# STAT 366 Spring 10 HW 5 Due 3/17/10 #6, Review Problems, page

```STAT 366 Spring 10 HW 5 Due 3/17/10
1) #6, Review Problems, page 305 ,Freedman et al.
For this box [1,1,2,3], AVG(box)=7/4= 1.75 and
SD(box)=Sqrt{(2*(1-1.75) + (2-1.75)^2 + (3-1.75)^2)/4}=.829
Observed value for sum of draws is 45+23*2 + 32*3 = 187
Observed value for the number of 3’s = 32
Observed value for the number of 1’s = 45
Expected value for the sum of draws 100*AVG(box)= 175
Expected value for the number of 3’s = 100*[proportion of 3’s]
= 100*.25 = 25
Expected value of the number of 1’s = 100*[proportion of 1’s]= 50
Chance error in the sum of draws= Observed sum – Expected sum
= 187-175=12
SE for the sum of 1’s= Sqrt(100)*SD[Box model for # of 1’s=[1,1,0,0]]
= 10*Sqrt{.5*(1-.5)} = 10*.5 = 50
2) #13, “
“
, page 306, “
The box model for the letter for each draw is [A,A,A,B,I,R]
The box model for # A’s for each draw is [1,1,1,0,0,0] with
AVG(boxA)= .5 and SD(boxA)= Sqrt{.5*(1-.5)}=.5
The box model for # B’s for each draw is [0,0,01,0,0] with
AVG(boxB)=.167 and SD(boxB)=.373.
For 1000 draws,
EV(sum of A’s)= 1000*.5=500,
SE(sum of A’s)= Sqrt(1000)*.5=15.81,
EV(sum of B’s)=1000*(.167)=167,
SE(sum of B’s)=Sqrt(1000)*.373=11.8
To find the chance of getting 10 or more A’s above the expected value, we need
the Zscore for 10 above the expected value = 10/SE(sum of A’s)=.633.
Similarly, the Zscore for 10 B’s above the expected value = 10/11.8 =.847
Since the Zscore for 10 B’s above expected is greater than the Zscore for 10 A’s
above the expected, we know that the chance of getting 10 or more A’s
above the expected is greater than the chance of getting 10 or more B’s
above the expected.
3) Hospital A has 100 births in one week, while hospital B has 200 births in the
same week. Assuming that there is an equal chance of having a girl or boy
baby, which hospital has a higher chance of having 55% or more of the births
being boys?
Have a male birth is like drawing from the box [1,0]. AVG(box)=SD(box)=.5.
Since at this point all we are certain of is that the sum of draws from a box is
normally distributed, we need to convert this question into something about
sums. 55% of 100 births is 55, and 55% of 200 births is 110. So the question
becomes “which event has a higher chance: 55 or more male births from
hospital A or 110 or more male births from hospital B?” The number of male
births from hospital A is like taking 100 draws from the box above, and the
number of male births from hospital B is like taking 200 draws from the box.
For 100 draws,
EV(sum)=100*AVG(box)=50
SE(box) = Sqrt(100)*SD(box)=5
Then Zscore for 55 = (55-50)/5=1 and so the probability of having 55 or more
male births at hospital A is about 16%.
For 200 draws,
EV(sum)=200*.5 = 100,
SE(sum)=Sqrt(200)*.5=7.07
Then the Zscore for 110= (110-100)/7.07=1.41.
I could calculate the % chance for 110 male births from hospital B, but I don’t
need to. Since the Zscore for 110 out of 200 is higher than the Zscore for 55 out
of 100, I know that the % chance of 110 or more out of 200 is less than the
% chance of 55 or more out of 100.
4) A medical procedure has a 70% success rate. Fill in the blanks below:
Out of 100 procedures, we expect to have _70___ successes, give or take_~5.
There is about 95% chance that the total # of successes, out of 100, will be
between _61__ and _79___.
# of successes for each procedure is like drawing from the box
[1,1,1,1,1,1,1,0,0,0]. The AVG(box)=p = .7 and SD(box)=Sqrt{.7*(1-.7)}=.46
EV(sum of 100 draws)=100*.7=70 and SD(sum of 100 draws)= 10*.46 = 4.6.
So we expect about 70 successes give or take 5 (hard to have a .6 success).
There is about 95% chance that the # of successes will be in the interval
70 +- 2*4.6 which is about 70+-9.
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