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.
<h3>
Answer:</h3>
18
<h3>
Step-by-step explanation:</h3>
Put 3 where x is, then do the arithmetic.
... f(x) = x(5x -9)
... f(3) = 3(5·3 -9) = 3(15 -9) = 3·6
... f(3) = 18
_____
Your calculator can do this, too.
The standardized probability of a normally distributed data is given by
P(X < x) = P(z < (x - mean) / standard deviation)
P(X > x) = 1 - P(X < x)
P(X > 125) = 1 - P(X < 125) = 1 - P(z < (125 - 100) / 15) = 1 - P(z < 25/15) = 1 - P(z < 1.667) = 1 - 0.95221 = 0.04779 = 4.8%
Though it cannot be said to be usual, at the same time it is not unusual because 125 is above the first standard deviation of the mean but below the second standard deviation of the mean.
First step is to mph into miles:
Turn 1hr and 40 min into a decimal:
1hr 40min ----> 1.6667hr
Then multiply:
300 x 1.6667 = 500.1
The distance is 500.1 miles.
Hope this helps!