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I'm measuring my backyard and I've ran into a problem with a bay window area. There are three segments, each being 3' 5". The middle segment sticks out from the main exterior wall by 2'. What formula could I use to find the total width of the bay, taking angles into consideration, and then find what degree the angles should be at (for the left and right segments)? 3' 8" * 3 obviously wouldn't work. Here's an image that might help: http://img354.imageshack.us/img354/5492/confusion.jpg
I'd go out and measure it myself, but it's too dark, and it's also a torrential rainfall out there right now.
-- 69.154.119.41 ( talk) 00:15, 31 July 2009 (UTC)
The floor plan can be viewed as a rectangle and two right triangles: the rectangle is 2' by 3'5". Each right triangle has a 3'5" hypotenuse and a 2' leg. The question is how long the other leg is. So use the Pythagorean theorem:
Thus the whole width of the bay is about
Michael Hardy ( talk) 02:47, 31 July 2009 (UTC)
Statistics, population mean - question that I'm 75% done with, but need one more:
I've determined that A, F, and G are TRUE. I'm also sure that D is FALSE (each sample will likely have a different mean and end points).
So that means that one of B, C, and E is true, and the remaining two are false. Any insights on which of B, C, and E is the true property? 70.169.186.78 ( talk) 03:02, 31 July 2009 (UTC)
-- 70.169.186.78 ( talk) 03:50, 31 July 2009 (UTC)
B is absolutely false. And I was surprised the first time I saw students interpreting it that way. D is false and F is true if you construe the words the way they're usually construed, but there I have some bones to pick. As someone pointed out, the question is whether we're talking about confidence intervals in general or about THIS interval. That's what "Igny" seems to be missing. F is true of confidence intervals in general, but not of any one particular confidence interval, unless we do some funny and somewhat elaborate revisions of our concept of probability. Where it gets involved is that maybe we SHOULD do some strange and elaborate revisions of the concept, but that's more than I could get into here. E is true, and it's just another way of restating the statement about what the 95% probability means. Michael Hardy ( talk) 23:56, 31 July 2009 (UTC)
Q2:
For a test of Ho: p = 0.5, the Z-test statistic equals -1.52. Find the p-value for Ha: p < 0.5.
I looked it up in the z-table and found a value of 0.0643 (A) for the area under the standard normal curve. But is this the solution, or is there something more? 70.169.186.78 ( talk) 20:03, 31 July 2009 (UTC)
Homework question, sorry. I just can't figure something out.
I'm assuming that I need to do a bilateral statistical hypothesis test.
Info: [...]In 1993-1994, a little more than half (52 %) of foreign exchange students were from Asia.[...]
Assuming that a random sample of 1,346 foreign students contains 756 students from Asia, in 1996.
a) With a significance level (not sure of the term in English) of 5 % can we conclude that the proportion of Asian foreign exchange students, in 1996, is 52 % ?
What I did:
1) H0: µ = 0,52
H1: µ = (not equal) 0,52
2) The significance level is 5 %
3) n = 756 > (or equal) 30
4) Since the size of the sample is n = 756 > (or equal) 30, we can use the normal distribution.
5) If Zx < -1.96 or if Zx > 1.96, the average proposed will be discarded and in consequence the alternative hypothesis H1 will be accepted. If -1.96 < (or equal) Zx < (or equal) 1.96, the average proposed by the null hypothesis H0 will not be discarded.
6) Calculations: http://i31.tinypic.com/xljhc9_th.jpg
However, the standard derivation is given NOWHERE. How am I supposed to calculate Zx ? Is there an error in the manual ? Rachmaninov Khan ( talk) 20:02, 31 July 2009 (UTC)
I can't say how much I'm thankful. Thanks again !!! Rachmaninov Khan ( talk) 15:08, 1 August 2009 (UTC)
We want to determine the true average number of drinks University of Michigan students have over a weeklong period. Assume the standard deviation is ~6.3. How many students should be sampled to be within 0.5 drink of population mean with 95% probability?
Is it 610? 70.169.186.78 ( talk) 04:39, 31 July 2009 (UTC)
The answer above misses the point completely and is entirely wrong.
The standard deviation of the sampling distribution of the sample average is σ/√n, where n is the sample size and σ is the population standard deviation, in this case 6.3. Since you've asked for 95%, you get 1.96 out of the normal table (that's the 97.5th percentile, so 2.5% is above that, and another 2.5% below −1.96, so that leaves 95% between ±1.96. So you need
That tells you that
and so on....... Michael Hardy ( talk) 00:12, 1 August 2009 (UTC)
OK, I see what you mean: no absolute value needed when you've got "2.5%" there. Michael Hardy ( talk) 17:32, 1 August 2009 (UTC)
Clarification: The statement we've derived is that
But now we want to rearrange it like this:
with the unobservable population mean μ in the middle, and only observable quantities as the endpoints of the interval. That observable interval is the 95% confidence interval for the unobservable population mean μ. Michael Hardy ( talk) 17:38, 1 August 2009 (UTC)
Mathematics desk | ||
---|---|---|
< July 30 | << Jun | July | Aug >> | August 1 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
I'm measuring my backyard and I've ran into a problem with a bay window area. There are three segments, each being 3' 5". The middle segment sticks out from the main exterior wall by 2'. What formula could I use to find the total width of the bay, taking angles into consideration, and then find what degree the angles should be at (for the left and right segments)? 3' 8" * 3 obviously wouldn't work. Here's an image that might help: http://img354.imageshack.us/img354/5492/confusion.jpg
I'd go out and measure it myself, but it's too dark, and it's also a torrential rainfall out there right now.
