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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

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.

The lumen output was determined for each of \(I = 3\)different brands of light bulbs having the same wattage, with \(J = 8\) bulbs of each brand tested. The sums of squares were computed as \(SSE = 4773.3\,and\,SS{T_r} = 591.2.\)state hypothese of intrest (including word definitions of parameters),and use the F test of ANOVA \(\left( {\alpha = .05} \right)\)to decide whether there are any differences in true average lumen outputs among the three brands for this type of bulb by obtaining as much information as possible about the p-values.

In an experiment to compare the quality of four different brands of magnetic recording tape, five 2400-ft reels of each brand (A鈥揇) were selected and the number of flaws in each reel was determined.

A:

10

5

12

14

8

B:

14

12

17

9

8

C:

13

18

10

15

18

D:

17

16

12

22

14

It is believed that the number of flaws has approximately a Poisson distribution for each brand. Analyse the data at level .01 to see whether the expected number of flaws per reel is the same for each brand.

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.

The following data refers to yield of tomatoes (kg/plot) for four different levels of salinity. Salinity level here refers to electrical conductivity (EC), where the chosen levels were EC = 1.6, 3.8, 6.0, and 10.2 nmhos/cm.

Use the F test at level\(\alpha \)=.05 to test for any differences in true average yield due to the different salinity levels.

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