High School

The body temperatures in degrees Fahrenheit of a sample of 7 adults in one small town are:

- 96.9
- 99.8
- 98.6
- 96.3
- 99.2
- 99.5
- 96.7

Assume body temperatures of adults are normally distributed. Based on this data, find the 80% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).

80% C.I. =

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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