Homework 8 Solutions

Ch. 22: 


12.  (a) True

     (b) True

     (c) False.  The normal curve will not work here, the data are too
skewed.  You need a bigger sample size when the data is so skewed.



Ch. 23:


3.  Model: There is a box with 50,000 tickets, one for each household
in the town.  The ticket shows the commute distance for the head of
household.  The data are like 1000 draws from the box.  The SD of the
box is unknown, but can be estimated by the SD of the data, as 9.0
miles.  The SE for the sum of the draws is estimated as (1000)^0.5 x
9, which is about 285 miles, and the SE for theaverage is estimated as
285/1000, about 0.3 miles.

    (a) The average comuute distance of all 50,000 heads of households
in the town is estimated as 8.7 miles; this estimate is likely to be
off by 0.3 miles or so.

    (b) 8.7 miles +/- 0.6 miles


8.  (a) True: the interval is "average +/- SE".

    (b) True: section 21.3

    (c) The data don't follow the normal curve, but the 68% might be
right; you need the data to tell.   (To see that the data don't follow
the curve: enrollments can't be negative, but the SD was a lot bigger
than the average, so there must have been a long right hand tail.)

    (d) False: 325 is not the SD.  (The data aren't normal, which is
another problem.)

    (e) False.  The normal curve is being used on the probability
histogram for the sample average, not the data (pp. 411 and 418-19).


The expected value is computed from the box; the observed value, from
the sample.



Ch. 26:


1.  (a) True (p. 481).

    (b) False.  The null says it's chance, the alternative says it's
real (pp. 478-79).


2.  The data are like 3800 draws made at random with replacement from
a box with "0"'s and "1"'s (how many of each is the subject of the
hypothesis), where "1" is red.

    (a) Null: The fraction of "1"'s in the box is 18/38.
        Alternative:  The fraction of "1"'s in the box is more than
                      18/38.

    (b) the expected number of reds (computed using the null) is 1800.
The SD of the box (also computed using the null) is nearly 0.5, so the
SE for the number of reds is (3800)^0.5 x 0.5 (about 31).  So
z = (obs-exp)/SE = (1890-1800)/31 (which is about 2.9), and p is about
2/1000.

    (c) Since p is so small, we have evidence to reject the null
hypothesis.  It seems as though the wheel is not balanced correctly,
causing there to be too many reds.

Comments: This problem is about the number of reds.  In the formula
for the z-statistic, "obs", "exp", and "SE" all refer to the number of
reds.  The expected, as always, is computed from the null.  In this
problem, the null gives the composition of the box, so the SD is
computed from the null; it is not estimated from the data (p. 487).