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Airida [17]
2 years ago
10

An experiment was carried out to compare electrical resistivity for six different low-permeability concrete bridge deck mixtures

. there were 26 measurements on concrete cylinders for each mixture; these were obtained 28 days after casting. the entries in the accompanying anova table are based on information in an article. fill in the remaining entries. (round your answer for f to two decimal places.)
SAT
1 answer:
bulgar [2K]2 years ago
7 0

Assuming  there were 26 measurements on concrete cylinders for each mixture. Alternative Hypothesis at α=0.01 will be rejected.

<h3>Hypothesis</h3>

Electrical resistivity(I)=6

Concrete cylinder measurement(J)=26

First step

Degree of freedom for  SSb=I-1=(6-1)=5

Degree of freedom for SSw=I(1-J)=6(1-26)=150

Degree of freedom for SSr=IJ-1=(6×26)-1=155

Second step

Mean square(MSw)=13.929

SSw=MSw×I(J-1)

SSw=13.929×150

SSw=2,089.35

Third step

SSb=SSt-SSw

SSb=5,664.415-2,089.35

SSb=3,575.065

Fourth step

MSb=SSb(I-1)

MSb=3,575.065/(1-6)

MSb=3,575.065/5

MSb=715.013

Fifth step

F=MSb/Msw

F=715.013/13.929

F=51.33

The complete table is:

Source   df      Sum of squares       Mean square      F

Mixture    5         3,575.065                715.013              51.33

Error        150       2,089.35                 13.929

Total       155       5664.415

Therefore, since 51.33 is greater than 11.97 we are going to reject the hypothesis at α=0.01.

Inconclusion Hypothesis at α=0.01 will be rejected.

Learn more about hypothesis here:brainly.com/question/11555274

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Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).

<h3>What is a t-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm t\frac{s}{\sqrt{n}}

In which:

  • \overline{x} is the sample mean.
  • t is the critical value.
  • n is the sample size.
  • s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed <em>95% confidence interval</em>, with 49 - 1 = <em>48 df</em>, is t = 2.0106.

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\overline{x} - t\frac{s}{\sqrt{n}} = 12.2 - 2.0106\frac{1.6}{\sqrt{49}} = 11.7

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The 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).

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