Respuesta :
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;
[tex]\overline d[/tex] ≈ 0.222857
Where;
d = X₁ - X₂
[tex]\overline d = \dfrac{1}{n} \times \sum_{i = 1}^{n} d_i[/tex]
Similarly, we have;
[tex]s_d^2=\dfrac{\sum \left (d-\overline d \right )^{2} }{n - 1}[/tex]
From which we get;
[tex]s_d[/tex] ≈ 0.160905
The critical-t at n - 1 = 7 - 1 = 6 degrees of freedom for α = 1% = 0.01, is given as follows;
[tex]t_{\alpha /2}[/tex] = [tex]t_{(0.005, \, 6)}[/tex] = 3.707
The test statistic, t = [tex]\overline d[/tex]/([tex]s_d[/tex]/√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
