Supplementary Exercise 6.7 of IPS7e
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Coverage of confidence intervals (i.e., whether a confidence interval (CI)
includes the true value) in an agricultural field trial. A total of n=7
CIs are calculated.


(a) The number (X) of CIs out of 7 that cover/include the true value will
follow a binomial distribution B(7,0.95). This is because with the
assumed independence of the CIs the situation corresponds to a binomial
setting. We can therefore find P(X=7) in Table 1 of S, or calculate
  P(X=7) = 0.95^7 = 0.698.

It is seen that the probability is substantially below the 0.95 for each
CI. The 0.698 (or 69.8%) is the simultaneous confidence level.


(b) The probability that at least 6 out of 7 intervals cover the true
value can similarly be calculated from B(7,0.95), using here the table:
  P(X>=6) = P(X=6)+P(X=7) = 0.257+0.698 = 0.955.
We could also have calculated,
  P(X=6) = 7*0.95^6*(1-0.95) = 0.257.


Added note (not part of question):
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The fact that the probability for at least 6 of the 7 intervals to cover
the true value is so close to 0.95 is a coincidence and does not reflect
any general rule. We need to use adjustments for multiple testing/comparisons 
to achieve a simultaneous coverage of 0.95. Due to the independence of
the intervals, in this case we simply need to solve
  0.95 = p_adj^7, or p_adj=0.95^(1/7) = 0.9927,
meaning that if each of the confidence intervals have coverage 99.27%,
then the simultaneous coverage is 0.95.

For independent intervals/tests, this is (slightly) better than using the
Bonferroni method, which will tell us to use an error level of 0.05/7=0.0071
in each interval, and hence a coverage of 1-0.0071=0.9929 or 99.29% in
each interval.