Answers to the Second Practice Midterm

SE for the average = SQRT(400) x .2
                     ------------ =   .01
                          400 

This is a Chapter 23 problem.

2. B.

3. B

This is like a box which is completely known (chapter 17)
You could solve this without a calculation -- in gambling
situations the expected value is almost always negative.
So only A or B are valid here, and no one would gamble if
they were guaranteed to lose -- so A is wrong, that leaves
B.

To solve it mathematically:

$200  1/2000
$50   1/2000
$10   10/2000
$5    10/2000
$0    1978/2000

The box average is (200 * 1/2000) + (50 * 1/2000) +
(10 * 10/2000) + (5 * 10/2000) + (0 * 1978/2000) =
.20  dollars or 20 cents.

To get the net gain, .20 - 1.00 = -.80 dollars or -80 cents.

You could do it this way too:

$199  1/2000
$49   1/2000
$9   10/2000
$4    10/2000
$-1    1978/2000

And you will also get -.80.  You've figure the dollar cost
into the box rather than after calculating the average.

4.  A.  

The next best answer is E, but if you have picked
a bad sample, it doesn't matter what the response rate is.
D does not work either, because again, if you pick a bad
sample, it doesn't matter how large it is.  C.  You don't
usually have control of the Standard Deviation. B is just
terrible.

5.A
Expected value for the sample percentage = 85%
knowing nothing else, you expect to get 85% saying "yes"

It's like a box with

1= says yes  = .85
0 = does not say yes  = .15

like a lopsided coin and you are interested in the "yeses"
so they take a value of "1".


Standard Error for the sample percentage

= root(16) * (1-0)*(root (.85 * .15))
  -----------------------------------  * 100 = 8.9268%
                   16

5.B
You'll need to calculate two Z scores to figure this out:

Z(for 89%) = (89 - 85)/8.9268%  = .448 so about .45
Z(for 70%) = (70 - 85)/8.9268% = 1.68 so about 1.65

the area associated with a Z = .45 (from table A 105) is
34.73, divide this in two to get 17.365%
the area associated with a Z = 1.65 (from table A 105) is
90.11, divide this in two to get 45.055%
add 17.365 and 45.055 to get 62.42%
this is the chance of getting between 70% and 89% saying "yes"

6.  
The true average IQ for all UCLA undergraduates is captured
by the interval 105 to 135.  If we could draw 100 samples
of the same size, 95 of them would cover the true average
IQ.  We hope that the 105 to 135 is one of those times
and that the true average is in there.  There is a 5%
chance that my interval is completely wrong.  There is
a 95% chance that my interval is right.

Note: this is not the same as saying there is a 5% chance
that the parameter is not in there and this is not the
same as saying there is a 95% chance that the parameter
is in there.  Remember, the parameter is a FIXED, UNCHANGING
value.  Samples change. 

7.
You are looking for  a sum here.  See chapter 17.

Your box looks like this

+1.00  .40 or 40% chance
-1.10  .60 or 60% chance

the box average is (1 * .40) + (-1.1 * .6) or -.26

after 10 games the sum is 10 * -.26 = -2.60

You would expect to lose $2.60 after playing 10 times on average.

8.  

The standard error of the sum is (see 17.4 for the formula):

root(10) * [(1 - (-1.1))*root(.4 * .6)] = 3.2533 dollars

9.  We're looking for the chance (think I need the Z score)
of getting $3 after 10 games given that you expect to lose 2.60.

z = 3 - (-2.60)
   ____________  =  1.72 so about 1.70 = Z
    3.2533

from the table, the area greater than Z = 1.70 is
(100 - 91.09) /2 = 4.455 or about 4.5%

So, your friend is pretty lucky or knows alot about sports
(we'll get to than in Chapter 26) if she was picking teams
at random, she only had a 5% chance of doing this well.

A side note, if you plan to visit Vegas, you best chance
for winning outside of live poker (not video poker) 
is probabably in the sports book room...because if you 
follow teams closely, you have an edge in knowing how to 
place your bets.  All of the other games are very well defined, 
but in sports, anything can (and does) happen.  

Like, while I'm there, even though UCLA is doing terribly
this year in football, I'll bet UCLA to beat USC...and I'll
bet Notre Dame beats USC...and...