Mathematics, 05.05.2020 06:53 mochoa4

A regular type of laminate is currently being used by a manufacturer of circuit boards. A special laminate has been developed to reduce the warpage. The regular laminate will be used on one sample of specimens and the special laminate on another sample, and the amount of warpage will then be determined for each specimen. The manufacturer will then switch to the special laminate only if it can be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate. State the relevant hypotheses.
a. H0: μregular > μspecial
b. Ha: μregular = μspecialH0: μregular = μspecial
c. Ha: μregular < μspecial H0: μregular = μspecial
d. Ha: μregular > μspecialH0: μregular = μspecial
e. Ha: μregular

Answers: 3

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Answer from: EllaLovesAnime

The correct option is (d).

Step-by-step explanation:

It is provided that a special laminate has been developed to reduce the warpage.

The will then switch to the special laminate only if it can be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate.

To determine whether he should switch tot he special type of laminate or not, the manufacturer uses the regular laminate on one sample of specimens and the special laminate on another sample, and the amount of warpage will then be determined for each specimen.

The hypothesis for this test can be defined as follows:

H₀: The true average amount of warpage for special type of laminate is same as that for the regular laminate, i.e. $\mu_{regular}=\mu_{special}$

Hₐ: The true average amount of warpage for special type of laminate is less than for the regular laminate, i.e. $\mu_{regular}\mu_{special}$

Thus, the relevant hypotheses is defined by (d).

Answer from: satwikap3304

A circuit board producer is now using a standardized sort of lamination. To reduce warpage, a special laminate was developed. A conventional laminate will be applied to one collection of the specimens.

Its special laminate will be applied to some other specimens, as well as the quantity of warpage for each specimen will be measured. This manufacturer will subsequently use the special laminate only if it can be proved that its genuine average number of warpage for that laminate is less than that of ordinary lamination.

Specify the hypothesis at issue.

$\to H_a: \mu_{regular} \mu_{special} \\\\ \to H_0: \mu_{regular} = \mu_{special}$

Inside this scenario, describe the type I error.Whenever the null value is true, Type I rejects it.This type I error would've claimed that the special laminate had reduced warpage whenever the true warpages were equal.In this instance, describe the type II error.Whenever the null is untrue, Type II fails to reject it.The kind II error would've been reporting the characteristics that can affect true warpage to be equal whenever the special lamination has so little warpage.

Therefore, the final answer is "Option D".

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Answer from: Quest

the answer is 3/2 and for decimal: 1.5

hope i

-lavira: )

Answer from: Quest

unless you made a typo, the answers are both a and c.

step-by-step explanation:

1.5/3 is .5 whereas 3/9 is .3333333 not proportional

22/33 is .66666 and 4/6 is also .66666 proportional

2/4 is .5 whereas 142/213 is .66666 not proportional

14/7 is 2 and 7/3.5 is also 2 proportional

you could've also checked it by doing this simple division:

if the larger fraction divides with the smaller fraction to equal 1, it is proportional. so let me show you what i mean:

when i divide 22/33 by 4/6 i get 1, so we know it is proportional, and it is the same with 14/7 and 7/3.5. now if i divide 3/9 by 1.5/3 i get .66666 so therefore we know it is not proportional.

there are many ways to do this so i wanted to show you some possible ways.