1. Identity the value of the test statistics Homework: HmWrk-26 Score: 3.33 of 2
ID: 3357649 • Letter: 1
Question
1. Identity the value of the test statistics Homework: HmWrk-26 Score: 3.33 of 20 pts 3 of 8 (8 complete) 8.3.15-T Hw Score: 59.09%, 97.5 of 165 pts In a test of the effectiveness of garlic for lowering cholesterol, 64 subjects were treated with raw garlic. Cholesterol levels the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.2 and a standard 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than O. What do the results suggest about he effectiveness of the garlic treatment? Assume that a simple random sample Question Help * were measured before and after P-value, and state the final conclusion that addresses the original claim What are the null and alternative hypotheses? has been selected. Identify the null and alternative hypotheses, test statistic, O C. Ho:H>0 mg/dl Determine the test statistic. H1 : > 0 mg/dL Round to two decimal places as needed.)Explanation / Answer
8.3.15.
Given that,
population mean(u)=0
standard deviation, =2.11
sample mean, x =0.2
number (n)=64
null, Ho: =0
alternate, H1: >0
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 0.2-0/(2.11/sqrt(64)
zo = 0.76
| zo | = 0.76
critical value
the value of |z | at los 5% is 1.64
we got |zo| =0.76 & | z | = 1.64
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : right tail - ha : ( p > 0.76 ) = 0.22
hence value of p0.05 < 0.22, here we do not reject Ho
ANSWERS
---------------
null, Ho: =0
alternate, H1: >0
test statistic: 0.76
critical value: 1.64
decision: do not reject Ho
p-value: 0.22
we do not have enough evidence to support the claim
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