My friend... My method wasn't wrong and I couldn't find any mistake on my procedure... Check it out and analyze it by yourself...
Given:
n = 20, sample size
xbar = 17.5, sample mean
s = 3.8, sample standard deiation
99% confidence interval
The degrees of freedom is
df = n-1 = 19
We do not know the population standard deviation, so we should determine t* that corresponds to df = 19.
From a one-tailed distribution, 99% CI means using a p-value of 0.005.
Obtain
t* = 2.8609.
The 99% confidence interval is
xbar +/- t*(s/√n)
t*(s/√n) = 2.8609*(3.8/√20) = 2.4309
The 99% confidence interval is
(17.5 - 2.4309, 17.5 + 2.4309) = (15.069, 19.931)
Answer: The 99% confidence interval is (15.07, 19.93)
Four hundred seventy and forty seven hundredth
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.
Answer:
d
Step-by-step explanation:
the answer is d because if you do 3+2 it equals 5 so therefore it is b
