91Ó°ÊÓ

Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of $0.05$.

Consider the data provided in the table13.26, to test mean numbers of times a person eat outs are same at $5\%$level of significance:

Consider the null hypothesis that there is no significance difference between the mean numbers of times a person eat outs. For this, use Ti-83 calculator. For this, click on STAT $→$press $1→$EDIT, then put the above given weights of different groups data into the list $L1,L2,L3andL4$. The screenshot is given as below:

The calculator displays :

$F=0.6388149$

$p=0.5437096$

Factor

$df=2$

$SS=20.17321$

$MS=10.0866$

Error

$df=13$

$MS=15.7895604$

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:

$H_0:{\mathrm{μ}}_o={\mathrm{μ}}_h={\mathrm{μ}}_f$

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 localid="1649157482240" $2$,

and the degree of freedom in the denominator - $df\left(denom\right)$is $13$.

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

e) The value of the test statistic $(F-value)$is $0.6388$.

f) The $P$-value for the test is $0.5437$.

g). The graph of the distribution is

h)

i. Level of significance $\mathrm{Î\pm }$ is $0.05$.

ii. Decision: We do not reject the null hypothesis

iii. Reason for decision: $P$-value is $0.5437$ which is greater than the $0.05$ level of significance.

iv. Conclusion: The mean scores of different class delivery are not different.

We accept the null hypothesis.