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Today, 20:16

A random sample of 40 students were asked how much money they spent on entertainment in the last week. They spent an average of $28, with a standard deviation of $18. What is the 90% confidence interval for mean amount spent on entertainment?

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  1. Today, 23:39
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    In between $23.3 and $32.7

    Explanation:

    standard deviation = $18

    sample mean = $28

    sample size = 40 students

    Significance level = 1 - confidence interval = 1 - 0.9 = 0.1

    Using the confidence interval calculator

    90% Confidence Interval: $28 ± $4.68

    ($23.3 to $32.7)

    With 90% confidence the population mean is between 23.3 and 32.7 based on 40 samples."
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