Question

A researcher used a new drug to treat 171 subjects with high sodium. For the patients in the study, after six (6) months of treatment the average decrease in sodum level was 5.879 milliequivalents per liter (mEq/L). Assume that the decrease in sodium in a human after six months of taking the drug follows a Normal distribution with an unknown mean and a standard deviation of 2.615 mEq/L.

In order to find the margin of error for a 96% confidence interval for the mean decrease in sodium level of the subjects in the study, what confidence coefficient z* should be used? Use Table C: t distribution Table

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 5.879

sample standard deviation = s = 2.615

sample size = n = 171

Degrees of freedom = df = n - 1 = 171 - 1 = 170

At 96% confidence level

= 1 - 96%

=1 - 0.96 =0.04

/2 = 0.02

t/2,df = t0.02,170 = 2.070

Margin of error = E = t/2,df * (s /n)

= 2.070 * ( 2.615 / 171)

Margin of error = E = 0.414