Question

A nutritionist says 72% of patients are trying to avoid trans fats in their diets. To check his claim, you randomly survey 200 patients and found that 128 patients stated that they are trying to avoid trans fats in their diets. Find the 95% confidence interval for the proportion of patients who are trying to avoid trans fats in their diets. state the formulas used and show your work. Does you study support his claim? Why?

Answer 1

Solution:

Confidence interval for Population Proportion is given as below:

Confidence Interval = p̂ ± Z* sqrt(p̂*(1 – p̂)/n)

Where, p̂ is the sample proportion, Z is critical value, and n is sample size.

We are given

Number of items of interest = x = 128

Sample size = n = 200

Sample proportion = p̂ = x/n = 128/200 = 0.64

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence Interval = p̂ ± Z* sqrt(p̂*(1 – p̂)/n)

Confidence Interval = 0.64 ± 1.96* sqrt(0.64*(1 – 0.64)/200)

Confidence Interval = 0.64 ± 1.96* 0.0339

Confidence Interval = 0.64 ± 0.0665

Lower limit = 0.64 - 0.0665 = 0.5735

Upper limit = 0.64 + 0.0665 = 0.7065

Confidence interval = (0.5735, 0.7065)

This confidence interval does not support the claim because the value for population proportion as 0.72 is not lies between the above confidence interval.