Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes​ (before minus​ after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 0.4 and a standard deviation of 2.38. Use a 0.10 significance level to test the claim that with garlic​ treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic​ treatment? Assume that a simple random sample has been selected. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim.

What are the null and alternative​ hypotheses?

A.

Upper H 0H0​:

muμequals=00

​mg/dL

Upper H 1H1​:

muμnot equals≠00

​mg/dL

B.

Upper H 0H0​:

muμequals=00

​mg/dL

Upper H 1H1​:

muμless than<00

​mg/dL

C.

Upper H 0H0​:

muμequals=00

​mg/dL

Upper H 1H1​:

muμgreater than>00

​mg/dL

D.

Upper H 0H0​:

muμgreater than>00

​mg/dL

Upper H 1H1​:

muμless than<00

​mg/dL

Determine the test statistic.

nothing

​(Round to two decimal places as​ needed.)

Determine the​ P-value.

nothing

​(Round to three decimal places as​ needed.)

State the final conclusion that addresses the original claim.

▼

Fail to reject

Reject

Upper H 0H0.

There is

▼

not sufficient

sufficient

evidence to conclude that the mean of the population of changes

▼

equal to

is greater than

is less than

is not

00.

Answer 1

Solution:

C)

This a right (One) tailed test.

The null and alternative hypothesis is,  

Ho: 0

Ha: 0

The test statistics,

t =( - )/ (s /n)

= ( 0.4 - 0 ) / (2.38 / 81 )

= 1.51

P-value = 0.067

The p-value is p = 0.067 < 0.10, it is concluded that the null hypothesis is rejected.

Reject H0. There is evidence to conclude that the mean of the population of changes is greater than 0.