A doctor would like to know what proportion of their patients are vitamin D deficient. They choose a random sample of
3
0
patients and find that
2
3
of them are vitamin D deficient. What is the margin of error with
9
5
%
confidence? Round your answer to the nearest hundredth.
(
1
point
)
A doctor would like to know what proportion of their patients are vitamin D deficient. They choose a random sample of
3
0
patients and find that
2
3
of them are vitamin D deficient. What is the margin of error with
9
5
%
confidence? Round your answer to the nearest hundredth.
(
1
point
)
To calculate the margin of error, we can use the formula:
Margin of Error = Z * sqrt[(p-hat)(1-p-hat)/n]
Where:
Z = Z-score corresponding to the confidence level (95% confidence corresponds to a Z-score of 1.96)
p-hat = Sample proportion (23/30 = 0.7667)
n = Sample size (30)
Plugging in the values:
Margin of Error = 1.96 * sqrt[(0.7667)(1-0.7667)/30]
Margin of Error ≈ 1.96 * sqrt[(0.7667)(0.2333)/30]
Margin of Error ≈ 1.96 * sqrt(0.179105/30)
Margin of Error ≈ 1.96 * sqrt(0.00597)
Margin of Error ≈ 1.96 * 0.07734
Margin of Error ≈ 0.1516
Therefore, the margin of error with 95% confidence is approximately 0.17, rounded to the nearest hundredth.
This answer is wrong.
I apologize for the mistake. Let's reevaluate the calculation:
Given:
Sample size (n) = 30
Proportion of patients vitamin D deficient (p-hat) = 23/30 = 0.7667
Confidence level = 95%
First, we need to find the standard error of the proportion, which is given by:
Standard Error = sqrt[(p-hat)(1 - p-hat) / n]
Standard Error = sqrt[(0.7667)(1 - 0.7667) / 30]
Standard Error = sqrt[0.178971 / 30]
Standard Error ≈ sqrt(0.005966) ≈ 0.07728
Next, we use the z-score for 95% confidence, which is approximately 1.96 for a large sample size:
Margin of Error = z * Standard Error
Margin of Error = 1.96 * 0.07728
Margin of Error ≈ 0.1516
Therefore, the correct margin of error with 95% confidence is approximately 0.15, rounded to the nearest hundredth. Thank you for pointing out the error.