Math 153 Elementary Statistics doaa shinab 12/12/17 5:40 PM Test: Chapters 8-10,
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Question
Math 153 Elementary Statistics doaa shinab 12/12/17 5:40 PM Test: Chapters 8-10, 11.1 & 12.1 Exam nent This Question: 1 pt Time Remaining: 01:19:19 Submit Test 140127 (o complete) This Test: 27 pts possible Question Help * In a test of the effectiveness of garlic for before and after the minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.1 and a standard d treatment. evel to test the claim that with garlic treatment, the mean change i conc deviation of 16.5. Use a 0.05 significance n LDL cholesterol is greater than 0 What do the results suggest about the effectiveness of the test statistic, P-value, and state the final H: H>0 mg/dl Determine the test statistic Round to two decimal places as needed) (Round to three decimal places as needed.) State the final conclusion that addresses the original claim evidence to conclude that the mean of the population of changes 1 0. Click to select your answer(s).Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: d< 0
Alternative hypothesis: d > 0
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).
s = sqrt [ ((di - d)2 / (n - 1) ]
s = 16.5
SE = s / sqrt(n)
S.E = 2.0625
DF = n - 1 = 64 -1
D.F = 63
t = [ (x1 - x2) - D ] / SE
t = 0.05
where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 63 degrees of freedom is greater than 0.05.
Thus, the P-value = 0.48
Interpret results. Since the P-value (0.48) is greater than the significance level (0.05), we have to reject the null hypothesis.