Question

Using a significance level of $5\%$, test the hypothesis that the four magazine types have the same mean length.

Short answer

We accept the null hypothesis.

Step-by-step solution

Given Information

A significance level of $5\%$, test the hypothesis that the four magazine types have the same mean length.

Explanation

To find these results on the calculator:

Press STAT. Press $1$EDIT. Put the data into the lists $L_1,L_2,L_3,L_4,L_5,L_6,L_7$

Press STAT, and arrow over to TESTS, and arrow down to ANOVA. Press ENTER, and then enter $L_1,L_2,L_3,L_4,L_5,L_6,L_7$. 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 :

$F=1.7056$

$p=0.12765$

Factor

$df=6$

$SS=519.2$

$MS=86.5333$

Error

$df=98$

$SS=4972$

$MS=50.7347$

Hypothesis test

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

The null hypothesis is:

$Ho:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5={\mu }_6={\mu }_7$

The alternate hypothesis is that at least any two of the mean lap times are different.

The degree of freedom in the numerator - df(num) is $6$,

and the degree of freedom in the denominator - $df\left(denom\right)$is $98$.

We use the $F$distribution.

The value of the test statistic (F-value) is $1.7$.

The $P$-value for the test is $0.12765$.

Explanation

The graph of the distribution.

Explanation

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

ii. Decision: We do not reject the null hypothesis

iii. Reason for decision: $P$-value is $0.12765$which is greater than the $0.05$level of significance.

iv. Conclusion: There is insufficient evidence to conclude that the mean lap times are different.