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Why is it that:
when:
and k1 is imaginary? How do you calculate the absolute value of t anyhow? -- 99.237.234.104 ( talk) 00:50, 25 April 2010 (UTC)
"In projective n-space over a field, a homogeneous multivariable polynomial of degree d-1 cannot have d roots on a line (projective line) without vanishing identically on the line." According to my textbook, this is not "too hard to see". I've been stuggling to see this. Do I need to do some heavy calculations, or am I supposed to visualise it? I can visualise why this is true somewhat, but I just can't see why a degree 2 multivariable polynomial for example cannot have 3 roots on a line algebraically, especially in projective n-space which I can't visualize for dimension greater than 3. How do I show that a d-1 degree multivariable homegeneous polynomial cannot have d roots on a given line in projective space without vanishing on the entire line? This is not homework (I'm just trying to understand the text). Thank you in advance. --Annonymous —Preceding unsigned comment added by 122.109.239.224 ( talk) 03:47, 25 April 2010 (UTC)
Given a table that yields the probability distribution for a term in a series given the n terms before it, is there an algorithm to invert the table so that it instead yields the distribution of a term given the n terms after it (i.e. if the table is empirically derived, to give the table we would have come up with had we reversed the order of the sample data)? Neon Merlin 04:09, 25 April 2010 (UTC)
I was trying to solve the QHO and I found a power series expansion for the each energy eigenstate (in the position basis). This was where with and E measured in units of . What prevents me from using and arbitrary value of E rather than only odd integers? 74.14.111.225 ( talk) 05:40, 25 April 2010 (UTC)
Dear Wikipedians:
I am doing a two sample t-test in R on two samples shown below:
x = (9, 10, 1, 2, 4, 2, 1, 7, 9, 2, 2, 5, 6, 7, 6, 3, 9, 2, 10, 4, 8, 5, 6, 9, 2) y = (20, 18, 19, 11, 10, 17, 12, 20, 12, 15, 11, 13, 19, 17, 12, 16, 15, 11, 14, 12, 19, 10, 12, 20, 17)
However, R outputs the following:
Welch Two Sample t-test data: x and y t = -10.383, df = 47.182, p-value = 8.962e-14 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -11.507589 -7.772411 sample estimates: mean of x mean of y 5.24 14.88
where my textbook shows the following output:
Two sample t-test: Assuming same variance Data: ER visits by asthma status t = -10.383, df = 48, p-value = 7.29 x 10^-14 Sample Estimates: Mean in Non-Asthmatics 5.24 Mean in Asthmatics 14.88 95 percent confidence interval: (0.000, 11.234) (Non-Asthmatics) (8.036, 21.724) (Asthmatics)
My question is: how do I make R output two confidence intervals?
Thanks,
174.88.242.191 ( talk) 16:13, 25 April 2010 (UTC)
> x <- c(9, 10, 1, 2, 4, 2, 1, 7, 9, 2, 2, 5, 6, 7, 6, 3, 9, 2, 10, 4, 8, 5, 6, 9, 2) > y <- c(20, 18, 19, 11, 10, 17, 12, 20, 12, 15, 11, 13, 19, 17, 12, 16, 15, 11, 14, 12, 19, 10, 12, 20, 17) > data.frame(asthma=rep(c("no","yes"),each=c(length(x),length(y))),d=c(x,y)) -> jj > t.test(jj$d~jj$asthma,var.equal=TRUE) Two Sample t-test data: jj$d by jj$asthma t = -10.383, df = 48, p-value = 7.294e-14 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -11.506753 -7.773247 sample estimates: mean in group no mean in group yes 5.24 14.88
Mathematics desk | ||
---|---|---|
< April 24 | << Mar | April | May >> | April 26 > |
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. |
Why is it that:
when:
and k1 is imaginary? How do you calculate the absolute value of t anyhow? -- 99.237.234.104 ( talk) 00:50, 25 April 2010 (UTC)
"In projective n-space over a field, a homogeneous multivariable polynomial of degree d-1 cannot have d roots on a line (projective line) without vanishing identically on the line." According to my textbook, this is not "too hard to see". I've been stuggling to see this. Do I need to do some heavy calculations, or am I supposed to visualise it? I can visualise why this is true somewhat, but I just can't see why a degree 2 multivariable polynomial for example cannot have 3 roots on a line algebraically, especially in projective n-space which I can't visualize for dimension greater than 3. How do I show that a d-1 degree multivariable homegeneous polynomial cannot have d roots on a given line in projective space without vanishing on the entire line? This is not homework (I'm just trying to understand the text). Thank you in advance. --Annonymous —Preceding unsigned comment added by 122.109.239.224 ( talk) 03:47, 25 April 2010 (UTC)
Given a table that yields the probability distribution for a term in a series given the n terms before it, is there an algorithm to invert the table so that it instead yields the distribution of a term given the n terms after it (i.e. if the table is empirically derived, to give the table we would have come up with had we reversed the order of the sample data)? Neon Merlin 04:09, 25 April 2010 (UTC)
I was trying to solve the QHO and I found a power series expansion for the each energy eigenstate (in the position basis). This was where with and E measured in units of . What prevents me from using and arbitrary value of E rather than only odd integers? 74.14.111.225 ( talk) 05:40, 25 April 2010 (UTC)
Dear Wikipedians:
I am doing a two sample t-test in R on two samples shown below:
x = (9, 10, 1, 2, 4, 2, 1, 7, 9, 2, 2, 5, 6, 7, 6, 3, 9, 2, 10, 4, 8, 5, 6, 9, 2) y = (20, 18, 19, 11, 10, 17, 12, 20, 12, 15, 11, 13, 19, 17, 12, 16, 15, 11, 14, 12, 19, 10, 12, 20, 17)
However, R outputs the following:
Welch Two Sample t-test data: x and y t = -10.383, df = 47.182, p-value = 8.962e-14 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -11.507589 -7.772411 sample estimates: mean of x mean of y 5.24 14.88
where my textbook shows the following output:
Two sample t-test: Assuming same variance Data: ER visits by asthma status t = -10.383, df = 48, p-value = 7.29 x 10^-14 Sample Estimates: Mean in Non-Asthmatics 5.24 Mean in Asthmatics 14.88 95 percent confidence interval: (0.000, 11.234) (Non-Asthmatics) (8.036, 21.724) (Asthmatics)
My question is: how do I make R output two confidence intervals?
Thanks,
174.88.242.191 ( talk) 16:13, 25 April 2010 (UTC)
> x <- c(9, 10, 1, 2, 4, 2, 1, 7, 9, 2, 2, 5, 6, 7, 6, 3, 9, 2, 10, 4, 8, 5, 6, 9, 2) > y <- c(20, 18, 19, 11, 10, 17, 12, 20, 12, 15, 11, 13, 19, 17, 12, 16, 15, 11, 14, 12, 19, 10, 12, 20, 17) > data.frame(asthma=rep(c("no","yes"),each=c(length(x),length(y))),d=c(x,y)) -> jj > t.test(jj$d~jj$asthma,var.equal=TRUE) Two Sample t-test data: jj$d by jj$asthma t = -10.383, df = 48, p-value = 7.294e-14 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -11.506753 -7.773247 sample estimates: mean in group no mean in group yes 5.24 14.88