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An experiment was carried out to compare electrical resitivity for six different low-permeability concrete bridge deck mixtures. There were \(26\)measurements on concrete cylinders for each mixture; these were obtained days \(28\)after casting.The entries in the accompanying ANOVA table are based on information in the article 鈥淚n-place Resitivity of Bridge Deck Concrete Mixtures鈥(ACI Matrerials j.,2009: 114-122).Fill in the remaining entries and test appropriate hypothese.

Short Answer

Expert verified

Fill in the remaining entries and test appropriate hypothese.

Given,

\(\begin{aligned}{l}I = 6\\{n_i} = 26\\SST = 5664.4.415\\MSE = 13.929\end{aligned}\)

Let us assume:

There is sufficient evidence to support the claim that the population means are not all equal.

Step by step solution

01

definition of hypotheses

Hypotheses are typically written in the form of if/then statements, such as if someone consumes a lot of sugar, they will develop cavities in their teeth.

Given,

\(\begin{aligned}{l}I = 6\\{n_i} = 26\\SST = 5664.4.415\\MSE = 13.929\end{aligned}\)

Let us assume:

\(\alpha = 0.05\)

The null hypothesis states that all population means are equal:

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

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

The alternative hypothesis states the opposite of the null hypothesis:

The degrees of freedom of the treatment are the number of groups decreased by 1:

\(df{T_r}{\rm{ }} = I - 1 = 6 - 1 = 5\)

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

\(d{f_E} = N - I = 6\left( {26} \right) - 6 = 150{\rm{ }}\)

The total degrees of freedom is the sum of the previous two degrees of freedom:

\(d{f_{Tot}} = df{T_r} + d{f_E} = 150 + 5 = 155\)

The error sum of squares is the product of the mean square error and the degrees of freedom for the error:

\(SSE = MSE \times d{f_E} = 13.929 + \times 150 = 2089.35\)

The treatment sum of squares is the total sum of squares decreased by the error sum of squares:

\(SS{T_r} = SST - SSE = 5664.415 - 2089.35 = 3575.065\)

02

divide the value

The treatment mean square is the treatment sum of squares divided by the degrees of freedom for the treatment:

\(MS{T_r} = \frac{{SS{T_r}}}{{d{f_{Tr}}}} = \frac{{3575.065}}{5} = 715`013\)

The ANOVA F statistic is the ratio of the MSTr and the MSE:

\(F = \frac{{MS{T_r}}}{{MSE}} = \frac{{715.013}}{{13.929}} \approx 51.3327\)

Source

Df

Sum of squares

Mean square

F

Mixture

\(5\)

\(3575.065\)

\(715.013\)

\(51.3327\)

Error

\(150\)

\(155\)

\(2089.35\)

\(13.929\)

Total

\(5664.415\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table \(8\)containing the F -value in the row\(d\,\,f\,n = I - 1 = {\rm{ 6}} - 1 = 5\)and\(d\,fd = N - I = {\rm{ 156}} - 4 = 150:\)

\(p < 0.001\)

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

\(P < 0.05 \to {\mathop{\rm Re}\nolimits} ject{H_0}\,\)

There is sufficient evidence to support the claim that the population means are not all equal.

Hence,

There is sufficient evidence to support the claim that the population means are not all equal.

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Most popular questions from this chapter

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\(\begin{aligned}{l}1:\,\,\,20.5\,\,\,\,\,\\\,\,\,\,\,\,25.2\\2:26.3\\\,\,\,\,\,\,34.0\\3:29.5\\\,\,\,\,\,\,\,26.2\\4:36.5\\\,\,\,\,\,\,33.1\end{aligned}\)\(\begin{aligned}{l}28.1\\25.3\\24.0\\17.1\\34.0\\29.9\\44.2\\34.1\end{aligned}\) \(\begin{aligned}{l}27.8\\27.1\\26.2\\26.8\\27.5\\29.5\\34.1\\32.9\end{aligned}\) \(\begin{aligned}{l}27.0\\20.5\\20.2\\23.7\\29.4\\30.0\\30.3\\36.3\end{aligned}\) \(\begin{aligned}{l}28.0\\31.3\\23.7\\24.9\\27.9\\35.6\\31.4\\25.5\end{aligned}\)

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{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}

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\(\begin{aligned}{*{20}{c}}{1:}&{85.06}&{85.25}&{84.87}\\{2:}&{84.99}&{84.28}&{84.88}\\{3:}&{84.48}&{84.72}&{85.10}\\{4:}&{84.10}&{84.55}&{84.05}\end{aligned}\)

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