Question

Boxes of raisins are labled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is reasonable to assume that the population is approximatly normal.
21.88 21.76 22.14 21.63 21.81 22.12 21.97 21.57 21.75 21.96 22.20 21.80

Required:
Construct a 99% confidence interval for the mean weight.

Answer 1

Answer:

The 99 % confidence interval on the basis of mean is ( 21.7393  ;   22.0257)

Step-by-step explanation:

Mean = Sum of observations / Number of observations

Mean = 21.88 +21.76 +22.14 +21.63+ 21.81 +22.12+ 21.97+ 21.57+ 21.75+ 21.96 +22.20 +21.80/ 12

Mean =x`= 262.59/12=  21.8825

Standard Deviation = s= ∑x²/n - ( ∑x/n)²

∑x²/n= 478.7344 +473.4976 + 490.1796+467.8569+ 475.6761 + 489.2944+ 482.6809+ 465.2649+ 473.0625+ 482.2416 +492.84 + 475.24/ 12

∑x²/n= 5746.5689/12= 478.8807 = 478.881

Standard Deviation = s= ∑x²/n - ( ∑x/n)²

s= 478.881- (21.8825)²= 478.881-478.843= 0.037

The confidence limit 99% for the mean will be determined by

x` ± α(100-1) √s/n

Putting the values in the above equation

= 21.8825 ± 2.58 √0.037/12

Solving the square root

= 21.8825 ± 2.58 (0.05549)

Multiplying the square root with 2.58

=21.8825 ± 0.1432

Adding and subtracting would give

21.7393  ;   22.0257,

Hence the 99 % confidence interval on the basis of mean is ( 21.7393  ;   22.0257)