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

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;

≈ 0.222857

Where;

d = X₁ - X₂

Similarly, we have;

From which we get;

≈ 0.160905

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

= = 3.707

The test statistic, t = /(/√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

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

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x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

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x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

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x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

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step-by-step explanation: