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.

Question

Madyson Mckee

Mathematics

13 August, 16:26

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.

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Lilia Manning · Answer 1

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