College

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

[tex]\[
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline
97.6 & 97.1 & 99.7 & 96.9 & 99.1 & 99.9 & 97 & 98.7 & 97.3 & 98.2 & 98.6 & 98.1 \\
\hline
\end{tabular}
\][/tex]

Assume body temperatures of adults are normally distributed. Based on this data, find the [tex]$80\%$[/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.

[tex]$80\%$[/tex] C.I. = (97.299, 98.529)

Answer :

The 80% confidence interval for the mean body temperature is (98.12, 98.646).

To calculate the 80% confidence interval for the mean body temperature, we follow these steps:

The sample data is given as 97.6, 97.1, 99.7, 96.9, 99.1, 99.9, 97, 98.7, 97.3, 98.2, 98.6, 98.1

From the above, the sample size (n) is 12.

Start by calculating the sample mean by dividing the sum of the samples by the number of sample

[tex]\bar x = \dfrac{97.6 + 97.1 + 99.7 + 96.9 + 99.1 + 99.9 + 97 + 98.7 + 97.3 + 98.2 + 98.6 + 98.1}{12}[/tex]

[tex]\bar x = \dfrac{1178.2}{12}[/tex]

[tex]\bar x = 98.183[/tex]

Next, we calculate the sample standard Deviation

[tex]s = \sqrt{ \dfrac{\sum(x - \bar x)\²}{n - 1}}[/tex]

[tex]s = \sqrt{\dfrac{(97.6 - 98.183)^2 + (97.1 - 98.183)^2 + (99.7 - 98.183)^2 +....+ (98.1 - 98.183)^2}{12 - 1}}[/tex]

[tex]s = \sqrt{\dfrac{11.876668}{11}[/tex]

[tex]s = \sqrt{1.07969709091[/tex]

s = 1.039

For an 80% confidence level and degrees of freedom (df = n - 1 = 11), the critical t-value (t*) is 0.876 (See attachment for table)

Calculate the Margin of Error: Margin of Error (E)

[tex]E = \dfrac{t* s}{\sqrt n}[/tex]

[tex]E = \dfrac{0.876 * 1.039}{\sqrt{12}}[/tex]

E = 0.263

To find the confidence interval, we use the following:

CI = (mean - E, mean + E).

CI = (98.383 - 0.263, 98.383 + 0.263).

CI = (98.12, 98.646).

Missing Data

97.6, 97.1, 99.7, 96.9, 99.1, 99.9, 97, 98.7, 97.3, 98.2, 98.6, 98.1

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