Asked by Jerome
According to the Journal of the American Medical Association, the doctors found that in the week before and the week after Thanksgiving, there were 12,000 total deaths and 6062 of them occurred in the week before Thanksgiving. Construct a 95% confidence interval of the proportion of deaths in the week before Thanksgiving to the total deaths in the week before and after Thanksgiving.
Answers
Answered by
MathGuru
Use a proportional confidence interval formula:
CI95 = p + or - (1.96)(√pq/n)
Note: + or - 1.96 represents 95% confidence interval.
For p in your problem: 6062/12000 = 0.505
For q: 1 - p = 1 - 0.505 = 0.495
n = 12000
I let you take it from here to calculate the interval.
CI95 = p + or - (1.96)(√pq/n)
Note: + or - 1.96 represents 95% confidence interval.
For p in your problem: 6062/12000 = 0.505
For q: 1 - p = 1 - 0.505 = 0.495
n = 12000
I let you take it from here to calculate the interval.
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