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5 July, 01:06

The height (in inches) of adult men in the United States is believed to be Normally distributed with mean μ. The average height of a random sample of 25 American adult men is found to be ¯ x = 69.72 inches, and the standard deviation of the 25 heights is found to be s = 4.15. A 90% confidence interval for μ is: a. 69.72 ± 1.37 b. 69.72 ± 1.42 c. 69.72 ± 1.09

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  1. 5 July, 01:21
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    Answer: a. 69.72 ± 1.37

    Step-by-step explanation:

    We want to determine a 90% confidence interval for the mean height (in inches) of adult men in the United States.

    Number of sample, n = 25

    Mean, u = 69.72 inches

    Standard deviation, s = 4.15

    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

    69.72 ± 1.645 * 4.15/√25

    = 69.72 ± 1.645 * 0.83

    = 69.72 ± 1.37

    The lower end of the confidence interval is 69.72 - 1.37 = 68.35

    The upper end of the confidence interval is 69.72 + 1.37 = 71.09
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