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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 \(\alpha = 0.1\)

b. Analysis of the data using a statistical computer package yielded P-value = 0.29. At level .01, what would you conclude, and why?

Short Answer

Expert verified

a. There is not sufficient evidence to reject the claim that the true average flow rates are the same for the four different types of nozzle

b. There is not sufficient evidence to reject the claim that the true average flow rates are the same for the four different types of nozzle.

Step by step solution

01

Stating null and alternative hypothesisPart (a)

\({n_1} = 5\)

\({n_2} = 6\)

\({n_3} = 7\)

\({n_4} = 6\)

\(\begin{aligned}{l}f = 3.68\\\alpha = 0.01\end{aligned}\)

The null hypothesis states that all population means are equal:

\({H_0}:{\mu _1} = {\mu _2} = {\mu _3} = {\mu _4}\)

The alternative hypothesis states the opposite of the null hypothesis:

\({H_1}:\)not all of\({\mu _1},{\mu _2},{\mu _3},{\mu _4}\)are equal

02

Finding degree of freedom for numerator

The degree of freedom for the numerator is the number of groups decreased by 1:

\(\begin{aligned}{c}df = I - 1\\ = 4 - 1\\ = 3\end{aligned}\)

03

Finding degree of freedom for denominator

The degrees of freedom for the denominator is the total sample size decreased by the number of groups:

\(\begin{aligned}{c}df = N - I\\ = (5 + 6 + 7 + 6) - 4\\ = 24 - 4\\ = 20\end{aligned}\)

04

Finding P-value

The P-value is the probability of obtaining the value of the test statistic, of a value more extreme. The P-value is the number (or interval) in the column title of the F-distribution table in the appendix containing the F-value in the row \(df = 3\)and\(dfd = 20\):

\(0.010 < P < 0.050\)

If the P-value is less than the significance level, then reject the null hypothesis

\(p > 0.01 \Rightarrow \)Fail to reject \({H_0}\)

There is not sufficient evidence to reject the claim that the true average flow rates are the same for the four different types of nozzle.

05

Stating null and alternative hypothesisPart (b)

\({n_1} = 5\)

\({n_2} = 6\)

\({n_3} = 7\)

\({n_4} = 6\)

\(\begin{aligned}{l}f = 3.68\\\alpha = 0.01\\P = 0.029\end{aligned}\)

The null hypothesis states that all population means are equal:

\({H_0}:{\mu _1} = {\mu _2} = {\mu _3} = {\mu _4}\)

The alternative hypothesis states the opposite of the null hypothesis:

\({H_1}:\)not all of\({\mu _1},{\mu _2},{\mu _3},{\mu _4}\)are equal

06

Finding P-value

If the P-value is less than the significance level, then reject the null hypothesis

\(P = 0.029 > 0.01 \Rightarrow \)Fail to reject \({H_0}\)

There is not sufficient evidence to reject the claim that the true average flow rates are the same for the four different types of nozzle.

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