Question

The amounts (in ounces) of juice in eight randomly selected juice bottles are:

15.3 15.3 15.7 15.7
15.3 15.9 15.3 15.9

Construct a 98% confidence interval for the mean amount of juice in all such bottles.

Answer 1

Solution:

Given that,

x	x2
15.3	234.09
15.3	234.09
15.7	246.49
15.7	246.49
15.3	234.09
15.9	252.81
15.3	234.09
15.9	252.81
x = 124.4	x2 = 1934.96

The sample mean is

Mean   = (x / n)

= (15.3+15.3+15.7+15.7+15.3+15.9+15.3+15.9 / 8 )

= 124.4 / 8

= 15.55

Mean   = 15.55

The sample standard is S

S   = ( x² ) - (( x )² / n ) / 1 -n )

= ( 1934.96 ( (- 124.4 )² / 8 ) / 7

   = (1934.96 -1934.42 / 7 )

= 0.54 / 7

= 0.0771

= 0.2777

The sample standard is = 0.28

Degrees of freedom = df = n - 1 = 8 - 1 = 7

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,7 = 2.998

Margin of error = E = t_/2,df * (s /n)

= 2.998 * (0.28 / 8)

= 0.30

Margin of error = 0.30

The 98% confidence interval estimate of the population mean is,

- E < < + E

15.55 - 0.30 < < 15.55 + 0.30

15.25 < < 15.85

(15.25, 15.85 )