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9 June, 15:08

Construct a 95% confidence interval for the population mean, µ. Assume the population has a normal distribution. A sample of 25 randomly selected students has a mean test score of 81.5 with a standard deviation of 10.2.

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  1. 9 June, 15:11
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    Step-by-step explanation:

    We want to determine a 95% confidence interval for the mean mean test score of students.

    Number of sample, n = 25

    Mean, u = 81.5

    Standard deviation, s = 10.2

    For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

    We will apply the formula

    Confidence interval

    = mean ± z score * standard deviation/√n

    It becomes

    81.5 ± 1.96 * 10.2/√25

    = 81.5 ± 1.96 / * 2.04

    = 81.5 ± 3.9984

    The lower end of the confidence interval is 81.5 - 3.9984 = 77.5

    The upper end of the confidence interval is 81.5 + 3.9984 = 85.5

    Therefore, with 95% confidence interval, the mean test score of students is between 77.5 and 85.5
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