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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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Most popular questions from this chapter

The critical flicker frequency \(\left( {cff} \right)\) is the highest frequency at which a person can detect the flicker in a flickering light source. At frequencies above the cff, the light source appear to be continuous even though it is actually flickering. An investigation carried out to see whether true average cff depends on iris color yielded the following data (based on the article 鈥淭he Effects of Iris Color on Critical Flicker Frequency鈥.

Iris color

1.Brown

2.Green

3.Blue

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\({J_i}\)

\({\bf{8}}\)

\({\bf{5}}\)

\({\bf{6}}\)

\({x_i}\)

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

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

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

\({\overline x _i}\)

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

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

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

\(n = 19,{x_{..}} = 508.3\)

  1. State and test the relevant hypotheses at significance level\(.05\)(Hint:\(\sum {\sum {{x_{ij}}^2} = 13659.67,CF = 13598.36} \))
  2. Investigate difference between iris colors with respect to mean cff.

A study of the properties of mental plate 鈥揷onnected trusses used for roof support (鈥淢odeling joints made with Light-Gauge metal connector plates,鈥 Forest products J., 1979:39-44) yielded the following observations on axial-stiffness index(kips/in.) for plate lengths \(4,6,8.10,12\)in:

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

Does variation in plate length have any effect on true average axial stiffness? state and test the relevant hypotheses using analysis of variance with Display your results in an ANOVA table.(Hint : \({\sum x ^2}_{ij.} = 5,241,420.79.)\)

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 \(\alpha = 0.5.\)

When sample sizes are equal\(\left( {{J_i} = J} \right)\), the parameters\({\alpha _1},{\alpha _2},...{\alpha _I}\,\)of the alternative parameterization are restricted by\(\Sigma {\alpha _i} = 0\). For unequal sample sizes, the most natural restriction is\(\Sigma {J_i}{\alpha _i} = 0\). Use this to show that

\(E\left( {MSTr} \right) = {\alpha ^2} + \frac{1}{{I - 1}}\Sigma {J_i}\alpha _i^2\)

What is\(E\left( {MSTr} \right)\)when\({H_0}\)is true? (This expectation is correct if\(\Sigma {J_i}{\alpha _i} = 0\,\)is replaced by the restriction\(\Sigma {\alpha _i} = 0\)(or any other single linear restriction on the ai 鈥檚 used to reduce the model to I independent parameters), but\(\Sigma {J_i}{\alpha _i} = 0\)simplifies the algebra and yields natural estimates for the model parameters (in particular,\({\mathop {\,\,\,\,\alpha }\limits^{\,\,\,\,\,\^} _i} = {\mathop X\limits^\_ _i}\, - \,\mathop X\limits^\_ ..\,\,\,\,{H_0}\)).)

The article 鈥淭he Effect of Enzyme Inducing Agents on the Survival Times of Rats Exposed to Lethal Levels of Nitrogen Dioxide鈥 (Toxicology and Applied Pharmacology, 1978: 169鈥174) reports the following data on survival times for rats exposed to nitrogen dioxide (70 ppm) via different injection regimens. There were J = 14 rats in each group.

a. Test the null hypothesis that true average survival time does not depend on an injection regimen against the alternative that there is some dependence on an injection regimen using a 5 .01.

b. Suppose that 100\(\left( {1 - \alpha } \right)\)% CIs for k different parametric functions are computed from the same ANOVA data set. Then it is easily verified that the simultaneous confidence level is at least 100\(\left( {1 - k\alpha } \right)\)%. Compute CIs with a simultaneous confidence level of at least 98% for\({\mu _1} - 1/5\left( {{\mu _2} + {\mu _3} + {\mu _4} + {\mu _5} + {\mu _6}} \right)\)and\(1/4\left( {{\mu _2} + {\mu _3} + {\mu _4} + {\mu _5}} \right) - {\mu _6}\).

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