A special diet is intended to reduce the cholesterol of patients at risk of heart disease. If the diet is effective, the target is to have the average cholesterol of this group be below 200. After six months on the diet, an SRS of 50 patients at risk for heart disease had an average cholesterol of x̅ = 192, with standard deviation s = 21. Is this sufficient evidence that the diet is effective in meeting the target? Assume the distribution of cholesterol for patients in this group is approximately Normal with mean μ.

Question

A special diet is intended to reduce the cholesterol of patients at risk of heart disease. If the diet is effective, the target is to have the average cholesterol of this group be below 200. After six months on the diet, an SRS of 50 patients at risk for heart disease had an average cholesterol of  x̅ = 192, with standard deviation s = 21. Is this sufficient evidence that the diet is effective in meeting the target? Assume the distribution of cholesterol for patients in this group is approximately Normal with mean μ.

nurali · Answer

You are comparing a sample mean with a target value so you use a t location test with sample mean equal to 192, sample standard deviation of 21, sample size of 50, and comparison mean of 200. 

t = (192 - 200) / (21/ sqrt(50)) = -2.69374, you can then compare this to a t distribution rejection table. 

To convert your t-value to a p-value, you can use the t-table in your book, for a one sided test with 49 degrees of freedom (sample size minus one ) your pvalue is: 

0.004833