He amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9, 15.5 construct a 98% confidence interval for the mean amount of juice in all such bottles. assume an approximate normal distribution.

Given the sample: 15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9, 15.5
mean=15.46
var=0.09410714
standard deviation=0.30676887~0.307
the 98% confidence interval will be:
+/-(0.307/sqrt(8))
=+/-0.1085
thus our answer will be:
15.46+/-0.1085
=15.5685 or15.3515