The data shown is the recorded body temperatures of $130$subjects as estimated from available histograms. Traditionally we are taught that the normal human body temperature is $98.6$F. This is not quite correct for everyone. Are the mean temperatures among the four groups different? Calculate $95\%$confidence intervals for the mean body temperature in each group and comment about the confidence intervals.

To find these results on the calculator:

Press STAT. Press 1 EDIT. Put the data into the lists $L_1,L_2,L_3$

Press STAT, and arrow over to TESTS, and arrow down to ANOVA. Press ENTER, and then enter $L_1,L_2,L_3$. Press ENTER. We will see that the values in the foregoing ANOVA table are easily produced by the calculator, including the test statistic and the $p$-value of the test.

The calculator displays :

$F=3.1363622$

$p=0.080746$

Factor

$df=2$

$SS=1468909.2$

$MS=734454.6$

Error

$df=12$

$SS=2819076$

$MS=234923.067$

We use a solution sheet to conduct the hypothesis tests, and we have:

a) The null hypothesis that three mean commuting mileages are the same is:

$H_0:{\mu }_p={\mu }_m={\mu }_h$

b) The alternate hypothesis is that at least any two of the means are different.

c) The degree of freedom in the numerator - $df\left(num\right)$is $2$,

and the degree of freedom in the denominator - $df\left(denom\right)$is $12$.

d) We use the $F$ distribution for the test.

e) The value of the test statistic ($F-$value) is $3.14$.

f) The $P$-value for the test is $0.08$.

g). The graph of the distribution is

h)

i. Level of significance $\alpha$ is $0.05$.

ii. Decision: We do not reject the null hypothesis

iii. Reason for decision: $P$-value is $0.08$ which is greater than the $0.05$ level of significance.

iv. Conclusion: There is not sufficient evidence to conclude that the mean numbers of daily visitors are different.

We accept the null hypothesis.