College

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

[tex]
\[
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline
96.9 & 99.7 & 98.4 & 97.5 & 96.8 & 97.4 & 96.6 & 98.7 & 99.6 & 99.8 & 96.5 & 97.2 \\
\hline
\end{tabular}
\]
[/tex]

Assume body temperatures of adults are normally distributed. Based on this data, find the [tex]$95\%$[/tex] 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. Assume the data is from a normally distributed population.

[tex]\[ 95\% \, \text{C.I.} = \square \][/tex]

Answer :

The 95% confidence interval for the mean body temperature is approximately: (97.038, 98.646)

To find the 95% confidence interval for the mean body temperature, we can use the formula for the confidence interval for a population mean when the population is normally distributed and the population standard deviation is unknown:

Confidence Interval Formula:

CI = x ± t * (s / √n)

Where:

x is the sample mean

s is the sample standard deviation

n is the sample size

t is the t-critical value for a 95% confidence level with n - 1 degrees of freedom

Step 1: Calculate the sample mean

96.9, 99.7, 98.4, 97.5, 96.8, 97.4, 96.6, 98.7, 99.6, 99.8, 96.5, 97.2

Sum of the values = 96.9 + 99.7 + 98.4 + 97.5 + 96.8 + 97.4 + 96.6 + 98.7 + 99.6 + 99.8 + 96.5 + 97.2

= 1174.1

Sample mean = Sum / n = 1174.1 / 12

= 97.8417

Step 2: Calculate the sample standard deviation (s)

The sample variance formula is:

s² = Σ(xi - x)² / (n - 1)

First, subtract the mean from each data point, square the result, and sum them up:

(96.9 - 97.8417)² = 0.8892 (99.7 - 97.8417)² = 3.4482 (98.4 - 97.8417)² = 0.3104 (97.5 - 97.8417)² = 0.1160 (96.8 - 97.8417)² = 1.0877 (97.4 - 97.8417)² = 0.1922 (96.6 - 97.8417)² = 1.5524 (98.7 - 97.8417)² = 0.7399 (99.6 - 97.8417)² = 3.0761 (99.8 - 97.8417)² = 3.8292 (96.5 - 97.8417)² = 1.7831 (97.2 - 97.8417)²

= 0.4062

Sum of squared differences = 17.5183

Sample variance (s²) = 17.5183 / (12 - 1) = 1.5962

Sample standard deviation (s) = √1.5962 = 1.264

Step 3: Find the t-critical value for a 95% confidence interval

Since we have 12 data points, the degrees of freedom (df) is 12 - 1 = 11. Using a t-table or calculator for a 95% confidence level and df = 11, the t-critical value (t) is approximately 2.201.

Step 4: Calculate the margin of error

Margin of error = t * (s / √n)

Margin of error = 2.201 * (1.264 / √12) = 2.201 * (1.264 / 3.464)

= 2.201 * 0.365

= 0.804

Step 5: Calculate the confidence interval

Lower bound = x - margin of error = 97.8417 - 0.804

= 97.0377

Upper bound = x + margin of error = 97.8417 + 0.804

= 98.6457

Complete question

The body temperatures (in degrees Fahrenheit) of a sample of adults from a small town are as follows:

96.9, 99.7, 98.4, 97.5, 96.8, 97.4, 96.6, 98.7, 99.6, 99.8, 96.5, 97.2

Assume that the body temperatures of adults in this town are normally distributed. Based on this data, calculate the 95% confidence interval for the mean body temperature of adults in the town.

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