Question

A recent study of 3100 children randomly selected found 24​% of them deficient in vitamin D.
​a) Construct the​ 98% confidence interval for the true proportion of children who are deficient in vitamin D.

Answer 1

Answer on Question #67671 – Math – Statistics and Probability

Question

A recent study of 3100 children randomly selected found 24% of them deficient in vitamin D.

a) Construct the 98% confidence interval for the true proportion of children who are deficient in vitamin D.

Solution

If $n\hat{p} \geq 10$ and $n(1 - \hat{p}) \geq 10$, we can use the following formula to compute the confidence interval for the true population proportion:

$$\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}},$$

where $\hat{p}$ is a sample proportion, $n$ is the sample size, $z^*$ is multiplier dependent on the level of confidence:

In our case, we have $\hat{p} = 0.24$, $n = 3100$, $z^* = 2.326$.

Conditions $n\hat{p} \geq 10$ and $n(1 - \hat{p}) \geq 10$ are met. Thus, a 98% confidence interval for the true proportion of children who are deficient in vitamin D is given by

$0.24 \pm 2.326 \sqrt{\frac{0.24(1 - 0.24)}{3100}} \approx 0.24 \pm 0.018 = (0.222, 0.258)$, hence in terms of percent it will be $24\% \pm 1.8\% = (22.2\%, 25.8\%)$

Answer: $0.24 \pm 0.018 = (0.222, 0.258)$.

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