In a survey of 2306 adult Americans aged 18 and older, it was found that 418 of them have donated blood in the past two years. Construct a 99% confidence interval estimate of the percentage of adult Americans aged 18 and older that have donated blood in the past two years. Please be sure to round each value to the nearest thousandth (third decimal place) when making calculations and submitting your solutions in the text entry boxes provided. Do not use commas.
Answer:
The 99% confidence interval estimate of the percentage of adult Americans aged 18 and older that have donated blood in the past two years is between 16.061% and 20.193%
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 – \frac{\alpha}{2}[/tex].
In a survey of 2306 adult Americans aged 18 and older, it was found that 418 of them have donated blood in the past two years.
This means that [tex]n = 2306, \pi = \frac{418}{2306} = 0.18127[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 – \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi – z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.18127 – 2.575\sqrt{\frac{0.18127*0.81873}{2306}} = 0.16061[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.18127 + 2.575\sqrt{\frac{0.18127*0.81873}{2306}} = 0.20193[/tex]
Confidence interval for the percentage:
Proportions multiplied by 100%. So
0.16061*100% = 16.061%
0.20193 = 20.193%
The 99% confidence interval estimate of the percentage of adult Americans aged 18 and older that have donated blood in the past two years is between 16.061% and 20.193%