Chapter 10, Problem 35SE

An experiment was carried out to compare flow rates for four different types of nozzle.

a. Sample sizes were 5, 6, 7, and 6, respectively, and calculations gave f = 3.68. State and test the relevant hypotheses using α = .01

b. Analysis of the data using a statistical computer package yielded P-value = .029. At level .01, what would you conclude, and why?

To determine

State and test the hypotheses at $\alpha =0.01$

Answer to Problem 35SE

The test hypotheses are:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4$

Alternative hypothesis:

$H_a:\text{not all population proportions are equal}$.

The test reveals that the flow rates for the 4 types of nozzle are not significantly different.

Explanation of Solution

Given info:

An experiment conducted to compare the flow rates for 4 types of nozzle considered respective sample sizes 5, 6, 7, 6 with F statistic value $f=3.68$.

Calculation:

Let the average flow rates for the 4 types of nozzle be ${\mu }_1$, ${\mu }_2$, ${\mu }_3$, ${\mu }_4$.

The test hypotheses are:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4$

That is, the flow rates for 4 types of nozzle are equal.

Alternative hypothesis:

$H_a:\text{not all population proportions are equal}$

That is, the flow rates for at least 2 types of nozzle are not equal.

The test statistic value is found to be $f=3.68$.

Degrees of freedom (df):

The number of treatments is $I=4$. Thus, the treatment df or the numerator df is:

$\begin{array}{c}I-1=4-1\\ =3.\end{array}$

The total number of observations is:

$\begin{array}{c}n=5+6+7+6\\ =24.\end{array}$

Thus, the total df is:

$\begin{array}{c}n-1=24-1\\ =23.\end{array}$

The error df for the one factor ANOVA or the denominator df is:

$\begin{array}{c}\text{Error df}=\text{Total df}-\text{Treatment df}\\ =23-3\\ =20.\end{array}$

Thus, the degrees of freedom are 3, 20.

Level of significance:

The given level of significance is $\alpha =0.05$.

Bounds of the P-value:

The Table A.9, the table for “Critical Values for F Distributions” shows that the F statistic value $f=3.68$ lies between the table values $F_{0.05,3,20}=3.10$ and $F_{0.01,3,20}=4.94$. As a result, the P-value lies between 0.01 and 0.05.

Thus, the lower bound of the P-value is 0.01 and the upper bound of the P-value is 0.05.

Rejection rule:

If the $P\text{-value}\le \alpha$, then reject the null hypothesis.

Conclusion:

Here, the P-value is greater than the level of significance.

That is, $0.01<P\text{-value}<0.05$.

Thus, the decision is “fail to reject the null hypothesis”.

Therefore, the data do not provide sufficient evidence to conclude that the flow rates vary for at least 2 types of nozzle.

That is, the flow rates for the 4 types of nozzle are not significantly different.

To determine

Give the conclusion at level 0.01, if analysis of the yielded $P\text{-value}=0.029$.

Answer to Problem 35SE

It can be concluded that the flow rates for the 4 types of nozzle are not significantly different.

Explanation of Solution

Calculation:

Level of significance:

The given level of significance is $\alpha =0.01$.

P-value:

From statistical computer package, the P-value is 0.029.

Rejection rule:

If the $p\text{-value}\le \alpha$, then reject the null hypothesis.

Conclusion:

Here, the p-value greater than the level of significance.

That is, $p\text{-value}\left(=0.029\right)>\alpha \left(=0.01\right)$

Thus, the decision is “fail to reject the null hypothesis”.

Therefore, the data do not provide sufficient evidence to conclude that the flow rates vary for at least 2 types of nozzle.

That is, the flow rates for the 4 types of nozzle are not significantly different.