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.
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.
To construct a 95% confidence interval for the proportion of deaths in the week before Thanksgiving to the total deaths in the week before and after Thanksgiving, you need to use the formula for the confidence interval for a proportion.
The formula is:
\[ \text{Confidence Interval} = \hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
where:
- \(\hat{p}\) is the sample proportion (the number of deaths in the week before Thanksgiving divided by the total number of deaths),
- \(n\) is the sample size (total number of deaths in the week before and after Thanksgiving), and
- \(Z\) is the z-score associated with the desired level of confidence. For a 95% confidence level, the z-score is approximately 1.96.
Given the information from the Journal of the American Medical Association, we know that there were 12,000 total deaths and 6,062 deaths in the week before Thanksgiving.
Step 1: Calculate the sample proportion, \(\hat{p}\):
\(\hat{p} = \frac{6062}{12000} \approx 0.5052\)
Step 2: Calculate the standard error:
\(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)
\(SE = \sqrt{\frac{0.5052 \times (1-0.5052)}{12000}} \approx 0.0099\)
Step 3: Determine the z-score based on the desired confidence level (95% confidence level):
For a 95% confidence level, the z-score is approximately 1.96.
Step 4: Calculate the confidence interval:
\(\text{Confidence Interval} = 0.5052 \pm 1.96 \times 0.0099\)
Calculate the lower bound:
\(0.5052 - (1.96 \times 0.0099) = 0.4869\)
Calculate the upper bound:
\(0.5052 + (1.96 \times 0.0099) = 0.5235\)
Therefore, the 95% confidence interval for the proportion of deaths in the week before Thanksgiving to the total deaths in the week before and after Thanksgiving is approximately 0.487 to 0.524.