Question

A national survey of companies included a question that asked whether the customers like the new flavor of a cola from company A. The sample results of 1000 customers, and 850 of them indicated that they liked the new flavor. The 98% confidence interval on the population proportion of people who like the new flavor is _______________.

Answer 1

Answer:

The 98% confidence interval on the population proportion of people who like the new flavor is (0.8237, 0.8763).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 1000, \pi = \frac{850}{1000} = 0.85$

98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$, so $Z = 2.327$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.85 - 2.327\sqrt{\frac{0.85*0.15}{1000}} = 0.8237$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.85 + 2.327\sqrt{\frac{0.85*0.15}{1000}} = 0.8763$

The 98% confidence interval on the population proportion of people who like the new flavor is (0.8237, 0.8763).