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it is common practice in many countries to destroy (shred)refrigerators at the end of their usefull lives.In this process material from insulating foam may be released into the atmosphere.The article 鈥淩elease of fluorocarbons from Insulation foam in Home Appliances During shredding鈥(J.of the Air and waste Mgmt.Assoc.,2007:1452-1460)gave the following data on foam density(g/L)For each of two refrigerators produced by four different manufactures:

\(\begin{aligned}{l}1.30.4,\,29.2\,\,\,\,\,2.27.7,27.1\\3.27.1,24.8\,\,\,\,\,\,4.25.5,28.8\end{aligned}\)

Does it appear that true average foam density is not the same for all these manufacture?carry out an appropriate test of hypotheses by obtaining as much p-value information as possible,and summarize your analysis in an ANOVA table.

Short Answer

Expert verified

carry out an appropriate test of hypotheses by obtaining as much p-value information as possible,and summarize your analysis in an ANOVA table.

\(I = 4\)

Let assume

\(\alpha = 0.05\)

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

There is not sufficient evidence to support the claim that the true average form density is not the same for all manufacturers.

Step by step solution

01

definition of hypothesis

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:

\(I = 4\)

Let assume

\(\alpha = 0.05\)

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

Data summary

Samples

\(1\)

\(2\)

\(3\)

\(4\)

Total

N

\(2\)

\(2\)

\(2\)

\(2\)

\(8\)

\(\sum x \)

\(59.6\)

\(54.8\)

\(51.9\)

\(54.3\)

\(220.6\)

Mean

\(29.8\)

\(27.4\)

\(25.95\)

\(27.15\)

\(27.575\)

\({\sum x ^2}\)

\(1776.8\)

\(1501.7\)

\(1349.45\)

\(1479.69\)

\(6107.64\)

Variance

\(0.72\)

\(0.18\)

\(2.645\)

\(5.445\)

\(3.5136\)

Std.Dev.

\(0.8485\)

\(0.4243\)

\(1.6263\)

\(2.335\)

\(1.8745\)

Std.Err.

\(0.6\)

\(0.3\)

\(1.15\)

\(1.65\)

\(0.6627\)

02

divide the number

The overall mean is the sum of all values divided by the number of values:

\(\overline x = \frac{{{n_1}{{\overline x }_1} + {n_2}{{\overline x }_2} + ...{n_k}{{\overline x }_k}}}{N}\)

\(\begin{aligned}{l} = \frac{{2\left( {29.8} \right) + 2\left( {27.4} \right) + 2\left( {25.95} \right) + 2\left( {27.15} \right)}}{{24}}\\ \approx 27.575\end{aligned}\)

The mean square for treatment is:

\(MS{T_r} = \frac{{{n_1}{{\left( {\overline {{x_1}} - \overline x } \right)}^2} + {n_2}{{\left( {\overline {{x_2}} - \overline x } \right)}^2} + ... + {n_k}{{\left( {\overline {{x_k}} - \overline x } \right)}^2}}}{{I - 1}}\)

\( = \frac{{2{{\left( {29.8 - 27.575} \right)}^2} + 2{{\left( {27.4 - 27.575} \right)}^2} + 2{{\left( {25.95 - 27.575} \right)}^2} + 2{{\left( {27.15 - 27.575} \right)}^2}}}{{4 - 1}}\)

\( = 5.2017\)

Mean square for error is:

\(MSE = \frac{{\left( {{n_1} - 1} \right){s_1}^2 + \left( {{n_2} - 1} \right){s_2}^2 + \left( {{n_k} - 1} \right){s_k}^2}}{{N - I}}\)

\( = \frac{{(2 - 1){{\left( {0.8485} \right)}^2} + (2 - 1){{\left( {0.4243} \right)}^2} + (2 - 1){{\left( {1.6263} \right)}^2} + (2 - 1){{\left( {2.3335} \right)}^2}}}{{8 - 4}}\)\( = 2.24\)

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

ANOVA summary Independent samples \(k = 4\)

Source

SS

Df

Ms

F

P

Treatment between groups

\(15.605\)

\(3\)

\(5.2017\)

\(2.31\)

\(0.218001\)

Error

\(8.99\)

\(4\)

\(2.2475\)

Graph maker

Ss/BI

Total

\(24.595\)

\(7\)

\(F = \frac{{MS{T_r}}}{{MSE}} = \frac{{12851.3678}}{{6234.3995}} \approx 2.06\)

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 -value in the row\(d\,\,f\,n = I - 1 = {\rm{ }}4 - 1 = 3\)and\(d\,fd = N - I = {\rm{ 8}} - 4 = 4:\)

\(p > 0.10\)

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

\(P{\rm{ }} > 0.05 \Rightarrow Fail\;to{\rm{ }}reject{\rm{ }}{H_0}\)

There is not sufficient evidence to support the claim that the true average form density is not the same for all manufacturers.

Hence,

There is not sufficient evidence to support the claim that the true average form density is not the same for all manufacturers.

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Suppose the compression strength observation on the fourth type of box in Example \(10.1\)had been \(655.1,\,748.7,\,662.4,\,679.0,\,706.9,\,and\,640.0\) and (obtained by adding \(120\) to each previous \({x_{4j}}\)). Assuming no change in the remaining observations, carry out an F test with \(\alpha = 0.5.\)

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\(\begin{aligned}{l}4:309.2\\6:402.1\\8:392.4\\10:346.7\\12:407.4\end{aligned}\) \(\begin{aligned}{l}409.5\\347.2\\366.2\\452.9\\441.8\end{aligned}\) \(\begin{aligned}{l}311.0\\361.0\\351.0\\461.4\\419.9\end{aligned}\) \(\begin{aligned}{l}326.5\\404.5\\357.1\\433.1\\410.7\end{aligned}\) \(\begin{aligned}{l}316.8\\331.0\\409.9\\410.6\\473.4\end{aligned}\) \(\begin{aligned}{l}349.8\\348.9\\367.3\\384.2\\441.2\end{aligned}\) \(\begin{aligned}{l}309.7\\381.7\\382.0\\362.6\\465.8\end{aligned}\)

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Iris color

1.Brown

2.Green

3.Blue

\({\bf{26}}.{\bf{8}}\)

\({\bf{26}}.{\bf{4}}\)

\({\bf{25}}.{\bf{7}}\)

\({\bf{27}}.{\bf{9}}\)

\({\bf{24}}.{\bf{2}}\)

\({\bf{27}}.{\bf{2}}\)

\({\bf{23}}.{\bf{7}}\)

\({\bf{28}}.{\bf{0}}\)

\({\bf{29}}.{\bf{9}}\)

\({\bf{25}}.{\bf{0}}\)

\({\bf{26}}.{\bf{9}}\)

\({\bf{28}}.{\bf{5}}\)

\({\bf{26}}.{\bf{3}}\)

\({\bf{29}}.{\bf{1}}\)

\({\bf{29}}.{\bf{4}}\)

\({\bf{24}}.{\bf{8}}\)

\({\bf{28}}.{\bf{3}}\)

\({\bf{25}}.{\bf{7}}\)

\({\bf{24}}.{\bf{5}}\)

\({J_i}\)

\({\bf{8}}\)

\({\bf{5}}\)

\({\bf{6}}\)

\({x_i}\)

\({\bf{204}}.{\bf{7}}\)

\({\bf{134}}.{\bf{6}}\)

\({\bf{169}}.{\bf{0}}\)

\({\overline x _i}\)

\({\bf{25}}.{\bf{59}}\)

\({\bf{26}}.{\bf{92}}\)

\({\bf{28}}.{\bf{17}}\)

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