Question 1015146
Assuming a normally distributed population where the sample sd is an adequate estimate of the population sd,
t df=48 is the test statistic.
95% CI is +/-2.01*s/sqrt (49), where s is sd
=+/-2.01*21/7
=+/-6.03
add to mean
(-5.63, 6.43) units are mg/dL.
The interval contains 0, so that 0 is a reasonable value to assume, and we cannot conclude that garlic has any difference upon the LDL cholesterol levels.