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26 November, 06:46

The mean caloric intake of an adult male is 2800 with a standard deviation of 115. To verify this information, a sample of 25 men are selected and determined to have a mean caloric intake of 2950. Determine the 98% confidence interval for mean caloric intake of an adult male. Solution: Since n < 30, a t-test must be used. 2950 - qt (1.98/2,24) * 115/sqrt (25) 2950 + qt (1.98/2,24) * 115/sqrt (25) [2892.68, 3007.32] What is wrong with this solution?

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  1. 26 November, 07:22
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    Step-by-step explanation:

    We would use the t - distribution.

    From the information given,

    Mean, μ = 2950

    Standard deviation, σ = 115

    number of sample, n = 25

    Degree of freedom, (df) = 25 - 1 = 24

    Alpha level,α = (1 - confidence level) / 2

    α = (1 - 0.98) / 2 = 0.01

    We will look at the t distribution table for values corresponding to (df) = 24 and α = 0.01

    The corresponding z score is 2.492

    We will apply the formula

    Confidence interval

    = mean ± z * standard deviation/√n

    It becomes

    2950 ± 2.492 * 115/√25

    = 2950 ± 2.492 * 23

    = 2950 ± 57.316

    The lower end of the confidence interval is 2950 - 57.316 = 2892.68

    The upper end of the confidence interval is 2950 + 57.316 = 3007.32

    The solution is correct.
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