In 2012 The American Journal of Clinical Nutrition reported

In 2012, The American Journal of Clinical Nutrition reported that 31% of Australian adults over age 25 have a Vitamin D deficiency. The data came from AusDiab study of 11218 Australians.
a)do these data meet the assumptions necessary for inference?
b)create a 95% confidence interval
c) interepret the inerval in context
d)Explain what 95% confidence means

Solution

a)
Yes it meets & can be assume that approximately normal
b)
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=3477.58
Sample Size(n)=11218
Sample proportion = x/n =0.31
Confidence Interval = [ 0.31 ±Z a/2 ( Sqrt ( 0.31*0.69) /11218)]
= [ 0.31 - 1.96* Sqrt(0) , 0.31 + 1.96* Sqrt(0) ]
= [ 0.301,0.319]

c)
Interpretations:                      
1) We are 95% sure that the interval [ 0.301,0.319] contains the                       
true population proportion
2) If a large number of samples are collected, and a confidence interval is created                      
for each sample, 95% of these intervals will contain true population proportion