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A survey of 25randomly selected customers found the ages shown (in years). The mean is 32.88years...
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A survey of 25randomly selected customers found the ages shown (in years). The mean is 32.88years and the standard deviation is10.67 years .a) Construct a80% b) How large is the margin of error? c) How would the confidence interval change if you had assumed that the standard deviation was known to be11.0years? |
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