9. A doctor noticed that many female patients were Vitamin D deficient. She woul
ID: 3361535 • Letter: 9
Question
9. A doctor noticed that many female patients were Vitamin D deficient. She would like to det what proportion of her patients are deficient. She chooses a random sample of 45 medical files of female patients and finds that 37 of these patients are Vitamin D deficient. (Round to 3 decimal pl (Write out all formulas used and show ALL work) ermi aces a. Find x, n, p, and q.(4 points) b. For a 95% Confidence level, identify and the zen (2 points) c. Find the margin of error, E for a 95% confidence level. write out the formula for E. (3 points) d. Construct and interpret a 95% Confidence Interval for the true population proportion of female patients who are Vitamin D deficient. Write out the formula used for a Confidence Interval for Estimating a Population Proportion p. (4 points)Explanation / Answer
TRADITIONAL METHOD
given that,
possibile chances (x)=37
sample size(n)=45
success rate ( p )= x/n = 0.8222
I.
sample proportion = 0.8222
standard error = Sqrt ( (0.8222*0.1778) /45) )
= 0.057
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.057
= 0.1117
III.
CI = [ p ± margin of error ]
confidence interval = [0.8222 ± 0.1117]
= [ 0.7105 , 0.9339]
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DIRECT METHOD
given that,
possibile chances (x)=37
sample size(n)=45
success rate ( p )= x/n = 0.8222
CI = confidence interval
confidence interval = [ 0.8222 ± 1.96 * Sqrt ( (0.8222*0.1778) /45) ) ]
= [0.8222 - 1.96 * Sqrt ( (0.8222*0.1778) /45) , 0.8222 + 1.96 * Sqrt ( (0.8222*0.1778) /45) ]
= [0.7105 , 0.9339]
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interpretations:
1. We are 95% sure that the interval [ 0.7105 , 0.9339] 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 contains the true population proportion
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