Supplementary Exercise 6.87 of IPS7e
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Mean nicotine content in cigarettes. The advertised value is 1.4 mg.
A study was carried out to determine whether the mean content was
higher. The study wants to test the null hypothesis
  H0: mu = 1.4mg
against the alternative hypothesis
  Ha: mu > 1.4mg
where mu is the cigarette population mean nicotine content.


(a) A z-test statistic was computed: z_obs=1.75. This value is
significant at the 5% level, because with a one-sided Ha we 
compute the P-value as
  P = P(Z>1.75) = 0.040. 
(For example, Table B of PSLS gives P(Z<1.75)=0.9599 and P(Z<-1.75)=0.0401.)
Therefore, the P-value is indeed less than 0.05.

Alternatively, we could note that the 5% cut-point (sometimes called
the critical value) for the test with this one-sided alternative is the 
95% percentile of N(0,1) because the 95% percentile has 95% to the left and 5%
to the right. We can use the normal distribution table (with probability 0.95) 
or the table of critical values/percentiles (with upper tail probability 0.05) 
to determine the 95% percentile as 1.645. Because z_obs > 1.645, the test is
significant at the 5% significance level.


(b) We already computed the P-value, and it was greater than 0.01.
Therefore, the test is not significant at the 1% level. Alternatively,
the 1% cut-point for the test is the 99% percentile of N(0,1) = 2.326,
and z_obs < 2.326.


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Minitab commands (to compute P-value and cut-offs) and output:

MTB > CDF 1.75;
SUBC>   Normal 0.0 1.0.
Cumulative Distribution Function 

Normal with mean = 0 and standard deviation = 1

   x  P( X <= x )
1.75     0.959941

MTB > CDF -1.75;
SUBC>   Normal 0.0 1.0.
Cumulative Distribution Function 

Normal with mean = 0 and standard deviation = 1

    x  P( X <= x )
-1.75    0.0400592

MTB > InvCDF .95;
SUBC>   Normal 0.0 1.0.
Inverse Cumulative Distribution Function 

Normal with mean = 0 and standard deviation = 1

P( X <= x )        x
       0.95  1.64485

MTB > InvCDF .99;
SUBC>   Normal 0.0 1.0.
Inverse Cumulative Distribution Function 

Normal with mean = 0 and standard deviation = 1

P( X <= x )        x
       0.99  2.32635