In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen concrete has a low value of ther- mal conductivity in order to minimize subsequent damage due to changing temperatures. Suppose two types of concrete, a graded aggregate and a no-fines aggregate, are being considered for a certain road. The table below summarizes data on thermal conductivity from an experiment carried out to compare the two types of concrete.

Ğ¢ÑÑеп ni xi si
Graded 42 0.486 0.187
No-fines Ñ 42 0.359 0.158

Assume that the conductivity for each sample of concrete is independent of that of all the others. Our interest is in determining if the mean conductivity for the graded concrete is significantly different than the mean conductivity for the no-fines concrete?

Required:
a. Formulate the above in terms of a hypothesis testing problem.
b. Give the test statistic and its reference distribution (under the null hypothesis).
c. Report the p-value of the test statistic and use it to assess the evidence that this sample provides on the scientific question of difference in mean conductivity of the two materials at the 5% level of significance.

Answers

Answer 1

Step-by-step explanation:

Given - In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen concrete has a low value of thermal conductivity in order to minimize subsequent damage due to changing temperatures. Suppose two types of concrete, a graded aggregate and a no-fines aggregate, are being considered for a certain road. The table below summarizes data on thermal conductivity from an experiment carried out to compare the two types of concrete.

Type            ni                xi                  si

Graded       42           0.486            0.187

No-fines      42           0.359           0.158

To find - a. Formulate the above in terms of a hypothesis testing problem.

b. Give the test statistic and its reference distribution (under the null hypothesis).

c. Report the p-value of the test statistic and use it to assess the evidence that this sample provides on the scientific question of difference in mean conductivity of the two materials at the 5% level of significance.

Proof -

a.)

Hypothesis testing problem :

H0 : There is significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.

H1 : There is no significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.

b)

Test statistic :

[tex]Z = \frac{x_{1} - x_{2} }{\sqrt{\frac{s_{1} ^{2} }{n_{1}} + \frac{s_{2}^{2} }{n_{2}} } }[/tex]

[tex]Z = \frac{0.486 - 0.359 }{\sqrt{\frac{0.03496 }{42} + \frac{0.02496 }{42} } }[/tex]

[tex]Z = \frac{0.127 }{\sqrt{0.001468}}[/tex]

[tex]Z = \frac{0.127 }{0.0377}[/tex]

⇒Z(cal) = 3.3687

Z(tab) = 1.96

As Z (cal) > Z(tab)

So, we reject H0 at 5% Level of significance

p-value = 0.99962

Hence

There is significant difference in mean conductivity at the two materials.


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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Answer:

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Step-by-step explanation:

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Answers

Answer:

a) 0.8802 = 88.02% probability that the shipment is accepted

b) 0.1756 = 17.56% probability that the shipment is accepted if 15% of the total shipment is defective.

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of any other item. This means that we use the binomial probability distribution to solve this question.

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The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

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3% defective means that [tex]p = 0.03[/tex]

Probability of at most 1 defective is:

[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]

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[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{20,0}.(0.03)^{0}.(0.97)^{20} = 0.5438[/tex]

[tex]P(X = 1) = C_{20,1}.(0.03)^{1}.(0.97)^{19} = 0.3364[/tex]

[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.5438 + 0.3364 = 0.8802[/tex]

0.8802 = 88.02% probability that the shipment is accepted.

b. Find the probability that the shipment is accepted if 15% of the total shipment is defective

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[tex]P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388[/tex]

[tex]P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368[/tex]

[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0388 + 0.1368 = 0.1756[/tex]

0.1756 = 17.56% probability that the shipment is accepted if 15% of the total shipment is defective.

Two investments totallng $47,500 produce an annual income of $3000. One investment ylelds 9% per year, while the other yields 6% per year. How much is invested at each rate?​

Answers

Answer:

The correct answer is "$5000".

Step-by-step explanation:

The given values are:

Two investments totaling,

= $47,500

Annual income,

= $3000

One investment yields per year,

= 9%

Other yield,

= 6%

Let,

The amount invested in 9% will be "x".The amount invested in 6% will be "[tex]47,500-x[/tex]".

Now,

⇒  [tex]0.09x+0.06(47,500-x)=3000[/tex]

⇒        [tex]0.09x+2850-0.06x=3000[/tex]

⇒                     [tex]0.03x+2850=3000[/tex]

⇒                                [tex]0.03x=150[/tex]

⇒                                       [tex]x=\frac{150}{0.03}[/tex]

⇒                                       [tex]x=5000[/tex]

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Answers

Answer:

Im only in middle school, sorry

Step-by-step explanation:

I have tips tho!

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