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23 July, 10:59

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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  1. 23 July, 13:27
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    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
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