The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 80 customers surveyed, 15 ordered cheese pizza. They want a 99% confidence interval of the true proportion of customers who order only cheese pizza.
What is the 99% confidence interval of the true proportion of customers who order only cheese pizza?
Question 26 options:
(0.075, 0.300)
(0.115, 0.260)
(0.086, 0.289)
(0.102, 0.273)
To calculate the confidence interval, we can use the formula for a proportion:
CI = p̂ ± Z * sqrt((p̂(1-p̂)/n))
where p̂ is the sample proportion, Z is the z-score corresponding to the desired confidence level, and n is the sample size.
In this case, the sample proportion (p̂) is 15/80 = 0.1875.
The z-score for a 99% confidence level is approximately 2.576 (taken from a standard normal distribution table).
Plugging in the values, we get:
CI = 0.1875 ± 2.576 * sqrt((0.1875 * (1-0.1875)/80))
= 0.1875 ± 0.0497
= (0.1378, 0.2372)
Therefore, the 99% confidence interval of the true proportion of customers who order only cheese pizza is (0.138, 0.237).