Chapter 19, Problem 19.55SE

a.

To determine

To make: A stem plot for the distribution of body temperature of 20 healthy adults.

Answer to Problem 19.55SE

Explanation of Solution

Given info:

The data represent average body temperatures for 20 healthy adults daily.

Calculation:

Software procedure:

Step-by-step software procedure to draw stemplot using the MINITAB software is as follows:

• Select Graph > Stem and leaf.
• Select the column of Degrees in Graph variables.
• Select OK.

Observation:

Symmetric distribution:

When the left and right sides of the distribution are approximately equal or mirror images of each other, then it is symmetric distribution.

In the stemplot, the observations of the data set are extended approximately equal to the left and to the right in bell shape. Thus, the distribution of daily average temperatures is approximately normal.

The $1.5\times IQR$ rule for outliers:

A observation is a suspected outlier, if it is more than $Q_3+\left(1.5\times IQR\right)$ or less than $Q_1-\left(1.5\times IQR\right)$.

Software procedure:

Step-by-step software procedure for first quartile and third quartile in the MINITAB software is as follows:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns Degrees.
• Choose option statistics, and select first quartile, third quartile, inter quartile range.
• Click OK.

Output using the MINITAB software is as follows:

From Minitab output, the first quartile is 97.847, third quartile is 98.690, and interquartile range is 0.843.

Substitute IQR in the $1.5\times IQR$ rule

$\begin{array}{c}{Q}_3+\left(1.5\times IQR\right)=98.690+\left(1.5\times 0.843\right)\\ =98.690+1.2645\\ =99.9545\end{array}$

$\begin{array}{c}{Q}_1-\left(1.5\times IQR\right)=97.847-\left(1.5\times 0.843\right)\\ =97.847-1.2645\\ =96.5825\end{array}$

The $1.5\times IQR$ rule suspects one outlier in the body temperatures of 20 healthy adults. Because there is a value not in between limits $Q_1-\left(1.5\times IQR\right)=96.5825$ and $Q_3+\left(1.5\times IQR\right)=99.9545$.

Justification:

Thus, the stemplot shows that data of times are symmetrically distributed with one oulier.

b.

To determine

To check: Whether the data give evidence that the mean body temperature for all healthy adults is not equal to the traditional 98.6℉.

Answer to Problem 19.55SE

Yes, the data give evidence that the mean body temperature for all healthy adults is not equal to 98.6℉.

Explanation of Solution

Given info:

The body temperature varies normally with standard deviation $\sigma =0.7°\text{F}$.

Calculation:

STATE:

Does the data give evidence that mean body temperature for all healthy adults is not equal to the traditional 98.6℉?

PLAN:

Parameter:

Define the parameter $\mu$ as the mean body temperature.

The hypotheses are given below:

Null Hypothesis:

$H_0:\mu =98.6℉.$

That is, the mean body temperature is equal to 98.6℉.

Alternative hypothesis:

$H_a:\mu \ne 98.6℉.$

That is, the mean body temperature is not equal to 98.6℉.

Hence, the alternative hypothesis is two sided.

SOLVE:

Conditions for a valid test:

A sample of 20 healthy adults’ body temperature is randomly selected and times follow a normal distribution with standard deviation $\sigma =0.7°\text{F}$.

Test statistic and P-value:

Software procedure:

Step-by-step procedure to obtain test statistic and P-value using the MINITAB software:

• Choose Stat > Basic Statistics > 1-Sample Z.
• In Samples in Column, enter the column of Degrees.
• In Standard deviation, enter 0.7.
• In Perform hypothesis test, enter the test mean 98.6.
• Check Options, enter Confidence level as 95.
• Choose not equal in alternative.
• Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

From the MINITAB output, the test statistic is –2.54 and the P-value is 0.011.

Decision criteria for the P-value method:

If $P\text{-value}\le \alpha \left(=0.05\right)$, then reject the null hypothesis $\left(H_0\right)$.

If $P\text{-value}>\alpha \left(=0.05\right)$, then fail to reject the null hypothesis $\left(H_0\right)$.

CONCLUDE:

Use a significance level, $\alpha =0.05$.

Here, P-value is 0.011, which is less than the value of $\alpha =0.05$.

That is, $P\text{-value}\left(=0.011\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

Thus, there is good evidence that is the mean body temperature for all healthy adults is not equal to the traditional 98.6°F.