Question

In a survey, 376 out of 1,078 US adults said they drink at least 4 cups of coffee a day. Find a point estimate (P) for the population proportion of US adults who drink at least 4 cups of coffee a day, then construct a 99% confidence interval for the proportion of adults who drink at least 4 cups of coffee a day.

Answer 1

Answer:

The point estimate is 0.3488.

The 99% confidence interval for the proportion of adults who drink at least 4 cups of coffee a day is (0.3114, 0.3862).

Step-by-step explanation:

In a sample with a number n of people surveyed with a point estimate 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:

$n = 1078, \pi = \frac{376}{1078} = 0.3488$

The point estimate is 0.3488.

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3488 - 2.575\sqrt{\frac{0.3488*0.6512}{1078}} = 0.3114$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3488 + 2.575\sqrt{\frac{0.3488*0.6512}{1078}} = 0.3862$

The 99% confidence interval for the proportion of adults who drink at least 4 cups of coffee a day is (0.3114, 0.3862).