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24 September, 02:50

From a random sample of 41 teens, it is found that on average they spend 43.1 hours each week online with a population standard deviation of 5.91 hours. What is the 90% confidence interval for the amount of time they spend online each week?

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  1. 24 September, 05:54
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    Step-by-step explanation:

    We want to determine a 90% confidence interval for the mean amount of time that teens spend online each week.

    Number of sample, n = 41

    Mean, u = 43.1 hours

    Standard deviation, s = 5.91 hours

    For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.

    We will apply the formula

    Confidence interval

    = mean + / - z * standard deviation/√n

    It becomes

    43.1 ± 1.645 * 5.91/√41

    = 43.1 ± 1.645 * 0.923

    = 43.1 ± 1.52

    The lower end of the confidence interval is 43.1 - 1.52 = 41.58

    The upper end of the confidence interval is 43.1 + 1.52 = 44.62

    Therefore, with 90% confidence interval, the mean amount of time that teens spend online each week is between 41.58 and 44.62
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