Question

i am stuck on the secomd step of this question. In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes in their LDL cholesterol had a mean of 0.4. Assume the population standard deviation is 21.0, construct a 98% confidence interval for the true mean change in LDL cholesterol after the garlic treatment.

Answer 1

n= 49, = 0.4, = 21,

c= 98%

since standard deviation for mean difference in not given and population standard deviation is given, we consider mean difference as sample mean. we use z interval as follows

formula for confidence inerval is

Where is the z critical value for c= 98%

using normal z table we get

= 2.33

-6.579  < <   7.379

Thus we get 98% confidence interval as follows,

(-6.579 , 7.379 )