Show Condition

A drug manufacturer wants to estimate the mean heart rate for patients with a certain heart condition. Becau?

A drug manufacturer wants to estimate the mean heart rate for patients with a certain heart condition. Because the condition is rare, the manufacturer can only find 18 people with the condition currently untreated. From this small sample, the mean heart rate is 93 beats per minute with a standard deviation of 6.
(a) Find a 98% confidence interval for the true mean heart rate of all people with this untreated condition. Show your calculations and/or explain the process used to obtain the interval.

ANSWER: 98% Resulting Confidence Interval for 'true mean': = [89, 97]

Why???

SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION
x-bar = Sample mean 93
s = Sample standard deviation6
n = Number of samples 18
df = degrees of freedom 17

significant digits0

Confidence Level98
"Look-up" Table 't-critical value'3
Look-up Table of t critical values for confidence and prediction intervals. Central two-side area = 98%
with df = 17. Another Look-up method is to utilize Microsoft Excel function:
TINV(probability,degrees_freedom) Returns the inverse of the Student's t-distribution

98% Resulting Confidence Interval for 'true mean':
x-bar +/- ('t critical value') * s/SQRT(n)= 93 +/- 3 * 6/SQRT(18) = [89, 97]