tiffany has breakfast at a restaurant and the cost of her meal is $27.00.because of the service.she wants to leave a 10% tip.wha
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Answer:
With garlic treatment, the mean change in LDL cholesterol is not greater than 0.
Step-by-step explanation:
The dependent <em>t</em>-test (also known as the paired <em>t</em>-test or paired samples <em>t</em>-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.
In this case a paired <em>t</em>-test is used to determine the effectiveness of garlic for lowering cholesterol.
A random sample of 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment.
The hypothesis for the test can be defined as follows:
<em>H₀</em>: With garlic treatment, the mean change in LDL cholesterol is not greater than 0, i.e. <em>d</em> ≤ 0.
<em>Hₐ</em>: With garlic treatment, the mean change in LDL cholesterol is greater than 0, i.e. <em>d</em> > 0.
The information provided is:
Compute the test statistic value as follows:
The test statistic value is 0.22.
Decision rule:
If the <em>p</em>-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.
Compute the <em>p</em>-value of the test as follows:
*Use a <em>t</em>-table.
The <em>p</em>-value of the test is 0.4132.
<em>p-</em>value= 0.4132 > <em>α</em> = 0.01
The null hypothesis was failed to be rejected.
Thus, it can be concluded that with garlic treatment, the mean change in LDL cholesterol is not greater than 0.
For this case we must indicate the cosine of A, of the figure shown. By definition:
Substituting the values we have:
So, we have to:
Answer:
Option D