Respuesta :
Answer:
With garlic treatment, the mean change in LDL cholesterol is not greater than 0.
Step-by-step explanation:
The dependent t-test (also known as the paired t-test or paired samples t-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 t-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:
H₀: With garlic treatment, the mean change in LDL cholesterol is not greater than 0, i.e. d ≤ 0.
Hₐ: With garlic treatment, the mean change in LDL cholesterol is greater than 0, i.e. d > 0.
The information provided is:
[tex]\bar d=0.40\\SD_{d}=16.2\\\alpha =0.01[/tex]
Compute the test statistic value as follows:
[tex]t=\frac{\bar d}{SD_{d}/\sqrt{n}}\\\\=\frac{0.40}{16.2/\sqrt{81}}\\\\=0.22[/tex]
The test statistic value is 0.22.
Decision rule:
If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.
Compute the p-value of the test as follows:
[tex]p-value=P(t_{n-1}>0.22)\\=P(t_{80}>0.22)\\=0.4132[/tex]
*Use a t-table.
The p-value of the test is 0.4132.
p-value= 0.4132 > α = 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.
