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Answer :
The 80% confidence interval for the mean body temperature of adults in the town is approximately (96.98, 99.60).
To calculate the 80% confidence interval of the mean body temperature of adults in the town based on the given sample data, we can use the t-distribution since the population standard deviation is unknown and the sample size is small (n = 7).
1. Calculate the sample mean (x) of the body temperatures:
x = (96.9 + 99.8 + 98.6 + 96.3 + 99.2 + 99.5 + 96.7) / 7
≈ 98.29
2. Calculate the sample standard deviation (s) of the body temperatures:
s = √[((96.9 - 98.29)^2 + (99.8 - 98.29)^2 + (98.6 - 98.29)^2 + (96.3 - 98.29)^2 + (99.2 - 98.29)^2 + (99.5 - 98.29)^2 + (96.7 - 98.29)^2) / (7 - 1)]
≈ 1.362
3. Calculate the critical value (t*) based on the desired confidence level (80%) and the degrees of freedom (n - 1 = 7 - 1 = 6). You can use a t-table or a calculator to find the value.
For an 80% confidence level and 6 degrees of freedom, t* ≈ 1.943
4. Calculate the margin of error (E):
E = t* * (s / √n)
= 1.943 * (1.362 / √7)
≈ 1.31
5. Calculate the lower and upper bounds of the confidence interval:
Lower bound = x - E
= 98.29 - 1.31
≈ 96.98
Upper bound = x + E
= 98.29 + 1.31
≈ 99.60
Therefore, the 80% confidence interval for the mean body temperature of adults in the town is approximately (96.98, 99.60).
Learn more about confidence interval here: https://brainly.com/question/15712887
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