Question

A researcher used a new drug to treat 183 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.

Find the margin of error for a 70% confidence interval for the mean decrease in sodium of the subjects. Round your answer to three (3) decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 5.879

sample standard deviation = s = 2.615

sample size = n = 183

Degrees of freedom = df = n - 1 = 183 - 1 = 182

At 70% confidence level

= 1 - 70%

=1 - 0.70 =0.30

/2 = 0.15

t/2,df = t0.15,182 = 1.039

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

= 1.039 * ( 2.615 / 183)

Margin of error = E = 0.201