-- 69.154.119.41 ( talk) 00:15, 31 July 2009 (UTC)
The floor plan can be viewed as a rectangle and two right triangles: the rectangle is 2' by 3'5". Each right triangle has a 3'5" hypotenuse and a 2' leg. The question is how long the other leg is. So use the Pythagorean theorem:
Thus the whole width of the bay is about
Michael Hardy ( talk) 02:47, 31 July 2009 (UTC)
Statistics, population mean - question that I'm 75% done with, but need one more:
I've determined that A, F, and G are TRUE. I'm also sure that D is FALSE (each sample will likely have a different mean and end points).
So that means that one of B, C, and E is true, and the remaining two are false. Any insights on which of B, C, and E is the true property? 70.169.186.78 ( talk) 03:02, 31 July 2009 (UTC)
-- 70.169.186.78 ( talk) 03:50, 31 July 2009 (UTC)
B is absolutely false. And I was surprised the first time I saw students interpreting it that way. D is false and F is true if you construe the words the way they're usually construed, but there I have some bones to pick. As someone pointed out, the question is whether we're talking about confidence intervals in general or about THIS interval. That's what "Igny" seems to be missing. F is true of confidence intervals in general, but not of any one particular confidence interval, unless we do some funny and somewhat elaborate revisions of our concept of probability. Where it gets involved is that maybe we SHOULD do some strange and elaborate revisions of the concept, but that's more than I could get into here. E is true, and it's just another way of restating the statement about what the 95% probability means. Michael Hardy ( talk) 23:56, 31 July 2009 (UTC)
Q2:
For a test of Ho: p = 0.5, the Z-test statistic equals -1.52. Find the p-value for Ha: p < 0.5.
I looked it up in the z-table and found a value of 0.0643 (A) for the area under the standard normal curve. But is this the solution, or is there something more? 70.169.186.78 ( talk) 20:03, 31 July 2009 (UTC)
Homework question, sorry. I just can't figure something out.
I'm assuming that I need to do a bilateral statistical hypothesis test.
Info: [...]In 1993-1994, a little more than half (52 %) of foreign exchange students were from Asia.[...]
Assuming that a random sample of 1,346 foreign students contains 756 students from Asia, in 1996.
a) With a significance level (not sure of the term in English) of 5 % can we conclude that the proportion of Asian foreign exchange students, in 1996, is 52 % ?
What I did:
1) H0: µ = 0,52
H1: µ = (not equal) 0,52
2) The significance level is 5 %
3) n = 756 > (or equal) 30
4) Since the size of the sample is n = 756 > (or equal) 30, we can use the normal distribution.
5) If Zx < -1.96 or if Zx > 1.96, the average proposed will be discarded and in consequence the alternative hypothesis H1 will be accepted. If -1.96 < (or equal) Zx < (or equal) 1.96, the average proposed by the null hypothesis H0 will not be discarded.
6) Calculations: http://i31.tinypic.com/xljhc9_th.jpg
However, the standard derivation is given NOWHERE. How am I supposed to calculate Zx ? Is there an error in the manual ? Rachmaninov Khan ( talk) 20:02, 31 July 2009 (UTC)
I can't say how much I'm thankful. Thanks again !!! Rachmaninov Khan ( talk) 15:08, 1 August 2009 (UTC)
We want to determine the true average number of drinks University of Michigan students have over a weeklong period. Assume the standard deviation is ~6.3. How many students should be sampled to be within 0.5 drink of population mean with 95% probability?
Is it 610? 70.169.186.78 ( talk) 04:39, 31 July 2009 (UTC)
The answer above misses the point completely and is entirely wrong.
The standard deviation of the sampling distribution of the sample average is σ/√n, where n is the sample size and σ is the population standard deviation, in this case 6.3. Since you've asked for 95%, you get 1.96 out of the normal table (that's the 97.5th percentile, so 2.5% is above that, and another 2.5% below −1.96, so that leaves 95% between ±1.96. So you need
That tells you that
and so on....... Michael Hardy ( talk) 00:12, 1 August 2009 (UTC)
OK, I see what you mean: no absolute value needed when you've got "2.5%" there. Michael Hardy ( talk) 17:32, 1 August 2009 (UTC)
Clarification: The statement we've derived is that
But now we want to rearrange it like this:
with the unobservable population mean μ in the middle, and only observable quantities as the endpoints of the interval. That observable interval is the 95% confidence interval for the unobservable population mean μ. Michael Hardy ( talk) 17:38, 1 August 2009 (UTC)