Final answer:
To calculate the 95% confidence interval for the proportion of African-American women with vitamin D deficiency, we use the sample proportion and the standard Z-value for 95% confidence. The resulting range is where we expect the true population proportion to fall with 95% certainty.
Step-by-step explanation:
The question asks us to calculate a 95% confidence interval for the proportion of African-American women with vitamin D deficiency based on a national nutrition study. To compute this, we use the formula for the confidence interval of a proportion, which is p ± Z*(sqrt(p(1-p)/n)), where p is the sample proportion, Z is the Z-value corresponding to the 95% confidence level (1.96 for 95%), and n is the sample size.
In this case, we have 649 out of 1546 African-American women with vitamin D deficiency. The sample proportion p is 649/1546, which approximately equals 0.420. The sample size n is 1546. Plugging these values into the formula gives us the confidence interval:
0.420 ± 1.96*(sqrt(0.420*(1-0.420)/1546))
Calculating the margin of error and then adding and subtracting it from the sample proportion gives us the bounds of the confidence interval.
A confidence interval is a range of values, derived from sample statistics, that is likely to contain the value of an unknown population parameter. In the context of this study, a 95% confidence interval implies that there is a 95% probability that the interval calculated from the sample data contains the true proportion of vitamin D deficiency among African-American women in the population.