A group of six individuals with high cholesterol levels were given a new diet designed to lower cholesterol levels. Cholesterol levels, in milligrams per deciliter, were measured before and after the Implementation of the diet for each individual, with the following results:

$\begin{array} { c c c }
\hline \text { Individual } & \text { Before } & \text { After } \
\hline 1 & 227 & 160 \
2 & 234 & 174 \
3 & 257 & 178 \
4 & 269 & 170 \
5 & 218 & 178 \
6 & 248 & 178
\end{array}$

A dietician claims that the mean reduction in cholesterol level is 51 milligrams per deciliter. Does the 98% confidence interval contradict this claim? (Hint: you need to find the 98% confidence interval For the mean reduction in cholesterol level.)

Question

A group of six individuals with high cholesterol levels were given a new diet designed to lower cholesterol levels. Cholesterol levels, in milligrams per deciliter, were measured before and after the Implementation of the diet for each individual, with the following results:
 
$\begin{array} { c c c } 
\hline \text { Individual } & \text { Before } & \text { After } \
\hline 1 & 227 & 160 \
2 & 234 & 174 \
3 & 257 & 178 \
4 & 269 & 170 \
5 & 218 & 178 \
6 & 248 & 178
\end{array}$

A dietician claims that the mean reduction in cholesterol level is 51 milligrams per deciliter. Does the 98% confidence interval contradict this claim? (Hint: you need to find the 98% confidence interval For the mean reduction in cholesterol level.)

Quizplus · Accepted Answer

We need to find the mean and standard deviation of the reduction in cholesterol levels.

Mean = (227-160) + (234-174) + (257-178) + (269-170) + (218-178) + (248-178) / 6 = 52.33

Standard deviation = √[Σ(X-mean)² / (n-1)] 
where X is the reduction in cholesterol level for each individual

= 31.16

Using a t-distribution with 5 degrees of freedom (n-1), we can find the margin of error for a 98% confidence interval:

Margin of error = t(0.01/2,5) x (s/√n) 
where t(0.01/2,5) is the critical value from the t-distribution with 0.01/2 = 0.005 as the tail probability and 5 degrees of freedom 
and s is the standard deviation of the reduction in cholesterol levels and n is the sample size which is 6

t(0.005,5) = 4.032 
Margin of error = 4.032 x (31.16 / √6) = 63.42

The 98% confidence interval for the mean reduction in cholesterol levels is therefore:

52.33 ± 63.42

This means that the true population mean reduction in cholesterol levels could be as low as -11.09 or as high as 115.75 with 98% confidence. Since the interval includes 51, we cannot contradict the dietician's claim. Therefore, the answer is B.

A Group of Six Individuals with High Cholesterol Levels Were