Question

A random sample of 100 patients at RWJ Hospital showed that 45 of them are middle income. Using 99% level of confidence, estimate the percentage of middle income patients of the hospital.

Answer 1

Confidence Interval Formula:

Note: The procedure below is used:
a) if sample size (n) is less than or equal to 5% of the population size (N); and
b) n(p?)(1 - p?) ? 10

Step 1: Find ?/2
Level of Confidence = 99%
? = 100% - (Level of Confidence) = 1%
?/2 = 0.5% = 0.005

Step 2: Find z_?/2
Calculate z_?/2 by using standard normal distribution (normal population with mean (?) = 0 and standard deviation (?) = 1)
with ?/2 = 0.005 as right-tailed area and left-tailed area.

z_?/2 = 2.5758 (Obtained using online z score calculator screenshot attached)

Step 3: Calculate Confidence Interval

Given p? = 45/100 = 0.45, z_?/2 = 2.5758, n = 100

Lower Bound = p? - z_?/2•?p?(1 - p?)/n = 0.45 - (2.5758)*(0.04975) = 0.3219

Upper Bound = p? + z_?/2•?p?(1 - p?)/n = 0.45 + (2.5758)*(0.04975) = 0.5781

Confidence Interval = (0.3219, 0.5781)