Body Temperature. A study by researchers at the University of Maryland addressed the question of whether the mean body temperature of humans is $98.6°F$. The results of the study by P. Mackowiak et al. appeared in the article "A Critical Appraisal of $98.6°F$F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich" (Journal of the American Medical Association, Vol. 268, pp. 1578-1580). Among other data, the researchers obtained the body temperatures of 93 healthy humans, which we provide on the WeissStats site. Use the technology of your choice to do the following.

a. Obtain a normal probability plot, boxplot, histogram, and stem and leaf diagram of the data.

b. Based on your results from part (a), can you reasonably apply the one-mean z-test to the data? Explain your reasoning.

c. At the 1% significance level, do the data provide sufficient evidence to conclude that the mean body temperature of healthy humans differs from $98.6°F$? Assume that $\sigma =0.63°F$.

given,

The results of the study by P. Mackowiak et al. appeared in the article "A Critical Appraisal of $98.6°F$ , the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich"

MINITAB procedure:
Step 1: Choose Graph > Probability Plot.
Step 2: Choose Single, and then click OK.
Step 3: In Graph variables, enter the column of TEMP.
Step 4: Click OK.

Observation:

From the probability plot of the period, the observations are closer to a straight line. Therefore, the distribution of temperature is approximately normally distributed.

MINITAB procedure:

Step 1: Choose Graph ≥ Boxplot or stat > EDA > Boxplot.

Step 2: Under Multiple Yrs, choose Simple. Click OK.

Step 3: in Graph variables, enter the data of TEMP.

Step 4: Click OK.

From the boxplot, it is clear that the distribution of temperature is approximately normally distributed.

MINITAB procedure:

Step 1: Choose Oraph > Histogram.

Step 2: Choose Simple, and then click OK.

Step 3: In Graph variables, enter the corresponding column of TEMP.

Step 4: Click OK.

From the histogram, It is clear that the space of the distribution is well-shaped.

MINITAB procedure:

Step 1: Select Graph > Stem and leaf.

Step 2: Select the column of variables in Graph variables.

Step 3: Select OK.

Stem-and-Leaf Display: TEMP

From the stem-and-leaf diagram, it is clear that the shape of the distribution is bell-shaped.

Yes, it is reasonable apples the one-mean z-test to the data because the distribution of temperature is approximately symmetric. That is, the shape of the distribution is bell-shaped.

State the null and alternative hypotheses:

Null hypothesis:

$H_0:\mu ={98.6}^{\circ }\mathrm{F}$

That is, the mean body temperature of healthy humans does not differ from $98.6°F$.

Alternative hypothesis:

$H_0:\mu \ne {98.6}^{\circ }\mathrm{F}$

That is, the mean body temperature of healthy humans differs from $98.6°F$.

Here, the significance level is, α=0.01.

MIN.TAB procedure:

Slep 1: Choose Stat > Basic Statistics > 1-Sample Z.

Step 2: in Samples in the Column, enter the column of Temperature.

Step 3- In Standard deviation, enter 0.63.

Step 4: In-Perform hypothesis test, enter the least mean as 98.6.

Step 5: Check Options, and enter your Confidence level as 99.

Step 6: Choose not equal in alternative.

Step 7: Click OK in all dialogue boxes.

One-Sample Z: TEMP

From the MINITAB output, the value of the test statistic is -7.29 and the P-value is 0 .

Rejection rule:

If P≤α, then reject the null hypothesis.

Here, the P-value is 0 which is less than the level of significance. That is, P(=0)≤α(=0.01).

Therefore, the null hypothesis is rejected at the 1% level.

Thus, it can be concluded that the test results are statistically significant at a 1% level of significance.

Interpretation:

The data provide sufficient evidence to conclude that the mean body temperature of healthy humans differs from $98.6°F$ at a 1% level.