

An educational psychologist wants to check claims that regular physical exercise improves academic achievement. To control for a

cademic aptitude, pairs of college students with similar GPAs are randomly assigned to either a treatment group that attends daily exercise classes or a control group. At the end of the experiment, the following scores were reported for the six pairs of participants:
Pair Number Phyasical excercise(X1) No Physical excercise(X2)
1 4.00 3.75
2 2.67 2.74
3 3.65 3.42
4 2.11 1.67
5 3.21 3.00
6 3.60 3.25
7 2.80 2.65

a. Using t, test the null hypothesis at the .01 level of significance.
b. Specify the p-value for this test result.
c. If appropriate (because the test result is statistically significant), use Cohen's d to estimate the effect size.
d. How might this test result be reported in the literature?
Mathematics
1 answer:
Eva8 [605]3 years ago
6 0

Answer:

a. We fail to reject the null hypothesis

b. p > 0.01

c. The test result is not statistically significant

d. There not is enough statistical evidence to suggest that regular exercise improves academic achievement

Step-by-step explanation:

No. Physical Exercise (X₂); 3.75, 2.74, 3.42, 1.67, 3, 3.25, 2.65

Physical Exercise (X₁); 4.00, 2.67, 3.65, 2.11, 3.21, 3.60, 2.80

GPAs Pair number; 1, 2, 3, 4, 5, 6, 7

The null hypothesis, H₀; μ₁ ≤ μ₂

The alternative hypothesis, Hₐ; μ₁ > μ₂

From the data, we have;

\overline d ≈ 0.222857

Where;

d = X₁ - X₂

\overline d = \dfrac{1}{n} \times \sum_{i = 1}^{n} d_i

Similarly, we have;

s_d^2=\dfrac{\sum \left (d-\overline d  \right )^{2} }{n - 1}

From which we get;

s_d ≈ 0.160905

The critical-t at n - 1 = 7 - 1 = 6 degrees of freedom for α = 1% = 0.01, is given as follows;

t_{\alpha /2} = t_{(0.005, \, 6)} = 3.707

The test statistic, t = \overline d/(s_d/√n)

∴ t = 0.222857/(0.160905/√7) ≈ 3.664

The test statistic , t ≈ 3.664

Therefore, given that the test statistic is smaller than the critical-t,

b. The p-value estimate is p > 0.01

c. The test result is not statistically significant

d. The test result can be reported in a literature as that there is not enough statistical evidence to reject the null hypothesis, therefore, there not is enough statistical evidence to suggest that regular exercise improves academic achievement

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