Thank you for visiting 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. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
The 90% confidence interval for the mean body temperature of adults in the town is approximately (97.299, 98.529).
How to solve for the confidence interval
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).
Read more on confidence interval here:https://brainly.com/question/15712887
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