Question

A local health center noted that in a sample of 1000 patients, 220 were referred to them by the local hospital.

a. Provide a 99% confidence interval for the proportion of all patients who are referred

Answer 1

The 99% confidence interval for the proportion of all patients who are referred is approximately 19.18% to 24.82%.

Step-by-step explanation:

To calculate the confidence interval for the proportion of patients referred, we use the formula:

`\[ \text{Confidence Interval} = \hat{p} \pm Z * \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]`

Where:

`\(\hat{p}\)` is the sample proportion (220/1000 = 0.22),

Z is the Z-score corresponding to the desired confidence level (for 99%, Z ≈ 2.576),

n is the sample size (1000).

Plugging in these values:

`\[ \text{Confidence Interval} = 0.22 \pm 2.576 * \sqrt{(0.22 * (1-0.22))/(1000)} \]`

After calculation, the confidence interval is approximately 0.1918 to 0.2482.

This means we can be 99% confident that the true proportion of patients referred lies within this interval. In other words, if we were to take many samples and compute a 99% confidence interval for each, we would expect about 99% of them to contain the true proportion of referred patients.

It's important to note that the width of the interval is influenced by both the sample proportion and the sample size. In this case, the relatively large sample size contributes to a narrower interval, providing a more precise estimate of the true proportion of referred patients in the population.