What is the common mutiple for 1/2 1/4 1/6 1/3 and 1/5
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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.