Answer :
The 99% confidence interval of the mean body temperature of adults in the town is (96.773, 99.927).
To calculate the confidence interval, we use the formula for a confidence interval of a population mean with a known standard deviation:
\[ \text{Confidence Interval} = \text{Sample Mean} \pm \text{Margin of Error} \]
First, calculate the sample mean (\(\bar{x}\)) of the given body temperatures:
\[ \bar{x} = \frac{96.7 + 98.5 + 97.5 + 98.7 + 99.7 + 97.6 + 98.2}{7} = 98.071 \]
Next, calculate the standard deviation (\(s\)) of the sample:
\[ s = \sqrt{\frac{\sum{(x - \bar{x})^2}}{n - 1}} = 0.936 \]
The margin of error (\(E\)) for a 99% confidence interval is calculated using the t-distribution with \(n - 1\) degrees of freedom and a significance level of \(\alpha = 0.01\):
\[ E = t_{\alpha/2, \, df} \times \frac{s}{\sqrt{n}} \]
Using a t-table or calculator, for \(df = 6\) and \(\alpha/2 = 0.005\), \(t_{\alpha/2, \, df} = 3.7074\).
Now, calculate the margin of error:
\[ E = 3.7074 \times \frac{0.936}{\sqrt{7}} = 1.598 \]
Finally, the confidence interval is:
\[ \text{Confidence Interval} = \bar{x} \pm E = 98.071 \pm 1.598 \]
Rounded to 3 decimal places, the 99% confidence interval is (96.773, 99.927).
In summary, the 99% confidence interval of the mean body temperature of adults in the town is (96.773, 99.927), calculated using the sample mean, standard deviation, t-distribution, and margin of error formula for confidence intervals.
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