The body temperatures in degrees Fahrenheit of a sample of adults in one small town are: 97.1, 98.1, 98.0, 97.7, 97.4, 99.3, 96.8. Assume body temperatures of adults are normally distributed.

Based on this data, find the 90% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open interval (i.e., parentheses) accurate to 3 decimal places.

90% C.I. =

Question

College

Answer :

ogorwyne ogorwyne · Answer 1 · 2025-04-13T18:09:25-04:00

The 90% confidence interval for the mean body temperature of adults in the town is approximately (97.299, 98.529).

The given data points are:

97.1, 98.1, 98, 97.7, 97.4, 99.3, 96.8

Step 1: Calculate the sample mean (µ):

µ = (97.1 + 98.1 + 98 + 97.7 + 97.4 + 99.3 + 96.8) / 7 ≈ 97.914

sample standard deviation (s):

≈ 0.839

Step 3: Calculate the standard error (SE):

SE = s / sqrt(n)

≈ 0.839 / sqrt(7) ≈ 0.317

Step 4: Find the t-value for a 90% confidence interval with 6 degrees of freedom (n - 1 = 7 - 1 = 6).

Using a t-distribution table we find that the t-value is approximately 1.943.

Step 5: Calculate the margin of error (ME):

ME = t-value * SE

≈ 1.943 * 0.317 ≈ 0.615

Step 6: Calculate the confidence interval:

Lower limit = µ - ME ≈ 97.914 - 0.615 ≈ 97.299

Upper limit = µ + ME ≈ 97.914 + 0.615 ≈ 98.529

So, the 90% confidence interval for the mean body temperature of adults in the town is approximately (97.299, 98.529).

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