91Ó°ÊÓ

Consider the data provided in the table13.27, to test mean numbers of daily visitors to a ski resort are same for three types of snow conditions at $5\%$level of significance.

Consider the null hypothesis that there is no significance difference between three types of snow conditions. For this, use Ti-83 calculator. For this, click on STAT $→$press $1→$EDIT, then put given data into the list $L1,L2,andL3$. The screenshot is given as below:

Now again press STAT $→$arrow over the TESTS $→$arrow down to ANOVA $→$Press var then select1, press var then select 2, press var then select 3 and press var then select 4 to fill the values of $L1,L2,L3andL4$. The screen shot is given as below:

a).The null hypothesis that three mean commuting mileages are same is given as below:

$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2={\mathrm{μ}}_3$

b). The alternate hypothesis will be at least any two of the means are different.

c). The degree of freedom in the numerator is $2$, and the degree of freedom in the denominator is $12$.

d). The $F$distribution is used for the test of three different means.

e). The value of the test statistic is$3.12$.

f). The $P$-value for the test is $\mathbf{0}\mathbf{.}\mathbf{08}$.

g.The graph of the distribution is given as below:

h).

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

ii). Decision: Null hypothesis will be accepted.

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

iv). The mean numbers of daily visitors to a ski resort are same for three types of snow conditions.

We accept the null hypothesis.