Sample Test2 Answer Key.                   Stat 401, Fall 2010
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#1. Non-numeric answers.


#2. (a) 7.1    (b)  7/9


#3. Chi-square test-statistic value = 8.4, with 2 degrees 
of freedom, so p-value is .015 and reject at alpha=.05.


#4. Using manufacturer spec's mu and sigma as 3200 and 1000,
find the expected number out of 60 observations respectively 
falling into the 4 designated intervals as
 0.8  13.7  32.8   12.7
leading to chi-square stat with 3 d.f. of 84.4, with 
astronomically small p-value. So definitely reject normal dist
with the spec values of mu and sigma.

Using estimated mu and sigma as 3500 and 1500, to test normal 
distribution without regard to the manufacturer's suggested mu 
and sigma, we find the expected number out of 60 observations 
respectively falling into the 4 designated intervals as
 2.3  12.3  22.7  22.2
leading to chi-square stat of 41.5, with d.f. between 1 and 3, 
but otherwise with same conclusion.

In these calculations, I rounded the expected number to one 
decimal place.

IN FACT, WITH THE VALUES 2344 FOR MU AND 1248 FOR SIGMA, THE FIT
WOULD HAVE BEEN NEARLY PERFECT. So here is an example where 
the method of estimating mu and sigma makes a huge difference !


#5, 34*s_x^2 = 2720, sumxy - 35*xbar*ybar = -2000, so 
bhat= -2000/2720 = -.735, ahat = 130 - (-.735)*40 = 159.4
Residual = 150 - (159.4 -.735*36) = 17.06
r = (-2000/34)/sqrt(80*300) = -.380,  
Coeff of determination = rsq = (-.38)^2 = .144



#6. A line does not fit well through even most of the points, 
so NO, you would probably not view these data as likely normal.
If you did fit a line through the middle 20-30 points, then 
the line lies above the points at either tail, suggesting that:
at both tails, the true quantiles are larger than normal theory 
would suggest.  So the distribution is skewed with fatter tail 
than normal toward the right and thinner toward the left.


#7. (a) CI is  (8/6)*(1/3.115, 3.115) = 0.4280364 4.153
(b) SSB = 14*3*var(c(27,25,30,28)) = 182, SSE = 13*(6+8+7+6) = 351,
F = (182/3)/(351/(4*13)) = 8.99, reject (P-value < .0001)
(c) Estimate sigma^2 as MSE = 351/52 = 6.75