Vitamin D, whether ingested as a dietary supplement or produced naturally when sunlight falls upon the skin, is essential for strong, healthy bones. The bone disease rickets was largely eliminated during the 1950s, but now there is concern that a generation of children more likely to watch TV or play computer games than spend time outdoors is at increased risk. A recent study of 3400 children randomly selected found 24% of them deficient in vitamin D.

a) Find a 98% confidence interval.
b) Explain carefully what your interval means.
c) Explain what "98% confidence" means.

b) This interval establish the limits on where we can expect the true value for the population proportion with deficient in vitamin D at 98% of confidence

c) The 98% represent the confidence level for the interval founded so we have a probability of 2% of comit error Type I.

Step-by-step explanation:

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. The confidence level is at 98% of confidence, our significance level would be given by [tex]\alpha=1-0.98=0.02[/tex] and [tex]\alpha/2 =0.01[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-2.33, z_{1-\alpha/2}=2.33[/tex]

The confidence interval for the mean is given by the following formula:

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.24 - 2.33\sqrt{\frac{0.24(1-0.24)}{3400}}=0.223[/tex]

[tex]0.24 + 2.33\sqrt{\frac{0.24(1-0.24)}{3400}}=0.257[/tex]

The 98% confidence interval would be given by (0.223;0.257)

Part b

This interval establish the limits on where we can expect the true value for the population proportion with deficient in vitamin D at 98% of confidence

Part c

The 98% represent the confidence level for the interval founded so we have a probability of 2% of comit error Type I.