Respuesta :
Answer:
There is enough evidence to support the claim
Step-by-step explanation:
We are conduction a hypothesis test for dependent samples. We want to see if there was a change in the test subjects cholesterol levels.
For our situation:
n = 64
d = 0.7
s = 1.72
µ(d) = 0
The hypothesis are:
H0: µ(d) = 0
Ha: µ(d) > 0
This is a right tailed test.
We are testing at the 1% level of significance. Our critical region is z > 2.325
If our test statistic is in this region, we will reject the null hypothesis
See attached photo for the calculation of the test statistic and conclusion of the test
Using the t-distribution, it is found that since the p-value of the test is of 0.0009 < 0.01, the results suggest that the garlic treatment is effective.
At the null hypothesis, we test if the mean change is not greater than 0, that is:
[tex]H_0: \mu \leq 0[/tex]
At the alternative hypothesis, we test if it is greater than 0, that is:
[tex]H_1: \mu > 0[/tex]
We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
The parameters are:
- [tex]\overline{x}[/tex] is the sample mean.
- [tex]\mu[/tex] is the value tested at the null hypothesis.
- s is the standard deviation of the sample.
- n is the sample size.
For this problem, the values of the parameters are: [tex]\overline{x} = 0.7, s = 1.72, \mu = 0, n = 64[/tex].
Hence, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{0.7 - 0}{\frac{1.72}{\sqrt{64}}}[/tex]
[tex]t = 3.26[/tex]
The p-value is found using a right-tailed test(test if the mean is greater than a value), with t = 3.26 and 64 - 1 = 63 df.
Using a calculator, the p-value of test is 0.0009.
Since the p-value of the test is of 0.0009 < 0.01, the results suggest that the garlic treatment is effective.
A similar problem is given at https://brainly.com/question/15229576
