#1

Hey everyone, I'm currently doing a Statistics assignment for uni and I'm absolutely stuck, I've been working on this damn question for hours and not getting anywhere. Does anyone think they might be able to help?

#2

you could start by posting the question

#3

I wanted to see if anyone was interested first before posting the question.

A company that markets user assembled furniture sells a computer desk that is advertised with the claim "less than an hour to assemble". However, through postpurchase surveys the company has learned that only 25% of their customers succeeded in building the desk in under an hour; 5% said it took them over 2 hours.

(a) Using this information and assuming that desk assembly time follows the normal

probability model, show that the mean and standard deviation for the assembly time

are 77.45 and 25.87 minutes respectively.

(b) One way the company could solve this problem would be to change the advertising claim. What assembly time should the company quote in order that 60% of customers succeed in finishing the desk within that time?

(c) Wishing to maintain the "less than an hour to assemble" claim, the company hopes that revising the instruction and labelling the parts more clearly can improve the 1 hour success rate to 60%. If the standard deviation stays the same, what new lower mean time does the company need to achieve?

(d) Months later, another postpurchase survey shows that this action did lower the mean assembly time but only to 55 minutes. Nevertheless, the company did achieve the 60% in an hour goal too. How was that possible?

(e) A consumer advocacy group has conducted their own test. Of the 10 people asked to assemble the desk only 4 managed to complete the assembly within the 1 hour period. One member of the team conducting this test says that clearly the company's claim, that 60% of people can complete the assembly within 1 hour, is incorrect as 4 out of 10 is not 60%. Another team member is not so sure. What do you think? Does this test provide sufficient evidence to refute the company's claim?

*Last edited by slickerthnsleek at Sep 8, 2009,*

#4

I've got all my books packed up atm but if you PM me later I can help

#5

I wanted to see if anyone was interested first before posting the question.

#6

^ And that's why I asked if anyone was interested before posting it. It's a c@#t of a question.

#7

hold on i did part a already its just really hard to type on the computer

EDIT:

p(x < 1) = 0.25

p(z < [1 - mean]/standard deviation] ) = 0.25

[1 - mean]/standard deviation = -.67449

p(x < 2) = 1-0.05=0.95

standardize and get [1 - mean]/standard deviation]=1.6449

1.6449/-.67449 = (2-mean)/(1- mean)

solve for mean and get 1.291 hrs which is 77.45 mins

sub mean = 1.291 hrs in any eqn above and get standard deviation = .43119 hrs which is 25.87 mins

EDIT:

p(x < 1) = 0.25

p(z < [1 - mean]/standard deviation] ) = 0.25

[1 - mean]/standard deviation = -.67449

p(x < 2) = 1-0.05=0.95

standardize and get [1 - mean]/standard deviation]=1.6449

1.6449/-.67449 = (2-mean)/(1- mean)

solve for mean and get 1.291 hrs which is 77.45 mins

sub mean = 1.291 hrs in any eqn above and get standard deviation = .43119 hrs which is 25.87 mins

*Last edited by xyber56 at Sep 8, 2009,*

#8

If I remembered any of my data management that I learned last year I would help you. The first question seems pretty easy, though; you should know how to calculate the mean and standard deviation of a normal distribution model. From there the other questions all kind of build on each other..

and here, I'll just give you the last one: the claim cannot be refuted based on that data as a sampling bias exists in that scenario; ten people isn't enough to prove or disprove a statistical theory. There needs to be way more people than that.

and here, I'll just give you the last one: the claim cannot be refuted based on that data as a sampling bias exists in that scenario; ten people isn't enough to prove or disprove a statistical theory. There needs to be way more people than that.

#9

Thanks heaps guys, that's brilliant. Hopefully I'll be able to work on the rest of the questions now based on that.

xyber56 & joemama9119, legends.

xyber56 & joemama9119, legends.

#10

One last question... Pretty easy one, I can probably do it, but my brain is fried atm.

[2 – u]/[1 – u] = x

How do I turn that left side into just u?

[2 – u]/[1 – u] = x

How do I turn that left side into just u?

#11

One last question... Pretty easy one, I can probably do it, but my brain is fried atm.

[2 – u]/[1 – u] = x

How do I turn that left side into just u?

[2 – u]/[1 – u] = x

2 – u = x[1 - u]

**Multiply both sides by [1 - u]**2 – u = x - xu

**Multiply out the right hand side**xu - u = x - 2

**put all the terms with u to the left**u[x-1] = x - 2

**take out the u as a common factor**u = [x - 2]/[x - 1]

**divide both sides by [x - 1]**
#12

Thanks heaps for that, I finally beat that f@#$ing question a. Now for the rest. ^_^ lol

#13

hahha

#14

Care to help with parts b-e? ^_^ lol

*Last edited by slickerthnsleek at Sep 9, 2009,*

#15

post it and i'll see what i can do. No promises though

#16

All questions from a-e are posted a few posts back.

Never mind with b-d, I'm onto e now, lol

Never mind with b-d, I'm onto e now, lol

*Last edited by slickerthnsleek at Sep 9, 2009,*

#17

what would happen if the examiner marking your work read this thread? just curious

#18

We're allowed to get help as long as we show our working and have actually put in the effort, which I think at this point is safe to say that I have.

I'm stuck on e at the moment, any ideas?

I think it's asking 'If 60% of customers can complete the desk in under an hour, what are the chances that only 4/10 selected people completed the desk in under an hour?' but it's a bit confusing.

I'm stuck on e at the moment, any ideas?

A consumer advocacy group has conducted their own test. Of the 10 people asked to assemble the desk only 4 managed to complete the assembly within the 1 hour period. One member of the team conducting this test says that clearly the company's claim, that 60% of people can complete the assembly within 1 hour, is incorrect as 4 out of 10 is not 60%. Another team member is not so sure. What do you think? Does this test provide suffcient evidence to refute the company's claim?

(Hint: If the company's claim is correct, what is the probability that of the 10 people selected, four or less completed the assembly within the hour?)

I think it's asking 'If 60% of customers can complete the desk in under an hour, what are the chances that only 4/10 selected people completed the desk in under an hour?' but it's a bit confusing.

*Last edited by slickerthnsleek at Sep 9, 2009,*

#19

i <3 pie

pi = 3.141592653589793238462643383

1+1=4

pi = 3.141592653589793238462643383

1+1=4

#20

^ Would've been so appropriate if you'd posted a picture of a Big Muff Pi then, lol

Still stuck. Does ANYONE have any ideas? I can go home when this one is done. =[

Anyone think this would work:

X ~ B (n = 10, p = 0.6)

P (X = 4) = [10!/4!(10 – 4)!](0.6)4(1 – 0.6)10 - 4

?

Still stuck. Does ANYONE have any ideas? I can go home when this one is done. =[

Anyone think this would work:

X ~ B (n = 10, p = 0.6)

P (X = 4) = [10!/4!(10 – 4)!](0.6)4(1 – 0.6)10 - 4

?

#21

Never mind, I finally finished. Thanks for the suggestions, they really helped steer me in the right direction.

#22

I wanted to see if anyone was interested first before posting the question.

haha! uni of newcastle!

i am also taking STAT1070

what degree are you doing, and which parts of the question are you stuck with? oh, and what time is your tut?

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