27.2	a) The SE for the difference is 5.9%, so z = 2.4/5.9 = .4; 
	which likely means that the difference is due to chance.
	b) Yes. SE(difference) = 1.8, z = 4.9/1.8 = 2.7, p = .003.

27.3	a) You need a two sample test (there are two samples!).
	b) Two boxes.  Each box has one ticket for each person in the
	population (marked with a 1 or 0).  Each sample is like 1000
	draws from the appropriate box.  Null hypothesis says that 
	the boxes have equal percentages of 1s (clergy >= high), the
	alternative hypothesis says the 1992 box has a lower percentage
	of 1s.
	c) The estimated SEs for the boxes are about 1.5%; SE of the 
	difference is about 2.2%.
		z = (67-54)/2.2= about 6, so p = about 0.
	The difference is real, but this test doesn't tell you what the
	cause might be.

27.5	The data say that the difference is real.  SE(difference)=4.2%.
	z= (46%-88%)/4.2% = 10!

27.10	The samples are not chosen at random... there is dependence
	between the first- and second-born children, so the test is
	not legitimate.

28.1	a) i; b) iii.  See p 541-2 (ex. 3-6).

28.2	ii.  

28.3	Using the method of section 4...
	  X^2 with 4 df is about 30... p <<.001  This is not chance
	variation.

28.4	a) it is a chance (probability) histogram.
	b) the chance that 5 <= X^2 < 5.2.
	c) ii
	d) 10%,

28.5 	a) ... bigger with 10 df than with 5 df.
	b) ... bigger with 15 than with 20.

28.6	Using the method of section 3... The observed frequencies are
	too close to the expected frequencies: X^2 = 2 on 10 df, so
	p < 1%.  Don't play dice with him.

28.9	a) Use the method of section 4.  X^2 = 11 on 2 df, p < .01.  
	Slightly older people are more likely to get married.
	b) Yes.