Posted by Mack on .
The mean serum cholesterol level in the population of males who do not develop heart disease is known to be 219 mg/100 ml. Suppose you perform a study to investigate whether the mean serum cholesterol levels in males who do develop heart disease is higher than that of the disease-free population. In a sample of 28 men who have heart disease, the sample mean serum cholesterol level is 232 mg/100 ml with a sample standard deviation of 38 mg/100 ml. Perform the appropriate test at the 0.10 level of significance.
a. Ho :
b. Ha :
c. Critical value(s) =
d. Test statistic =
e. Approximate P-value =
Interpret this decision:
a non-normal population is determined to have a mean of 60 and a standard deviation of 4. Ninety-six percent of all observed values will occur in what range?
Assume a normal distribution.
Ho: mean = 219 mg/100 ml
Ha: mean ≠ 219 mg/100 ml
c. Z > ± 1.645
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion
Statistics anon. -
First, if you have a question, it is much better to put it in as a separate post in <Post a New Question> rather than attaching it to a previous question, where it is more likely to be overlooked.
If it is not normal, the assumptions applied to a normal distribution will be in error. However, for a normal distribution, this is what you would want.
Mean ± 2.055 SD