A soft-drink bottler sells “one-liter” bottles of ginger ale. The production manager

wishes to monitor the content of the bottles. He takes a random sampleof 30

bottles, and obtains the following contents, in milliliters:

1025 977 1018 975 977 990 959 957 1031 964

986 914 1010 988 1028 989 1001 984 974 1017

1060 1030 991 999 997 996 1014 946 995 987

(a) Find the meanand standard deviationof this sample, and compute the 95%

confidence interval for the true mean content. State in words what this interval

means.

(b) Find the 95% tolerance interval for 99% of the production. State in words

what this interval means.

a.

quantity

q[1:n] = 1025 977 1018 975 977 990 959 957 1031 964 986 914 1010 988 1028 989 1001 984 974 1017 1060 1030 991 999 997 996 1014 946 995 987

Here, n = 30 (sample size)

mean (q_bar) = sum(q[i])/30 = 992.63

SD = sqrt ( sum((q – mean)^2)/n) = 29.22

Let us assume true mean is mu.

Using t-distribution:

degree of freedom, d = n – 1 = 30 – 1 = 29,

Because,

1-alpha = 0.95

=> alpha = 0.05

=> alpha/2 – 0.025

=> 1 – alpha/2 = 0.975

Hence,

P(-t(n-1,1-alpha/2) < (mu-q_bar)/(SD/sqrt(n)) < t(n-1, 1-alpha/2) = 0.95

P(-t(29, 0.975) < (mu – …

Using given data, in first problem an interval with 95% confidence is estimated in which volume of soft-drink is expected to be. In second part 95% tolerance interval for given percentage of production is estimated.