Question

n a test of the effectiveness of garlic for lowering​ cholesterol, 8181 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 0.30.3 and a standard deviation of 18.218.2. Use a 0.010.01 significance level to test the claim that with garlic​ treatment, the mean change in LDL cholesterol is greater than 00. 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​: muμgreater than>00 ​mg/dL C. Upper H 0H0​: muμgreater than>00 ​mg/dL Upper H 1H1​: muμless than<00 ​mg/dL D. Upper H 0H0​: muμequals=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. ▼ Reject Fail to reject Upper H 0H0. There is ▼ not sufficient sufficient evidence to conclude that t

Answer 1

Given the sample size, n=8181.

Since, we are testing the cholestrol levels before and after the treatment, it is sufficient to use " paired t - test."

Let the difference or change in the cholestrol level before and after the treatment be denoted by "d".

Then, mean of d, and the standard deviation of d,

Level of significance,

• The null and alternative hypothesis is option B:  
• The test statistic is given by
• The table value is given by   [NOTE: since the degrees of freedom is huge, i have found the table value using R program by the following code. ​​]

.> qt(0.01,8180)
[1] -2.326804

We can see that t_cal > t_0.01,8180, hence we reject H₀ and conclude that .

• P-value can also be found by R since the degrees of freedom is too large. Here, P-value =0.068131 and the result is not significant for p<0.01.