(a) A 100(1-alpha )% confidence interval for the mean mu of a normal population when the value of sigma is known is given by
((ar{x} )-z_((alpha )/(2))*(sigma )/(sqrt(n)),(ar{x} )+z_((alpha )/(2))*(sigma )/(sqrt(n))).
Under the same conditions as those leading to the interval,
p[(((ar{x} )-mu ))/(((sigma )/(sqrt(n))))<1.645]=0.95.
Use this to derive a one-sided interval for mu that has infinite width and provides a lower confidence bound on mu .
@((ar{x} )+1.645(sigma )/(sqrt(n)),infty )
â—¯((ar{x} )-1.645(sigma )/(sqrt(n)),infty )
â—¯(infty ,(ar{x} )+1.645(sigma )/(n))
O((ar{x} )-1.645(sigma )/(n),infty )
@(infty ,(ar{x} )-1.645(sigma )/(sqrt(n)))
(
(b) Generalize the result of part (a) to obtain a lower bound with confidence level 100(1-alpha )%.
@(-(ar{x} )+z_(alpha )(sigma )/(sqrt(n)),infty )
â—¯(-infty ,(ar{x} )-z_(alpha )(sigma )/(sqrt(n)))
@(-infty ,(ar{x} )+z_(alpha )(sigma )/(sqrt(n)))
@((ar{x} )-z_(alpha )(sigma )/(sqrt(n)),infty )
@((ar{x} )-z_(alpha ^('))(ar{x} )+z_(alpha )(sigma )/(sqrt(n)))
(c) What is an analogous interval to that of part (b) that provides an upper bound on mu ?
@(-(ar{x} )+z_(alpha )(sigma )/(sqrt(n)),infty )
@(-infty ,(ar{x} )-z_(alpha )(sigma )/(sqrt(n)))
@(-infty ,(ar{x} )+z_(alpha )(sigma )/(sqrt(n)))
@((ar{x} )-z_(alpha )(sigma )/(sqrt(n)),infty )
@((ar{x} )-z_(alpha ^(')),(ar{x} )+z_(alpha )(sigma )/(sqrt(n)))
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