Question

A cocoa packaging machine fills bags so that the bag contents have a standard deviation of 3.5g. Weights of contents of bags are normally distributed.

(a) If a random sample of 20 bags gives a mean of 102.0g, what are the 99% confidence limits for the mean weight of the population (i.e., all bags)?

(b) How many bags would have to be taken so that the sample mean would be within ±0.9 of the sample mean with a 95% level of confidence?

Answer 1

a)

99% Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.01 /2) = 2.576
102 ± Z (0.01/2 ) * 3.5/√(20)
Lower Limit = 102 - Z(0.01/2) 3.5/√(20)
Lower Limit = 99.984
Upper Limit = 102 + Z(0.01/2) 3.5/√(20)
Upper Limit = 104.016
99% Confidence interval is ( 99.984 , 104.016 )

b)

Sample size = (Z(α/2) * σ / E)2

=( 1.96 * 3.5 / 0.9)2

= 58.1

Sample size = 59 (Rounded up to nearest integer)