Question

In a recent survey of 4276 randomly selected households showed that 4020 of them had telephones. Using these results, construct a 99% confidence interval estimate of the true proportion of households
with telephones.

• P – PARAMETER
  • Describe the parameter in context. This should be a complete sentence

• A – ASSUMPTIONS
  • Are we allowed to construct an interval and why?

• N – NAME THE INTERVAL
  • State what command you will be using the compute the interval along with the values you need.

• I – INTERVAL CALCULATION
  • Find the interval

• C – CONCLUSION
  • Write a sentence summarizing what you have found.

Answer 1

Given that 4020 of them had telephones out of the total sample size of 4276, therefore the sample proportion here is computed as:
p = x/n = 4020 / 4276 = 0.9401

From standard normal tables, we have here:
P( -2.576 < Z < 2.576) = 0.99

np = x = 4020 >= 5
n(1-p) = n - x = 4276 - 4020 >= 5
Therefore we are allowed to use the assumption here that the normal approximation is appropriate for the proportion.

Therefore the confidence interval here is obtained as:

This is the required 99% confidence interval here.