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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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

the answer is (-8) because x=4, y=-6, z=11

These consists of more than 1 equation with unknown variables. The value that is not a solution in (x,y, z)is -8

System of equations

These consists of more than 1 equation with unknown variables. Given the set of equations

x+y+z=9

x-y-z=-1

x-y+z= 21

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2x = 9 - 1

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-2z = -22

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4 + y + 11 = 9

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Hence the value that is not a solution in (x,y, z)is -8

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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%

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Answers

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10. A large shipment of video streaming devices is accepted upon delivery if an inspection of 20 randomly selected video streaming devices yields no more than 1 defective item. a. Find the probability that the shipment is accepted if 3% of the total shipment is defective. b. Find the probability that the shipment is accepted if 15% of the total shipment is defective.

Answers

Answer:

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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.

Binomial probability distribution

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20 randomly selected video streaming devices

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Answer:

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Answer:

Unit rate is often a useful means for comparing ratios and their associated rates when measured in different units. The unit rate allows us to compare varying sizes of quantities by examining the number of units of one quantity per one unit of the second quantity. This value of the ratio is the unit rate.

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Step-by-step explanation:

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Ethan's mom tried a new bean soup recipe. She used 2 cups of black beans and 5 cups of kidney beans. The family loved it, so she made a larger batch to freeze with 4 cups of black beans and 12 cups of kidney beans. Did the two batches have the same ratio of black beans to kidney beans?

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Answers

Answer:

no the amount of Kidney beans is not proportional to the amount of black beans

Step-by-step explanation:

Answer:

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Step-by-step explanation:

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Answer:

B

Step-by-step explanation:

x% of 6 = 5.25

[tex]\frac{x}{100}[/tex] x 6 = 5.25

[tex]\frac{(5.25)(100)}{6}[/tex] = x

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Answer:

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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:

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6) Four bacteria are placed in a petri dish. The population will double every day. The formula for the
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N(t) = 4(2)^t
where t is the number of days after the four bacteria are placed in the dish. How many bacteria are
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Answers

Answer:

128

Step-by-step explanation:

N(t) = 4(2)^t

N(t) = 4(2)^5

N(t) = 4(32)

N(t) = 128

The number of bacteria that are

in the dish five days after the four bacteria are placed in the dish is; 128 bacteria

We are given the formula for the

number of bacteria in the dish on day as;

N(t) = 4(2)^t

Where t is number of days after the four bacteria are placed in the dish.

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N(5) = 4(32)

N(5) = 128

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