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anonymous
 2 years ago
In a sample of 7 cars each car was tested for nitrogenoxide emissions (in grams per mile) and the following results are obtained
0.18, 0.16, 0.08, 0.06,0.17,0.12,0.15 assuming that this sample is representative of the cars in use. cosntruct a 98% confidence interval estimate of the mean amount of nitrogen=oxide emissions for all cars. if the epa requires that nitrogen oxide emissions can be less than 0.165g.mi we can safely conclude that requirement is being met
A)what is the confidence interval estimate of the mean amount of nitrogenoxide emissions for all cars
___g/mi<u<___g/mi
anonymous
 2 years ago
In a sample of 7 cars each car was tested for nitrogenoxide emissions (in grams per mile) and the following results are obtained 0.18, 0.16, 0.08, 0.06,0.17,0.12,0.15 assuming that this sample is representative of the cars in use. cosntruct a 98% confidence interval estimate of the mean amount of nitrogen=oxide emissions for all cars. if the epa requires that nitrogen oxide emissions can be less than 0.165g.mi we can safely conclude that requirement is being met A)what is the confidence interval estimate of the mean amount of nitrogenoxide emissions for all cars ___g/mi<u<___g/mi

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I have this same question! lol

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2The first step is to calculate the sample mean by adding the seven values and dividing by 7: \[\bar{x}=\frac{0.18+0.16+0.08+0.06+0.17+0.12+0.15}{7}=\frac{0.92}{7}=0.13143\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@kropot72, then how do you get the __g/mi <u< __g/mi ??

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2The next step is to calculate the sample standard deviation, s.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0not sure how to do that..

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2You can use a calculator with statistical functions to calculate the sample standard deviation, s.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0x= i got a totally different answer than you did

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0and the sd i think is 0.046

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0how did you get the sd @andijo76

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2@andijo76 what did you get for the sum of the seven data values?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i found a website back when i was doing pre algebra that helps me figuring out some calculations

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0the sum i got was 0.842/7=0.1203

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0does anyone know how to input this into the Ti84?

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2@andijo76 The correct sum of the seven data values is 0.92.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0im trying to figure out how i got that lol

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2My calculator gives the value for the sample standard deviation of s = 0.4634

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@kropot72 how did you input the sd in the calc?

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2My calculator is a Casio fx82 AU PLUS. I just followed the instructions in the User's Guide.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0im sorry you are correct i dont know what i did but i fixed it i think i rounded one decimal or something

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0there is a website that gives you sd i will paste what it said

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Simplify the result. σ=0.046344719225 The standard deviation should be rounded to one more decimal place than the original data. If the original data were mixed, round to one decimal place more than the least precise. 0.046

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@andijo76 can you paste the link to the website?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0www.mathway.com it costs 20 a month though i keep it for my kids it helps you figure out things step by step

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2The final calculation uses the following formula: \[\bar{x}2.576\frac{s}{\sqrt{n}}<\mu<\bar{x}+2.576\frac{s}{\sqrt{n}}\] Substituting values gives the confidence interval for the population mean as follows: \[0.131432.576\frac{0.046}{\sqrt{7}}<\mu<0.13143+2.576\frac{0.046}{\sqrt{7}}\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0where did you get the 2.576?

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2The value 2.576 is determined by an inverse normal technique. However it can be found in tables for use in calculating confidence intervals.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i know i am going to sound stupid but how do i figure _g/mi<u<_g/mi was that the answer

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2When you have calculated the two values for the confidence interval, just put the lower value where the first underscore is and the higher value where the right hand underscore is.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i got for the first one 44.6558 so i am thinking it is i just have to put round and put it in 3 decimal places

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so is this the answer then: .0866 g/mi< .1762 g/mi ?! @kropot72

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2Looks correct to me :)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0yeah i forgot the point haha

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0it amazing how that works hey

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2It is a very powerful technique!

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0thank you so much for your help
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