In: Statistics and Probability
The 49 subjects treated with raw garlic had LDL cholesterol measurements with a mean of 151 and a standard deviation of 15, while 48 subjects given placebos had LDL cholesterol measurements with a mean of 149 and a standard deviation of 14.
Use a 0.05 significance level to test the claim that there is no difference between the mean LDL cholesterol levels of subjects treated with raw garlic and subjects given placebos.
Here we have given two groups which are LDL cholesterol levels of subjects treated with raw garlic and subjects given placebos.
n1 = 49
X1bar = 151
sigma1 = 15;
n2 = 48
X2bar = 149
sigma2 = 14
Here we will use two sample z-test because :
The samples from each population must be independent of one another. • The populations from which the samples are taken must be normally distributed and the population standard deviations must be know, or the sample sizes must be large (i.e. n1≥30 and n2≥30.
1. Hypothesis for the test is,
H0 : mu1 = mu2 Vs H1 : mu1 mu2
where mu1 and mu2 are two population means.
2. Assume alpha = level of significance = 0.05
3. For two tailed test critical value = -1.96 , 1.96
4. Test statistic Z = 0.68
5. P-value = 0.4971
6. P-value > alpha
Accept H0 at 5% level of significance.
7. Conclusion : There is no difference between the mean LDL cholesterol levels of subjects treated with raw garlic and subjects given placebos.