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21 August, 11:18

The following data represent a random sample of the ages of players in a baseball league. Assume that the population is normally distributed with a standard deviation of 1.8 years. Find the 95% confidence interval for the true mean age of players in this league. Round your answers to two decimal places and use ascending order.

Age: 32, 24, 30,34,28, 23,31,33,27,25

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  1. 21 August, 12:03
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    Step-by-step explanation:

    n = 10

    Mean, m = (32 + 24 + 30 + 34 + 28 + 23 + 31 + 33 + 27 + 25) / 10

    = 28.7

    From the information given,

    Standard deviation, s = 1.8

    For a confidence level of 95%, the corresponding z value is 1.96.

    We will apply the formula

    Confidence interval

    = mean ± z * standard deviation/√n

    It becomes

    28.7 ± 1.96 * 1.8/√10

    = 28.7 ± 1.12

    The lower end of the confidence interval is 28.7 - 1.12 = 27.58 years

    The upper end of the confidence interval is 28.7 + 1.12 = 29.82 years
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