Testing whether the slope (B1) is 0 is equivalent to testing whether correlation coefficient is 0. T/F & why?
368727
If you give only obscure details (B1, T/F) but tell nothing more about the problem, you are not likely to get an answer....I do a lot of statistical correlations, but this really doesn't mean much.\n\nNow, my guess on what you mean.\nYou have a series of data points (x_n, y_n). If you make a scatterplot and the points are rougly on a line you have perfect correlation, correlation coefficient is +1 or -1.\n\nIf on the other hand Y doesn't depend on X, i.e. correlation coefficient 0, then when you do a linear regression the slope of the regression line will be close to zero, as the data points (x_n, y_n) are scattered randomly.
802801
i think -1; y=mx+b, "b" is the 3 and "m"(slope) is negative one.
746991
the slope is 4\nwhen an equation is in standard form:\ny=mx+b\nm (which is the coefficient of x) is the slope
41162
Two sided test is for coefficient to be equal to a number or another coefficient. One sided test is for coefficient being greater or less than another coefficient or a number. Categorical variables can have confidence intervals. If you find that your estimated coefficient is outside confidence interval (-inf, number) then it's significant, you explain it as any other variable, but comparison group is the other gender. For example, if sex =0 for males, =1 for females, its negative and significant, and the dependent variable is income, then females earn less than males and the amount of loss is estimated by the coefficient.\n\nPS. Your question is more for statistics or econometrics than math.
903169
1. Undefined (No Slope)\n2. Undefined again.
371794
slope = (-1-2)/(-1-0) = -3/-1 = 3
24979
The simple answer is no.  The K-S (Kolmogorov-Smirnov) test is actually used to test if two data sets have the same distribution, so the previous answer is not correct.  \n\nA T-Test is used when the data comes from an underlying normal population (you can assess that with K-S), and assumes continuous data, and a common unknown variance for both samples.  \n\nWhat exactly are you trying to test?
784879
3/4 is the slope if u put it in the format of y = mx + b\n\nor \n\ny = .75x + 1
531486
A positive slope would indicate that the dependent variable is increasing as time passes, while a negative slope indicates a decrease in the dependent variable.  A zero slope indicates no change with the passage of time.  An infinite slope indicates that time has stopped!!  8^)\n\nIf the dependent variable is something like the price of gasoline, then a positive slope would indicate that gas prices will be higher in the future.  If the dependent variable is your salary, a positive slope would indicate that you are getting paid more as time passes.  One is good, one is not so good.
333704
1. False, a polynomial has no = sign.\n2. False, the log of zero (or any negative number) does not exist.
1180441
Excel is a great tool.... I'm sorry I don't remember all the specifics but I know that there is more than one way to do it.  If you can narrow down which variables you expect to be most related that would be a great start.  If you don't have a physical (real-life) basis for that, you can try graphing all 6 variables against a column just numbered with integers from 1 to n (the number of data in each column).  Then you can plot pairs of columns against each other and find equations to fit the curves (polynomial up to 6th order, exponential, logarithmic) and use the R squared values to determine how well the equation represents the correlation of the data.  (R squared = 1 is a perfect match.)  Now when you start getting into cases where there are multiple independent variables effecting the dependent variables, it gets a little more complicated, and unfortunately I don't have Excel on this computer and I don't have my old notes or textbooks with me.  I hope this at least helps you get started.
979117
From a table -- see which y-value corresponds to x=0\nFrom a graph -- see where the graph crosses the y-axis\nFrom an equation -- get the equation in y=mx+b form; the y-interecept is b
733792
Yes it is a function because by the verticle line test it only intersects that line at 1 point.  \n\n\n It is a pretty simple test.  if a verticle line hits the graph at more than 1 point it is not a function.
789216
Try to find a trend in the table.  For example if the table were to tbe like x: 5, 6, 7 for y: 8,10,12, respectively.  Then, go from y to x and try to find some eqation that fits.  For the example, it would be 8 = b + 5m (in which be is some number that gets added to y and m is the constant slope).  The other equation would be 10 = b + 6m, in which b = 10 -6m.  Thus, since the b in the first equation and the second are the same, substitute.  8 = (10 - 6m) + 5m.  8 = 10 - m.  m = 2.\nFind b now by plugging in m.  8 = b + 5(2).  b = -2.\nThus, the equation is y = 2x - b.
839252
You must use this formula to find the Coefficient of Correlation for the above equation and the data set that it represents:\n\nr = (X-Xm)(Y-Ym)/(n-1)((Sy)(Sx)\n\nAround 1900, Karl Pearson, developed a technique that describes the relationship between two sets of interval-scaled variables.  This mathematical technique measures the relationship between linear variables.  If all of the data points lie on straight line, then the correlation coefficient is +1.00 or -1.00, depending on the slope of the line of the data set.  And of course, if there is no linear relationship, then the correlation will be zero.\nI like to plot the data set using MS Excel, and then do a linear regression using the built in data tools.  Remember, that you need a data set, X and Y values, to plot a graph of data points.\nHope this helps.
866301
because y- x > 0\n\n|y-x | = y- x (positive value)\n\nx-y < 0\n\nso | x-y| = y-x (as absolute value positive)
437570
You're probably talking about the vertical line test.  A relation is a function iff you can draw a vertical line anywhere and intersect the function at just one point.\n\nSo, something like y = x is a function because a vertical line always hits just one point.  y^2 = x is not a function because if you draw a vertical line at x=4, it hits the graph at both (4,2) and (4,-2).