Answer:
(4,3)
the distance between x1 and x2 then y1 then y2
Answer:
a. N=25
b. X[bar]= 60.52
c. Y[bar]= 106.72
d. SSx= 115.24
e. ∑X*∑Y = 4036684
f. SSxy= 202020.3296
g. √(SSx*SSy)= 449.46
Step-by-step explanation:
Hello!
Using the attached data you need to calculate some statistics.
a) N
The sample size is listed under the first column "subject" You can see that 25 subjects qhere studied so N=25.
b.
The mean of set X is equal to X[bar]= ∑X/n= 1513/25= 60.52
∑X is listed in the second table.
c.
The mean of ser Y is Y[bar]= ∑Y/n= 2668/25= 106.72
∑Y is listed in the second table.
d.
Sum of Squares of set X SSx= ∑X²-[(∑X)²/n]= 91682-[(1513)²/25]= 115.24
e.
∑X*∑Y =1513*2668= 4036684
f.
SSxy= (∑X²-[(∑X)²/n]) * (∑Y²-[(∑Y)²/n])= (91682-[(1513)²/25]) * (286482*[(2668)²/25])= 202020.3296
g.
√(SSx*SSy)= √(115.24*1753)= 449.46
I hope you have a SUPER day!
Answer:

False
Step-by-step explanation:

You substitute
for the second equation.

Then you have to distribute
.

Combine like terms.

The answer too your question is 65 it is equal to the square root of 65
Answer:
Step-by-step explanation:
In the model
Log (salary) = B0 + B1LSAT +B2GPA +B3log(libvol) +B4log(cost)+B5 rank+u
The hypothesis that rank has no effect on log (salary) is H0:B5 = 0. The estimated equation (now with standard errors) is
Log (salary) = 8.34 + .0047 LSAT + .248 GPA + .095 log(libvol)
(0.53) (.0040) (.090) (.033)
+ .038 log(cost) – .0033 rank
(.032) (.0003)
n = 136, R2 = .842.
The t statistic on rank is –11(i.e. 0.0033/0.0003), which is very significant. If rank decreases by 10 (which is a move up for a law school), median starting salary is predicted to increase by about 3.3%.
(ii) LSAT is not statistically significant (t statistic ≈1.18) but GPA is very significance (t statistic ≈2.76). The test for joint significance is moot given that GPA is so significant, but for completeness the F statistic is about 9.95 (with 2 and 130 df) and p-value ≈.0001.