Homework 9 Solutions

Ch. 26: 


4.  (55-63)/[20/(30^0.5)] is about -2.19.

   So, we would expect that a TA section would do this badly (or
worse) due to chance only about (100-97.22)/2 = 1.39%.  Probably not a
good defense!



Ch. 27:


3.  percentage who rated clergyman "very high or high"
1985: 67%   1992: 54%

Results are based on independent simple random sample of 1000 persons
in each year.

(a)  Should make a 2-sample z-test because these results
used a different sample each year (independent of the other year).  We
need to compare the means of the sample.

(b)  Need 2 box models, one for each year.  Each box represents a
different year's state of public opinion.  Tickets show "1" or "0",
with "1" meaning "very high" or "high" opinion of clergymen.  It
doesn't matter how many tickets - just need the right proportion of
"1"'s and "0"'s (67% in 1985 box and 54% in 1992).

null hypothesis:  same proportion rating clergymen "high" or "very high"
in both years - same percentage of "1"'s in both boxes

alternative hypothesis: percentage of "1"'s is lower in the 1992 box

(c)  z-score:  (0-13)/2.17 = -5.99 --> p-value almost 0

SE%(1985) = [(0.67)(0.33)]^0.5/(1000^0.5) x 100% = 1.49%
SE%(1992) = [(0.54)(0.46)]^0.5/(1000^0.5) x 100% = 1.58%

SE for difference of %: [(1.49)(1.49) + (1.58)(1.58)]^0.5 = 2.17%

Since p-value is so low, chance is not a good explanation fo rthis
result.  The test does not prove that scandals were the cause of the
change in opinion, however.


7.  (a) 592 assigned to treatment, 48.3% re-arrested
        154 assigned to control group, 49.4% re-arrested

null hypothesis: treatment group no better than control

alternative hypothesis: treatment group does better (less arrests)

SE%(T): [(0.483)(0.517)]^0.5/(592^0.5) x 100% = 2.05%
SE%(C): [(0.494)(0.506)]^0.5/(154^0.5) x 100% = 4.03%

z-score: [(48.3 - 49.4) - 0]/[(2.05)(2.05) + (4.03)(4.03)]^0.5 = -0.24

p-value = (100-19.74)/2 = 40.13  --> p-value = 0.4

So, there is no significant difference.

    (b)  null hypotheis: no difference in the averages for the 2 groups
         alternative hypothesis: income support group has lower average

SE for treatment average: 0.7 wks
SE for control average: 1.4 wks

SE for the difference: 1.6 wks
observed difference: -7.5 

z-score: -7.5/1.6 is about -4.7  --> p-value almost 0

So, we reject the null hypothesis in favor of the idea that income
support makes the released prisoners work less.



Ch. 28:


8.  You should use the chi-squared test.  The die has 6 possible
"states", and we're interested in whether the die exhibits observed
frequencies of outcomes that are sufficiently close to those we
expect.  We're not interested in the average outcome (which could
still be 3.5 even if the die is unfair) or the proportion of times a
certain side comes up (it's possible that an ace may still show 1/6 of
the time, but the other sides show up a disproportionate number of
times).  The chi-squared test considers the expected vs. observed
outcome for each possible result, and so is a better test to use to
see if the die is likely to be fair.